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Vasile Cîrtoaje
C
j
MATHEM
MATICAL
INEQUA
Q ALITIES
Volume 4
EXTENSIONS AND REFINEMENTS
OF JENSEN’SS INEQUALITY
OF JENSEN
UNIVERSITY OF PLOIESTI, ROMANIA
2015
Contents
1 Half Convex Function Method
1
1.1 Theoretical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2 Half Convex Function Method for Ordered Variables
97
2.1 Theoretical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
2.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
3 Partially Convex Function Method
141
3.1 Theoretical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
3.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
3.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
4 Partially Convex Function Method for Ordered Variables
193
4.1 Theoretical Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
4.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
4.3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
5 Bibliography
215
i
ii
Vasile Cîrtoaje
Chapter 1
Half Convex Function Method
1.1
Theoretical Basis
Half Convex Function Theorem (Vasile Cirtoaje, 2004). Let f (u) be a function defined
on a real interval I and convex on Iu≥s or Iu≤s , where s ∈ I. The inequality
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
a + a + ··· + a 1
2
n
n
holds for all a1 , a2 , . . . , an ∈ I satisfying a1 + a2 + · · · + an = ns if and only if
f (x) + (n − 1) f ( y) ≥ n f (s)
for all x, y ∈ I such that x + (n − 1) y = ns.
Proof (for the right convexity). The necessity is obvious. Without loss of generality,
assume that a1 ≤ a2 ≤ · · · ≤ an . If a1 ≥ s, then the required inequality follows from
Jensen’s inequality for convex functions. Otherwise, there exists k ∈ {1, 2, . . . , n − 1}
such that
a1 ≤ · · · ≤ ak < s ≤ ak+1 ≤ · · · ≤ an .
Since f is convex on Iu≥s , we can apply Jensen’s inequality to get
f (ak+1 ) + · · · + f (an ) ≥ (n − k) f (z),
z=
ak+1 + · · · + an
.
n−k
Thus, it suffices to show that
f (a1 ) + · · · + f (ak ) + (n − k) f (z) ≥ n f (s).
Let b1 , . . . , bk ∈ I defined by
ai + (n − 1)bi = ns,
1
i = 1, . . . , k.
2
Vasile Cîrtoaje
We claim that
z ≥ b1 ≥ · · · ≥ bk > s.
Indeed, we have
b1 ≥ · · · ≥ bk ,
bk − s =
s − ak
> 0,
n−1
(n − 1)b1 = ns − a1 = (a2 + · · · + ak ) + ak+1 + · · · + an ≤ (k − 1)s + ak+1 + · · · + an
= (k − 1)s + (n − k)z ≤ (n − 1)z.
By hypothesis, we have
f (a1 ) + (n − 1) f (b1 ) ≥ n f (s),
···
f (ak ) + (n − 1) f (bk ) ≥ n f (s),
hence
f (a1 ) + · · · + f (ak ) + (n − 1)[ f (b1 ) + · · · + f (bk )] ≥ kn f (s).
Consequently, it suffices to show that
kn f (s) − (n − 1)[ f (b1 ) + · · · + f (bk )] + (n − k) f (z) ≥ n f (s),
which is equivalent to
p f (z) + (k − p) f (s) ≥ f (b1 ) + · · · + f (bk ),
p=
n−k
≤ 1.
n−1
By Jensen’s inequality, we have
p f (z) + (1 − p) f (s) ≥ f (w),
w = pz + (1 − p)s ≥ s.
Thus, we only need to show that
f (w) + (k − 1) f (s) ≥ f (b1 ) + · · · + f (bk ).
Since the decreasingly ordered vector A~k = (w, s, . . . , s) majorizes the decreasingly ordered vector B~k = (b1 , b2 , . . . , bk ), this inequality follows from Karamata’s inequality for
convex functions.
The proof for the left convexity is similar.
Remark 1. Let us denote
g(u) =
f (u) − f (s)
,
u−s
h(x, y) =
g(x) − g( y)
.
x−y
Half Convex Function Method
3
In many applications, it is useful to replace the hypothesis f (x) + (n − 1) f ( y) ≥ n f (s)
in Half Convex Function Theorem (HCF Theorem) by the equivalent condition
h(x, y) ≥ 0 for all x, y ∈ I such that x + (n − 1) y = ns.
This equivalence is true since
f (x) + (n − 1) f ( y) − n f (s) = [ f (x) − f (s)] + (n − 1)[ f ( y) − f (s)]
= (x − s)g(x) + (n − 1)( y − s)g( y)
n−1
(x − y)[g(x) − g( y)]
=
n
n−1
=
(x − y)2 h(x, y).
n
Remark 2. Assume that f is differentiable on I, and let
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
Then, the desired inequality of Jensen’s type in HCF Theorem holds true by replacing
the hypothesis
f (x) + (n − 1) f ( y) ≥ n f (s)
with the more restrictive condition
H(x, y) ≥ 0 for all x, y ∈ I such that x + (n − 1) y = ns.
To prove this claim, we will show that the new condition implies f (x)+(n−1) f ( y) ≥
n f (s) for all x, y ∈ I such that x + (n − 1) y = ns. Write this inequality as f1 (x) ≥ n f (s),
where
ns − x f1 (x) = f (x) + (n − 1) f ( y) = f (x) + (n − 1) f
.
n−1
From
ns − x n
= f 0 (x) − f 0 ( y) =
(x − s)H(x, y),
f10 (x) = f 0 (x) − f 0
n−1
n−1
it follows that f1 is decreasing for x ≤ s and increasing for x ≥ s; therefore,
f1 (x) ≥ f1 (s) = n f (s).
Remark 3. The inequality in HCF Theorem becomes an equality for
a1 = a2 = · · · = an = s
and also for
a1 = x,
a2 = · · · = an = y
4
Vasile Cîrtoaje
(or any cyclic permutation), where x, y ∈ I satisfy the equations
x + (n − 1) y = ns,
f (x) + (n − 1) f ( y) = n f (s).
For x 6= y, these equations are equivalent to
x + (n − 1) y = ns, h(x, y) = 0.
Remark 4. HCF Theorem is also valid in the case when I = [a, b] \ {u0 } or I = (a, b) \
{u0 }, where a, b, u0 are real numbers such that a < u0 < b. Clearly, two cases are
possible:
(1) u0 < s - when f is right convex, i.e. convex on Iu≥s ;
(2) u0 > s - when f is left convex, i.e. convex on Iu≤s .
Remark 5. In HCF Theorem for the right convexity (when HCF Theorem is called RHCF
Theorem), it suffices to consider
x ≤ s ≤ y.
This claim is true because in the proof of RHCF Theorem, the hypothesis
f (x) + (n − 1) f ( y) ≥ n f (s)
is used to get the inequalities
f (ai ) + (n − 1) f (bi ) ≥ n f (s),
i = 1, 2, · · · , k,
where ai ≤ s ≤ bi .
Similarly, in HCF Theorem for the left convexity (when HCF Theorem is called LHCF
Theorem), it suffices to consider
x ≥ s ≥ y.
The following theorem (LCRCF-Theorem) is also useful to prove some symmetric inequalities.
Left Convex-Right Concave Function Theorem (Vasile Cirtoaje, 2004). Let a ≤ c be
real numbers, let f be a continuous function on I = [a, ∞), strictly convex on [a, c] and
strictly concave on [c, ∞), and let
E(a1 , a2 , . . . , an ) = f (a1 ) + f (a2 ) + · · · + f (an ).
If a1 , a2 , . . . , an ∈ I such that
a1 + a2 + · · · + an = S = const ant,
then
(a) E is minimum for a1 = a2 = · · · = an−1 ≤ an ;
Half Convex Function Method
5
(b) E is maximum for either a1 = a or a < a1 ≤ a2 = · · · = an .
Proof. Without loss of generality, assume that a1 ≤ a2 ≤ . . . ≤ an . Since E(a1 , a2 , . . . , an )
is a continuous function on the compact set
Λ = {(a1 , a2 , . . . , an ) : a1 + a2 + · · · + an = S, a1 , a2 , . . . , an ∈ I},
E attains its minimum and maximum values.
(a) For the sake of contradiction, suppose that E is minimum at a point (b1 , b2 , . . . , bn )
with b1 ≤ b2 ≤ . . . ≤ bn and b1 < bn−1 . For bn−1 ≤ c, by Jensen’s inequality for strictly
convex functions, we have
b1 + bn−1
,
f (b1 ) + f (bn−1 ) > 2 f
2
while for bn−1 > c, by Karamata’s inequality for strictly concave functions, we have
f (bn−1 ) + f (bn ) > f (c) + f (bn−1 + bn − c).
The both results contradict the assumption that E is minimum at (b1 , b2 , . . . , bn ).
(b) For the sake of contradiction, suppose that E is maximum at a point (b1 , b2 , . . . , bn )
with a < b1 ≤ b2 ≤ . . . ≤ bn and b2 < bn . There are three cases to consider.
Case 1: b2 ≥ c. By Jensen’s inequality for strictly concave functions, we have
b2 + bn
f (b2 ) + f (bn ) < 2 f
.
2
Case 2: b2 < c and b1 + b2 −a ≤ c. By Karamata’s inequality for strictly convex functions,
we have
f (b1 ) + f (b2 ) < f (a) + f (b1 + b2 − a).
Case 3: b2 < c and b1 + b2 −c ≥ a. By Karamata’s inequality for strictly convex functions,
we have
f (b1 ) + f (b2 ) < f (b1 + b2 − c) + f (c).
Clearly, the result of each case contradicts the assumption that E is msximum at (b1 , b2 , . . . , bn ).
Remark 6. The part (a) in LCRCF-Theorem is also true in the case where I = (a, ∞)
and lim x→a f (x) = ∞.
6
Vasile Cîrtoaje
Half Convex Function Method
1.2
7
Applications
1.1. If a, b, c are real numbers such that a + b + c = 3, then
3(a4 + b4 + c 4 ) + a2 + b2 + c 2 + 6 ≥ 6(a3 + b3 + c 3 ).
1.2. If a1 , a2 , . . . , an ≥
1 − 2n
such that a1 + a2 + · · · + an = n, then
n−2
a13 + a23 + · · · + an3 ≥ n.
1.3. If a1 , a2 , . . . , an ≥
−n
such that a1 + a2 + · · · + an = n, then
n−2
a13 + a23 + · · · + an3 ≥ a12 + a22 + · · · + an2 .
1.4. If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then
(n2 − 3n + 3)(a14 + a24 + · · · + an4 − n) ≥ 2(n2 − n + 1)(a12 + a22 + · · · + an2 − n).
1.5. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
(n2 + n + 1)(a13 + a23 + · · · + an3 − n) ≥ (n + 1)(a14 + a24 + · · · + an4 − n).
1.6. If a, b, c are real numbers such that a + b + c = 3, then
(a)
(b)
p
a4 + b4 + c 4 − 3 + 2(7 + 3 7)(a3 + b3 + c 3 − 3) ≥ 0;
p
a4 + b4 + c 4 − 3 + 2(7 − 3 7)(a3 + b3 + c 3 − 3) ≥ 0.
1.7. Let k ≥ 3 and n ≥ 3 be integer numbers such that k ≤ n + 1. If a1 , a2 , . . . , an are
nonnegative real numbers such that a1 + a2 + · · · + an = n, then
a1k + a2k + · · · + ank − n
a12 + a22 + · · · + an2 − n
≥ (n − 1)
n k−1
−1 .
n−1
8
Vasile Cîrtoaje
1.8. Let k ≥ 3 and n ≥ 3 be integer numbers. If a1 , a2 , . . . , an are nonnegative real
numbers such that a1 + a2 + · · · + an = n, then
a1k + a2k + · · · + ank − n
a12 + a22 + · · · + an2 − n
≤
nk−1 − 1
.
n−1
1.9. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n, then
1
1
1
2
n
+
+ ··· +
− n ≥ 4(n − 1)(a12 + a22 + · · · + an2 − n).
a1 a2
an
1.10. If a1 , a2 , . . . , a8 are positive real numbers such that a1 + a2 + · · · + a8 = n, then
1
a12
+
1
a22
+ ··· +
1
a82
≥ a12 + a22 + · · · + a82 .
1.11. If a1 , a2 , . . . , an are positive real numbers such that
a12
+ a22
+ · · · + an2
p
−n≥2 1+
1
1
1
+
+ ··· +
= n, then
a1 a2
an
n−1
(a1 + a2 + · · · + an − n).
n
1.12. If a, b, c are nonnegative real numbers, no two of which are zero, then
1
1
2
1
1
1
1
+
+
≤
+
+
.
3a + b + c 3b + c + a 3c + a + b
5 b+c c+a a+b
1.13. If a, b, c, d ≥ 3 −
p
7 such that a + b + c + d = 4, then
1
1
1
1
4
+
+
+
≥ .
2
2
2
2
2+a
2+ b
2+c
2+d
3
1.14. If a1 , a2 , . . . , an ∈ [0, n − 2] such that a1 + a2 + · · · + an = n, then
1
n + a12
+
1
n + a22
+ ··· +
1
n
≤
.
2
n + an
n+1
Half Convex Function Method
9
1.15. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
3− b
3−c
3
3−a
+
+
≥ .
2
2
2
9+a
9+ b
9+c
5
1.16. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
1
1
1
3
+
+
≥ .
2
2
2
5+a+a
5+ b+ b
5+c+c
7
1.17. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then
1
1
1
1
1
+
+
+
≤
.
10 + a + a2 10 + b + b2 10 + c + c 2 10 + d + d 2
3
1.18. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
1
k ≥ 1 − , then
n
1
n
1
1
≥
.
+
+ ··· +
1 + kan2
1+k
1 + ka12 1 + ka22
1.19. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n. If 0 < k ≤
n−1
, then
2
n −n+1
1
n
1
1
≤
.
+
+ ··· +
2
2
2
1 + kan
1+k
1 + ka1 1 + ka2
1.20. Let a1 , a2 , . . . , an be nonnegative numbers such that a1 + a2 + · · · + an = n. If
n2
k≥
, then
4(n − 1)
a1 (a1 − 1)
a12
+k
+
a2 (a2 − 1)
a22
+k
+ ··· +
an (an − 1)
≥ 0.
an2 + k
1.21. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n,
then
1 − an
1 − a1
1 − a2
+
+ ··· +
≤ 0.
2
2
(n − 2a1 )
(n − 2a2 )
(n − 2an )2
10
Vasile Cîrtoaje
1.22. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
n
, then
k ≥1+ p
n−1
a12 − 1
(a1 + k)2
+
a22 − 1
(a2 + k)2
+ ··· +
an2 − 1
(an + k)2
≥ 0.
1.23. Let a1s
, a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
2n − 1
0< k ≤1+
, then
n−1
a12 − 1
(a1 + k)2
+
a22 − 1
(a2 + k)2
+ ··· +
an2 − 1
(an + k)2
≤ 0.
1.24. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
(n − 1)(2n − 1)
k≥
, then
n2
1
1 + ka13
+
1
1 + ka23
+ ··· +
1
n
≥
.
3
1 + kan
1+k
1.25. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
n−1
, then
0<k≤ 2
n − 2n + 2
1
1 + ka13
+
1
1 + ka23
+ ··· +
n
1
≤
.
1 + kan3
1+k
1.26. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
n2
, then
k≥
n−1
v
v
v
t a1
t a2
t an
n
+
+ ··· +
≤p
.
k − a1
k − a2
k − an
k−1
1.27. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n,
then
2
2
2
n−a1 + n−a2 + · · · + n−an ≥ 1.
Half Convex Function Method
11
1.28. If a, b, c, d, e are nonnegative real numbers such that a + b + c + d + e = 5, then
(a2 + 1)(b2 + 1)(c 2 + 1)(d 2 + 1)(e2 + 1) ≥ (a + 1)(b + 1)(c + 1)(d + 1)(e + 1).
1.29. If a1 , a2 , . . . , a10 are nonnegative real numbers such that a1 + a2 + · · · + a10 = 10,
then
2
(1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a10 + a10
) ≥ 1.
1.30. If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 + a2 + · · · + an = 1,
then
p
p
p
1
1
1 n
1
p
−
a
−
a
−
a
n
−
.
·
·
·
≥
p
p
p
p
1
2
n
a1
a2
an
n
1.31. Let a1 , a2 , . . . , an be positive real numbers such that a1 + a2 + · · · + an = n. If
2
p
2 n−1
k ≤ 1+
,
n
then
ka1 +
1
a1
ka2 +
1
1
· · · kan +
≥ (k + 1)n .
a2
an
1.32. If a, b, c, d are nonzero real numbers such that
a, b, c, d ≥
then
−1
,
2
a + b + c + d = 4,
1
1
1
1
1 1 1 1
3 2 + 2 + 2 + 2 + + + + ≥ 16.
a
b
c
d
a b c d
1.33. If a1 , a2 , . . . , an are nonnegative real numbers such that a12 + a22 + · · · + an2 = n,
then
s
n
a13 + a23 + · · · + an3 − n +
(a1 + a2 + · · · + an − n) ≥ 0.
n−1
12
Vasile Cîrtoaje
1.34. If a, b, c, d, e are nonnegative real numbers such that a2 + b2 + c 2 + d 2 + e2 = 5,
then
1
1
1
1
1
+
+
+
+
≤ 1.
7 − 2a 7 − 2b 7 − 2c 7 − 2d 7 − 2e
1.35. If 0 ≤ a1 , a2 , . . . , an < k such that
then
a12 +a22 +· · ·+an2
= n, where 1 < k ≤ 1+
1
1
n
1
+
+ ··· +
≥
.
k − a1 k − a2
k − an
k−1
1.36. If a, b, c are nonnegative real numbers, no two of which are zero, then
v
v
v
t
48a t
48b t
48c
1+
+ 1+
+ 1+
≥ 15.
b+c
c+a
a+b
1.37. If a, b, c are nonnegative real numbers, then
v
v
v
t
t
t
3a2
3b2
3c 2
+
+
≤ 1.
7a2 + 5(b + c)2
7b2 + 5(c + a)2
7c 2 + 5(a + b)2
1.38. If a, b, c are nonnegative real numbers, then
v
v
v
t
t
t
a2
b2
c2
+
+
≥ 1.
a2 + 2(b + c)2
b2 + 2(c + a)2
c 2 + 2(a + b)2
1.39. Let a, b, c be nonnegative real numbers, no two of which are zero. If
k ≥ k0 ,
then
2a
b+c
k
+
k0 =
ln 3
− 1 ≈ 0.585,
ln 2
2b
c+a
k
+
2c
a+b
k
≥ 3.
p
1.40. If a, b, c ∈ [1, 7 + 4 3], then
v
v
v
t 2b
t 2a
t 2c
+
+
≥ 3.
b+c
c+a
a+b
s
n
,
n−1
Half Convex Function Method
13
1.41. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If
0 < k ≤ k0 ,
k0 =
ln 2
≈ 1.71,
ln 3 − ln 2
then
a k (b + c) + b k (c + a) + c k (a + b) ≤ 6.
1.42. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
n2
k≥
, then
4(n − 1)
(a12 + k)(a22 + k) · · · (an2 + k) ≥ (1 + k)n .
1.43. Let a, b, c be nonnegative real numbers such a + b + c = 3. If k ≥ k0 , where
p
p
p
6−2
k0 = p
= (2 + 2)(2 + 3) ≈ 12.74 ,
p
6− 2−1
then
p
a+
p
b+
p
c−3≥ k
v
ta + b
2
+
v
tb+c
2
+
s
c+a
−3 .
2
1.44. Let a, b, c be nonnegative real numbers such a + b + c = 3. If k ≤ k1 , where
p
p
p
k1 = ( 3 − 1)( 3 + 2) ≈ 2.303 ,
then
p
a+
p
b+
p
c−3≤ k
v
ta + b
2
+
v
tb+c
2
+
s
c+a
−3 .
2
1.45. If a, b, c are positive real numbers such that a bc = 1, then
a2 + b2 + c 2 − 3 ≥ 18(a + b + c − a b − bc − ca).
1.46. If a, b, c are positive real numbers such that a bc = 1, then
p
p
p
a2 − a + 1 + b2 − b + 1 + c 2 − c + 1 ≥ a + b + c.
14
Vasile Cîrtoaje
1.47. If a, b, c, d ≥
1
p such that a bcd = 1, then
1+ 6
1
1
1
4
1
+
+
+
≤ .
a+2 b+2 c+2 d +2 3
1.48. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
p
1
1
1
.
a12 + a22 + · · · + an2 − n ≥ 6 3 a1 + a2 + · · · + an −
−
− ··· −
a1 a2
an
1.49. If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that a1 a2 · · · an = 1, then
(n − 1)(a12 + a22 + · · · + an2 ) + n(n + 3) ≥ (2n + 2)(a1 + a2 + · · · + an ).
1.50. Let a, b, c, d be positive real numbers such that a bcd = 1. If p and q are nonnegative real numbers such that p + q = 3, then then
1
1
1
1
+
+
+
≥ 1.
1 + pa + qa3 1 + p b + q b3 1 + pc + qc 3 1 + pd + qd 3
1.51. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 . . . an = 1. If p
and q are nonnegative real numbers such that p + q ≥ n − 1, then
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
n
1
≥
.
2
1 + pan + qan
1+p+q
1.52. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If k ≥ n2 − 1,
then
1
1
n
1
+p
+ ··· + p
≥p
.
p
1+k
1 + kan
1 + ka1
1 + ka2
1.53. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then
1
1 + a1 + · · · + a1n−1
+
1
1 + a2 + · · · + a2n−1
+ ··· +
1
≥ 1.
1 + an + · · · + ann−1
Half Convex Function Method
15
1.54. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If p, q ≥ 0
1
, then
such that 0 < p + q ≤
n−1
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
n
1
≤
.
1 + pan + qan2
1+p+q
1.55. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 . . . an = 1. If
n 2
0<k≤
− 1,
n−1
then
1
p
1 + ka1
+p
1
1 + ka2
+ ··· + p
1
n
≤p
.
1+k
1 + kan
1.56. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then
1
1
1
n−1
n−1
n−1
a1 + a2 + · · · + an + n(n − 2) ≥ (n − 1)
+
+ ··· +
.
a1 a2
an
1.57. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If k ≥ n, then
1
1
1
k
k
k
a1 + a2 + · · · + an + kn ≥ (k + 1)
+
+ ··· +
.
a1 a2
an
1.58. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then
1 a1
1 a2
1 an
1−
+ 1−
+ ··· + 1 −
≤ n − 1.
n
n
n
1.59. If a, b, c are positive real numbers such that a bc = 1, then
1
1
1
+
+
≤ 1.
p
p
p
1 + 1 + 3a 1 + 1 + 3b 1 + 1 + 3c
1.60. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
1
1+
p
1 + 4n(n − 1)a1 1 +
1
p
1 + 4n(n − 1)a2
+ ··· +
1
1+
p
1 + 4n(n − 1)an
≥
1
.
2
16
Vasile Cîrtoaje
1.61. If a, b, c are positive real numbers such that a bc = 1, then
b6
c6
a6
+
+
≥ 1.
1 + 2a5 1 + 2b5 1 + 2c 5
1.62. If a, b, c are positive real numbers such that a bc = 1, then
p
p
p
25a2 + 144 + 25b2 + 144 + 25c 2 + 144 ≤ 5(a + b + c) + 24.
1.63. If a, b, c are positive real numbers such that a bc = 1, then
p
p
p
16a2 + 9 + 16b2 + 9 + 16c 2 + 9 ≥ 4(a + b + c) + 3.
1.64. If a, b, c, d are real numbers such that a + b + c + d = 4, then
a2
a
b
c
d
+ 2
+ 2
+ 2
≤ 1.
−a+4 b − b+4 c −c+4 d −d +4
1.65. If a, b, c are nonnegative real numbers such that a + b + c = 12, then
(a2 + 10)(b2 + 10)(c 2 + 10) ≥ 13310.
1.66. Let a, b, c be nonnegative real numbers. If
k0 ≤ k ≤ 3,
k0 =
ln 2
≈ 1.71,
ln 3 − ln 2
then
a k (b + c) + b k (c + a) + c k (a + b) ≤ 2
a+b+c
2
k+1
.
1.67. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n, then
1
1
1
2
(n + 1)
+
+ ··· +
≥ 4(n + 2)(a12 + a22 + · · · + an2 ) + n(n2 − 3n + 6).
a1 a2
an
Half Convex Function Method
1.3
17
Solutions
P 1.1. If a, b, c are real numbers such that a + b + c = 3, then
3(a4 + b4 + c 4 ) + a2 + b2 + c 2 + 6 ≥ 6(a3 + b3 + c 3 ).
(Vasile Cirtoaje, 2006)
Solution. Write the inequality as
f (a) + f (b) + f (c) ≥ 3 f (s),
s=
a+b+c
= 1,
3
where
f (u) = 3u4 − 6u3 + u2 ,
u ∈ R.
From
f 00 (u) = 2(18u2 − 18u + 1),
it follows that f 00 (u) > 0 for u ≥ 1, hence f is convex for u ≥ 1. By HCF Theorem, it
suffices to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3. Using
Remark 1, we only need to show that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
We have
g(u) = 3(u3 + u2 + u + 1) − 6(u2 + u + 1) + u + 1 = 3u3 − 3u2 − 2u − 2,
h(x, y) = 3(x 2 + x y + y 2 ) − 3(x + y) − 2 = (3 y − 4)2 ≥ 0.
From x + 2 y = 3 and h(x, y) = 0, we get x = 1/3, y = 4/3. Therefore, in accordance
with Remark 3, the equality holds for a = b = c = 1, and also for a = 1/3 and b = c =
4/3 (or any cyclic permutation).
Remark. Similarly, we can prove the following generalization:
• If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
(a12 − a1 )2 + (a22 − a2 )2 + · · · + (an2 − an )2 ≥
n2
n−1
(a2 + a22 + · · · + an2 − n),
− 3n + 3 1
with equality for a1 = a2 = · · · = an = 1, and also for a1 =
· · · = an = 1 +
n2
n−2
(or any cyclic permutation).
− 3n + 3
1
and a2 = a3 =
n2 − 3n + 3
18
Vasile Cîrtoaje
P 1.2. If a1 , a2 , . . . , an ≥
1 − 2n
such that a1 + a2 + · · · + an = n, then
n−2
a13 + a23 + · · · + an3 ≥ n.
(Vasile Cirtoaje, 2000)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
where
f (u) = u3 ,
u≥
a1 + a2 + · · · + an
= 1,
n
1 − 2n
.
n−2
From f 00 (u) = 6u, it follows that f is convex for u ≥ 1. According to HCF Theorem
1 − 2n
and Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥
such that
n−2
x + (n − 1) y = n. We have
g(u) =
h(x, y) =
f (u) − f (1)
= u2 + u + 1,
u−1
g(x) − g( y)
(n − 2)x + 2n − 1
= x + y +1=
≥ 0.
x−y
n−1
1 − 2n
n+1
, y=
. Therefore, in
n−2
n−2
accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also for
From x + (n − 1) y = n and h(x, y) = 0, we get x =
a1 =
1 − 2n
,
n−2
a2 = · · · = a n =
n+1
n−2
(or any cyclic permutation).
P 1.3. If a1 , a2 , . . . , an ≥
−n
such that a1 + a2 + · · · + an = n, then
n−2
a13 + a23 + · · · + an3 ≥ a12 + a22 + · · · + an2 .
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
where
f (u) = u3 − u2 ,
u≥
a1 + a2 + · · · + an
= 1,
n
−n
.
n−2
Half Convex Function Method
19
From f 00 (u) = 6u − 2, it follows that f is convex for u ≥ 1. According to HCF Theorem
1 − 2n
such that
and Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥
n−2
x + (n − 1) y = n. We have
g(u) =
f (u) − f (1)
= u2 ,
u−1
g(x) − g( y)
(n − 2)x + n
=x+y=
≥ 0.
x−y
n−1
−n
n
From x + (n − 1) y = n and h(x, y) = 0, we get x =
, y =
. Therefore, in
n−2
n−2
accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also for
h(x, y) =
a1 =
−n
,
n−2
a2 = · · · = an =
n
n−2
(or any cyclic permutation).
P 1.4. If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then
(n2 − 3n + 3)(a14 + a24 + · · · + an4 − n) ≥ 2(n2 − n + 1)(a12 + a22 + · · · + an2 − n).
(Vasile Cirtoaje, 2009)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = (n2 − 3n + 3)u4 − 2(n2 − n + 1)u2 , u ∈ R.
For u ≥ 1, we have
1 00
f (u) = 3(n2 − 3n + 3)u2 − (n2 − n + 1) ≥ 3(n2 − 3n + 3) − (n2 − n + 1) = 2(n − 2)2 ≥ 0.
4
By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ∈ R such
that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
We have
g(u) = (n2 − 3n + 3)(u3 + u2 + u + 1) − 2(n2 − n + 1)(u + 1)
20
Vasile Cîrtoaje
and
h(x, y) = (n2 −3n+3)(x 2 +x y+ y 2 +x+ y+1)−2(n2 −n+1) = [(n2 −3n+3) y−n2 +n+1]2 ≥ 0.
In accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also
2
2n − 4
for a1 = −1 + 2
and a2 = a3 = · · · = an = 1 + 2
(or any cyclic
n − 3n + 3
n − 3n + 3
permutation).
P 1.5. If a1 , a2 , · · · , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
(n2 + n + 1)(a13 + a23 + · · · + an3 − n) ≥ (n + 1)(a14 + a24 + · · · + an4 − n).
(Vasile Cirtoaje, 2009)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = (n2 + n + 1)u3 − (n + 1)u4 , u ∈ [0, n].
For u ∈ [0, 1], we have
f 00 (u) = 6u[n2 + n + 1 − 2(n + 1)u] ≥ 6u[n2 + n + 1 − 2(n + 1)] = 6(n2 − n − 1)u ≥ 0.
By HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such
that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
We have
g(u) = (n2 + n + 1)(u2 + u + 1) − (n + 1)(u3 + u2 + u + 1) = −(n + 1)u3 + n2 (u2 + u + 1)
and
h(x, y) = −(n + 1)(x 2 + x y + y 2 ) + n2 (x + y + 1)
= y[−(n + 1)(n2 − 3n + 3) y + n(n2 + n − 3)]
h
i
n
≥ y −(n + 1)(n2 − 3n + 3)
+ n(n2 + n − 3)
n−1
2n2 (n − 2) y
=
≥ 0.
n−1
In accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also
for a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation).
Half Convex Function Method
21
P 1.6. If a, b, c are real numbers such that a + b + c = 3, then
(a)
(b)
p
a4 + b4 + c 4 − 3 + 2(7 + 3 7)(a3 + b3 + c 3 − 3) ≥ 0;
p
a4 + b4 + c 4 − 3 + 2(7 − 3 7)(a3 + b3 + c 3 − 3) ≥ 0.
(Vasile Cirtoaje, 2009)
Solution. Setting
p
p = 2(7 ± 3 7),
we can write the desired inequalities as
f (a) + f (b) + f (c) ≥ 3 f (s),
s=
a+b+c
= 1,
3
where
f (u) = u4 + pu3 ,
u ∈ R.
From
f 00 (u) = 6u(2u2 + p),
it follows that f 00 (u) > 0 for u ≥ 1, hence f is convex for u ≥ 1. By HCF Theorem, it
suffices to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3. Using
Remark 1, we only need to show that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
We have
g(u) = u3 + u2 + u + 1 + p(u2 + u + 1) + u + 1 = u3 + (p + 1)(u2 + u + 1),
h(x, y) = x 2 + x y + y 2 ) + (p + 1)(x + y + 1) = 3 y 2 − (10 + p) y + 13 + 4p
10 + p 2
=3 y−
≥ 0.
6
In accordance with Remark 3, the equality holds for a = b = c = 1, and also for a =
−(1 + p)/3 and b = c = (10 + p)/6 (or any cyclic permutation).
p
(a) Since
p
=
2(7
+
3
7),p the equality holds for a = b = c = 1, and also for
p
a = −5 − 2 7 and b = c = 4 + 7 (or any cyclic permutation).
p
(b) Since
p p = 2(7 − 3 7),
p the equality holds for a = b = c = 1, and also for
a = −5 + 2 7 and b = c = 4 − 7 (or any cyclic permutation).
Remark. Similarly, we can prove the following generalization:
22
Vasile Cîrtoaje
• If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then
a14 + a24 + · · · + an4 − n + p(a12 + a22 + · · · + an2 − n) ≥ 0,
where
p=
2(n2 − n + 1) ± 2
p
3(n2 − n + 1)(n2 − 3n + 3)
.
(n − 2)2
The equality holds for a1 = a2 = · · · = an = 1, and also for
a1 =
2(n2 − n − 1) + (n − 2)p
−2(n2 − 3n + 1) − (n − 1)(n − 2)p
,
a
=
a
=
·
·
·
=
a
=
2
3
n
2(n2 − 3n + 3)
2(n2 − 3n + 3)
(or any cyclic permutation).
P 1.7. Let k ≥ 3 and (n ≥ 3) be integer numbers such that k ≤ n + 1. If a1 , a2 , . . . , an are
nonnegative real numbers such that a1 + a2 + · · · + an = n, then
a1k + a2k + · · · + ank − n
a12 + a22 + · · · + an2 − n
≥ (n − 1)
n k−1
−1 .
n−1
(Vasile Cirtoaje, 2012)
Solution. Denote
m = (n − 1)
n k−2 n k−3
n k−1
−1 =
+
+ · · · + 1,
n−1
n−1
n−1
and write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = uk − mu2 ,
u ∈ [0, n].
We will show that f (u) is convex for u ∈ [1, n]. Since
f 00 (u) = k(k − 1)uk−2 − 2m ≥ k(k − 1) − 2m,
we need to show that
k(k − 1) n k−2 n k−3
≥
+
+ · · · + 1.
2
n−1
n−1
Half Convex Function Method
23
Since n ≥ k − 1, this inequality is true if
k(k − 1)
k − 1 k−2
k − 1 k−3
≥
+
+ · · · + 1.
2
k−2
k−2
By Bernoulli’s inequality, we have
k−1
k−2
j
=
1
1−
1
k−1
j ≤
1
1−
j
k−1
=
k−1
,
k− j−1
0 ≤ j ≤ k − 2.
Therefore, it suffices to show that
k(k − 1)
1
1
≥ (k − 1) 1 + + · · · +
.
2
2
k−1
This is true if
k
1
1
≥ 1 + + ··· +
,
2
2
k−1
which can be easily proved by induction. According to HCF Theorem and Remark 1, we
only need to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
We have
(uk − 1) − m(u2 − 1)
= (uk−1 + uk−2 + · · · + 1) − m(u + 1),
u−1
k−1
x
− y k−1 x k−2 − y k−1
h(x, y) =
+
+ ··· + 1 − m
x−y
x−y
k−1
k−2
2
x
− y k−2 n k−3
x − y2
x
− y k−1 n k−2
n
+
+· · ·+
=
−
−
−
.
x−y
n−1
x−y
n−1
x−y
n−1
g(u) =
It suffices to show that
x j+1 − y j+1 n j
≥
x−y
n−1
n i
for x =
6 y and j = 1, 2, · · · , k − 2. This is true if f j ( y) ≥ 0 for y ∈ 0,
, where
n−1
n j
f j ( y) = x j + x j−1 y + · · · + x y j−1 + y j −
, x = n − (n − 1) y.
n−1
h
Since x 0 = −(n − 1) and n − 1 ≥ k − 2 ≥ j, we get
f j0 ( y) = −(n − 1)[ j x j−1 + ( j − 1)x j−2 y + · · · + y j−1 ] + x j−1 + 2x j−2 y + · · · + j y j−1
24
Vasile Cîrtoaje
≤ − j[ j x j−1 + ( j − 1)x j−2 y + · · · + y j−1 ] + x j−1 + 2x j−2 y + · · · + j y j−1
= −( j · j − 1)x j−1 − [ j · ( j − 1) − 2]x j−2 y − · · · − ( j · 2 − j + 1)x y j−2 ≤ 0.
n
Therefore, f j ( y) is decreasing, hence f j ( y) is minimal for y =
, when x = 0. Thus,
n−1
n = 0.
f j ( y) ≥ f j
n−1
This completes the proof. From x + (n − 1) y = n and h(x, y) = 0, we get x = 0,
n
y=
. Therefore, in accordance with Remark 3, the equality holds for a1 = 0 and
n−1
n
a2 = a3 = · · · = an =
(or any cyclic permutation).
n−1
Remark. For k = 3 and k = 4, we get the following statements (Vasile Cirtoaje, 2002):
• If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
(n − 1)(a13 + a23 + · · · + an3 − n) ≥ (2n − 1)(a12 + a22 + · · · + an2 − n),
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 0 and a2 = a3 = · · · = an =
n
(or any cyclic permutation).
n−1
• If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
(n − 1)2 (a14 + a24 + · · · + an4 − n) ≥ (3n2 − 3n + 1)(a12 + a22 + · · · + an2 − n),
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 0 and a2 = a3 = · · · = an =
n
(or any cyclic permutation).
n−1
P 1.8. Let k ≥ 3 and n ≥ 3 be integer numbers. If a1 , a2 , . . . , an are nonnegative real
numbers such that a1 + a2 + · · · + an = n, then
a1k + a2k + · · · + ank − n
a12 + a22 + · · · + an2 − n
≤
nk−1 − 1
.
n−1
(Vasile Cirtoaje, 2012)
Solution. Denote
m=
nk−1 − 1
= nk−2 + nk−3 + · · · + 1,
n−1
and write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
Half Convex Function Method
25
where
f (u) = mu2 − uk ,
u ∈ [0, n].
We will show that f (u) is convex for u ∈ [0, 1]. Since
f 00 (u) = 2m − k(k − 1)uk−2 ≥ 2m − k(k − 1),
we need to show that
nk−2 + nk−3 + · · · + 1 ≥
k(k − 1)
.
2
3k−2 + 3k−3 + · · · + 1 ≥
k(k − 1)
.
2
This is true if
By Bernoulli’s inequality, we have
3k−2 + 3k−3 + · · · + 1 ≥ [1 + 2(k − 2)] + [1 + 2(k − 3)] + · · · + [1 + 2 · 0]
k(k − 1)
.
2
According to HCF Theorem and Remark 1, we only need to show that h(x, y) ≥ 0 for
x, y ≥ 0 such that x + (n − 1) y = n, where
= (k − 1) + (k − 2)(k − 1) = (k − 1)2 >
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
We have
g(u) =
m(u2 − 1) − (uk − 1)
= m(u + 1) − (uk−1 + uk−2 + · · · + 1),
u−1
x k−1 − y k−1 x k−2 − y k−1
−
− ··· − 1
x−y
x−y
x k−1 − y k−1
x k−2 − y k−2
x2 − y2
k−2
k−3
= n
−
+ n
−
+ ··· + n −
.
x−y
x−y
x−y
h(x, y) = m −
It suffices to show that
nj ≥
x j+1 − y j+1
x−y
for x 6= y and j = 1, 2, · · · , k − 2. We will show that
n j ≥ (x + y) j ≥
x j+1 − y j+1
.
x−y
The left inequality is true since
n − (x + y) = x + (n − 1) y − (x + y) = (n − 2) y ≥ 0.
26
Vasile Cîrtoaje
The right inequality is also true since
j j−1
j
j
j
(x + y) = x +
x
y + ··· +
x y j−1 + y j
1
j−1
and
x j+1 − y j+1
= x j + x j−1 y + · · · + x y j−1 + y j .
x−y
In accordance with Remark 3, the equality holds for a1 = n and a2 = a3 = · · · = an = 0
(or any cyclic permutation).
Remark. For k = 3 and k = 4, we get the following statements (Vasile Cirtoaje, 2002):
• If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
a13 + a23 + · · · + an3 − n ≤ (n + 1)(a12 + a22 + · · · + an2 − n),
with equality for a1 = a2 = · · · = an = 1, and also for a1 = n and a2 = a3 = · · · = an = 0
(or any cyclic permutation).
• If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
a14 + a24 + · · · + an4 − n ≤ (n2 + n + 1)(a12 + a22 + · · · + an2 − n),
with equality for a1 = a2 = · · · = an = 1, and also for a1 = n and a2 = a3 = · · · = an = 0
(or any cyclic permutation).
P 1.9. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n, then
1
1
1
n2
+
+ ··· +
− n ≥ 4(n − 1)(a12 + a22 + · · · + an2 − n).
a1 a2
an
(Vasile Cirtoaje, 2004)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
s=
a1 + a2 + · · · + an
= 1,
n
n2
− 4(n − 1)u2 , u ∈ (0, n).
u
For u ∈ (0, 1], we have
f 00 (u) =
2n2
− 8(n − 1) ≥ 2n2 − 8(n − 1) = 2(n − 2)2 ≥ 0.
u3
Half Convex Function Method
27
Thus, f is convex on (0, 1]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y > 0 such that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
We have
g(u) =
and
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
−n2
− 4(n − 1)(u + 1)
u
[n − (2n − 2) y]2
n2
− 4(n − 1) =
≥ 0.
xy
xy
In accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also
for a1 = n/2 and a2 = a3 = · · · = an = n/(2n − 2) (or any cyclic permutation).
P 1.10. If a1 , a2 , . . . , a8 are positive real numbers such that a1 + a2 + · · · + a8 = n, then
1
a12
+
1
a22
+ ··· +
1
a82
≥ a12 + a22 + · · · + a82 .
(Vasile Cirtoaje, 2007)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (a8 ) ≥ 8 f (s),
where
f (u) =
s=
a1 + a2 + · · · + a8
= 1,
8
1
− u2 , u ∈ (0, n).
u2
For u ∈ (0, 1], we have
f 00 (u) =
6
− 2 ≥ 6 − 2 > 0.
u4
Thus, f is convex on (0, 1]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y > 0 such that x + 7 y = 8, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
We have
g(u) = −u − 1 −
f (u) − f (1)
.
u−1
1
1
− 2
u u
28
Vasile Cîrtoaje
and
h(x, y) = −1 +
From 8 = x + 7 y ≥ 2
p
x+y
1
+ 2 2.
xy
x y
7x y, we get x y ≤ 16/7. Therefore,
7(x + y) 112 y 2 − 170 y + 72
1
+
=
xy
16x y
16x y
2
2
112 y − 176 y + 72 14 y − 22 y + 9
>
=
> 0.
16x y
2x y
h(x, y) ≥ −1 +
The equality holds for a1 = a2 = · · · = a8 = 1.
Remark. Similarly, we can prove the following generalization.
• If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that a1 + a2 + · · · + an = n,
then
1
1
1
8 2
a1 + a22 + · · · + an2 .
+ 2 + ··· + 2 + 8 − n ≥
2
an
n
a1 a2
P 1.11. If a1 , a2 , . . . , an are positive real numbers such that
a12
+ a22
+ · · · + an2
p
−n≥2 1+
1
1
1
+
+ ··· +
= n, then
a1 a2
an
n−1
(a1 + a2 + · · · + an − n).
n
(Vasile Cirtoaje, 2006)
Solution. Replacing each ai by 1/ai , we need to prove that
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
s=
a1 + a2 + · · · + an
= 1,
n
p
1
n−1 1
f (u) = 2 − 2 1 +
, u ∈ (0, n).
u
n
u
For u ∈ (0, 1], we have
p
6 − 4 1 + n−1
u 6−4 1+
n
f 00 (u) =
≥
u4
u4
p
n−1
n
p
2( n − 1 − 1)2
=
≥ 0.
nu4
Thus, f is convex on (0, 1]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y > 0 such that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
Half Convex Function Method
We have
29
p
−1
2 n−1 1
g(u) = 2 + 1 +
u
n
u
and
1
h(x, y) =
xy
We only need to show that
p
2 n−1
1 1
+ −1−
.
x
y
n
p
2 n−1
1 1
+ ≥1+
.
x
y
n
Indeed, using the Cauchy-Schwarz inequality, we get
p
p
p
(1 + n − 1)2
(1 + n − 1)2
2 n−1
1 1
+ ≥
=
=1+
.
x
y
x + (n − 1) y
n
n
In accordance
p with Remark 3, the equality holds forpa1 = a2 = · · · = an = 1, and also for
1+ n−1
n−1+ n−1
a1 =
and a2 = a3 = · · · = an =
(or any cyclic permutation).
n
n
P 1.12. If a, b, c are nonnegative real numbers, no two of which are zero, then
1
1
1
2
1
1
1
+
+
≤
+
+
.
3a + b + c 3b + c + a 3c + a + b
5 b+c c+a a+b
(Vasile Cirtoaje, 2006)
Solution. Due to homogeneity, we may assume that a + b + c = 3. So, we need to show
that
a+b+c
f (a) + f (b) + f (c) ≥ 3 f (s), s =
= 1,
3
where
2
5
−
, u ∈ [0, 3).
f (u) =
3 − u 2u + 3
For u ∈ [1, 3), we have
f 00 (u) =
4
40
36[2u3 + 3u2 + 9(u − 1)(3 − u)]
−
=
> 0;
(3 − u)3 (2u + 3)3
(3 − u)3 (2u + 3)3
therefore, f is convex on [1, 3). By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + 2 y = 3, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
30
Vasile Cîrtoaje
We have
g(u) =
2
1
+
3 − u 2u + 3
and
9(2x + 2 y − 3)
1
2
−
=
(3 − x)(3 − y) (2x + 3)(2 y + 3) (3 − x)(3 − y)(2x + 3)(2 y + 3)
9x
≥ 0.
=
(3 − x)(3 − y)(2x + 3)(2 y + 3)
h(x, y) =
The equality holds for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 1.13. If a, b, c, d ≥ 3 −
p
7 such that a + b + c + d = 4, then
1
1
1
1
4
+
+
+
≥ .
2
2
2
2
2+a
2+ b
2+c
2+d
3
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
where
f (u) =
s=
a+b+c+d
= 1,
4
p
1
, u ≥ 3 − 7.
2
2+u
For u ≥ 1, we have
f 00 (u) =
3(3u2 − 2)
> 0.
(2 + u2 )3
Thus, f is convex for u ≥ 1. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + 3 y = 4. We have
g(u) =
and
h(x, y) =
since
f (u) − f (1)
−1 − u
=
u−1
3(2 + u2 )
xy + x + y −2
g(x) − g( y)
=
≥0
x−y
3(2 + x 2 )(2 + y 2 )
−x 2 + 6x − 2 (3 +
xy + x + y −2=
=
3
p
p
7 − x)(x − 3 + 7)
≥ 0.
3
Half Convex Function Method
31
In accordance with Remark 3, the
p equality holds for a = b = c = d = 1, and also for
p
1+ 7
(or any cyclic permutation).
a = 3 − 7 and b = c = d =
3
Remark. Similarly, we can prove the following generalization.
p
• If a1 , a2 , . . . , an ≥ n − 1 − n2 − 3n + 3 such that a1 + a2 + · · · + an = n, then
1
+
2 + a12
1
2 + a22
+ ··· +
n
1
≥ ,
2 + an2
3
with equality for a1 = a2 p= · · · = an = 1, and also for a1 = n − 1 −
1 + n2 − 3n + 3
a2 = a3 = · · · = an =
(or any cyclic permutation).
n−1
p
n2 − 3n + 3 and
P 1.14. If a1 , a2 , . . . , an ∈ [0, n − 2] such that a1 + a2 + · · · + an = n, then
1
n + a12
+
1
n + a22
+ ··· +
n
1
.
≤
2
n + an
n+1
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
s=
a1 + a2 + · · · + an
= 1,
n
−1
, u ∈ [0, n − 2], n ≥ 3.
n + u2
For u ∈ [0, 1], we have
f 00 (u) =
2(n − 3u2 )
≥ 0.
(n + u2 )3
Thus, f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have
g(u) =
and
h(x, y) =
f (u) − f (1)
u+1
=
u−1
(n + 1)(n + u2 )
g(x) − g( y)
n− x − y − xy
=
.
x−y
(n + 1)(n + x 2 )(n + y 2 )
We need to show that
n − x − y − x y ≥ 0.
32
Vasile Cîrtoaje
Indeed, we have
n− x − y − xy =
(n − x)(n − 2 − x)
≥ 0.
n−1
The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = n − 2 and a2 = a3 =
2
· · · = an =
(or any cyclic permutation).
n−1
P 1.15. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
3−a
3− b
3−c
3
+
+
≥ .
9 + a2 9 + b2 9 + c 2
5
(Vasile Cirtoaje, 2013)
Solution. Write the inequality as
f (a) + f (b) + f (c) ≥ 3 f (s),
where
f (u) =
s=
a+b+c
= 1,
3
3−u
, u ∈ [0, 3].
9 + u2
For u ∈ [1, 3], we have
1 00
u2 (9 − u) + 27(u − 1)
f (u) =
> 0.
2
(9 + u2 )3
Thus, f is convex on [1, 3]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + 2 y = 3, where
h(x, y) =
g(x) − g( y)
,
x−y
We have
g(u) =
and
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
−(6 + u)
5(9 + u2 )
x y + 6x + 6 y − 9
x(9 − x)
=
≥ 0.
5(9 + x 2 )(9 + y 2 ) 10(9 + x 2 )(9 + y 2 )
The equality holds for a = b = c = 1, and also for a = 0 and b = c = 3/2 (or any cyclic
permutation).
Remark. Similarly, we can prove the following generalization.
Half Convex Function Method
33
• If a1 , a2 , . . . , an (n ≥ 3) are nonnegative real numbers such that a1 + a2 + · · · + an = n,
then
n − a1
n2
+ (n2
− 3n + 1)a12
+
n − a2
n2
+ (n2
− 3n + 1)a22
+ ··· +
n2
n − an
n
≥
,
2
− 3n + 1)an
2n − 1
+ (n2
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 0 and a2 = a3 = · · · = an =
n
(or any cyclic permutation).
n−1
P 1.16. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
1
1
3
1
+
+
≥ .
2
2
2
5+a+a
5+ b+ b
5+c+c
7
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a) + f (b) + f (c) ≥ 3 f (s),
where
f (u) =
s=
a+b+c
= 1,
3
1
, u ∈ [0, 3].
5 + u + u2
For u ≥ 1, we have
f 00 (u) =
2(3u2 + 3u − 4)
> 0.
(5 + u + u2 )3
Thus, f is convex on [1,3]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + 2 y = 3. We have
g(u) =
f (u) − f (1)
−2 − u
=
u−1
7(5 + u + u2 )
and
h(x, y) =
g(x) − g( y)
x y + 2(x + y) − 3
x(5 − x)
=
=
≥ 0.
2
2
x−y
7(5 + x + x )(5 + y + y ) 14(5 + x + x 2 )(5 + y + y 2 )
In accordance with Remark 3, the equality holds for a = b = c = 1, and also for a = 0
and b = c = 3/2 (or any cyclic permutation).
Remark. Similarly, we can prove the following generalization.
34
Vasile Cîrtoaje
• Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
2(2n − 1)
0<k≤
, then
n−1
1
k + a1 + a12
+
1
k + a2 + a22
+ ··· +
1
n
≥
,
2
k + an + an
k+2
2(2n − 1)
with equality for a1 = a2 = · · · = an = 1. If k =
, then the equality holds also
n−1
n
(or any cyclic permutation).
for a1 = 0 and a2 = a3 = · · · = an =
n−1
P 1.17. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then
1
1
1
1
1
+
+
+
≤ .
2
2
2
2
10 + a + a
10 + b + b
10 + c + c
10 + d + d
3
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
where
f (u) =
s=
a+b+c+d
= 1,
4
−1
, u ∈ [0, 4].
10 + u + u2
For u ∈ [0, 1], we have
f 00 (u) =
6(3 − u − u2 )
> 0.
(10 + u + u2 )3
Thus, f is convex on [0,1]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + 3 y = 4. We have
g(u) =
and
h(x, y) =
f (u) − f (1)
2+u
=
u−1
12(10 + u + u2 )
g(x) − g( y)
8 − 2(x + y) − x y
=
.
x−y
12(10 + x + x 2 )(10 + y + y 2 )
We need to show that
8 − 2(x + y) − x y ≥ 0.
Indeed, we have
8 − 2(x + y) − x y = 3 y 2 ≥ 0.
Half Convex Function Method
35
The equality holds for a = b = c = d = 1, and also for a = 4 and b = c = d = 0 (or any
cyclic permutation).
Remark. Similarly, we can prove the following generalization.
• Let a1 , a2 , . . . , an (n ≥ 4) be nonnegative real numbers such that a1 + a2 +· · ·+ an = n.
If k ≥ 2n + 2, then
1
k + a1 + a12
1
+
k + a2 + a22
+ ··· +
1
n
≤
,
2
k + an + an
k+2
with equality for a1 = a2 = · · · = an = 1. If k = 2n + 2, then the equality holds also for
a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation).
P 1.18. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
1
k ≥ 1 − , then
n
1
n
1
1
≥
.
+
+ ··· +
2
2
2
1
+
ka
1
+
k
1 + ka1 1 + ka2
n
(Vasile Cirtoaje, 2005)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
a1 + a2 + · · · + an
= 1,
n
1
, u ∈ [0, n].
1 + ku2
For u ∈ [1, n], we have
f 00 (u) =
since
s=
2k(3ku2 − 1)
> 0,
(1 + ku2 )3
1
3
3ku − 1 ≥ 3k − 1 ≥ 3 1 −
− 1 = 2 − > 0.
n
n
2
Thus, f is convex on [1, n]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have
g(u) =
and
h(x, y) =
f (u) − f (1)
−k(u + 1)
=
u−1
(1 + k)(1 + ku2 )
g(x) − g( y)
k2 (x + y + x y) − k
=
.
x−y
(1 + k)(1 + k x 2 )(1 + k y 2 )
36
Vasile Cîrtoaje
We need to show that
k(x + y + x y) − 1 ≥ 0.
Indeed, we have
1
x(2n − 2 − x)
≥ 0.
k(x + y + x y) − 1 ≥ 1 −
(x + y + x y) − 1 =
n
n
1
The equality holds for a1 = a2 = · · · = an = 1. If k = 1 − , then the equality holds also
n
n
for a1 = 0 and a2 = a3 = · · · = an =
(or any cyclic permutation).
n−1
P 1.19. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n. If 0 < k ≤
n−1
, then
n2 − n + 1
n
1
1
1
≤
.
+
+ ··· +
2
2
2
1
+
ka
1
+
k
1 + ka1 1 + ka2
n
(Vasile Cirtoaje, 2005)
Solution. Replacing all negative numbers ai by −ai , we need to show the same inequality for a1 , a2 , . . . , an ≥ 0 such that a1 + a2 + · · · + an ≥ n. Since the left side of
the desired inequality is decreasing with respect to each ai , is sufficient to consider that
a1 + a2 + · · · + an = n. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
s=
a1 + a2 + · · · + an
= 1,
n
−1
, u ∈ [0, n].
1 + ku2
For u ∈ [0, 1], we have
f 00 (u) =
2k(1 − 3ku2 )
≥ 0,
(1 + ku2 )3
since
1 − 3ku2 ≥ 1 − 3k ≥ 1 −
3(n − 1)
(n − 2)2
=
≥ 0.
n2 − n + 1
n2 − n + 1
Thus, f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have
g(u) =
f (u) − f (1)
k(u + 1)
=
u−1
(1 + k)(1 + ku2 )
Half Convex Function Method
and
h(x, y) =
37
k − k2 (x + y + x y)
g(x) − g( y)
=
.
x−y
(1 + k)(1 + k x 2 )(1 + k y 2 )
We need to show that
1 − k(x + y + x y) ≥ 0.
Indeed, we have
1 − k(x + y + x y) ≥ 1 −
n−1
(x − n + 1)2
(x
+
y
+
x
y)
=
≥ 0.
n2 − n + 1
n2 − n + 1
The equality holds for a1 = a2 = · · · = an = 1. If k =
also for a1 = n − 1 and a2 = a3 = · · · = an =
n2
n−1
, then the equality holds
−n+1
1
(or any cyclic permutation).
n−1
P 1.20. Let a1 , a2 , . . . , an be nonnegative numbers such that a1 + a2 + · · · + an = n. If
n2
, then
k≥
4(n − 1)
a1 (a1 − 1)
a12
+k
+
a2 (a2 − 1)
a22
+k
+ ··· +
an (an − 1)
≥ 0.
an2 + k
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
From
f 0 (u) =
u2 + 2ku − k
,
(u2 + k)2
s=
a1 + a2 + · · · + an
= 1,
n
u(u − 1)
, u ∈ [0, n].
u2 + k
f 00 (u) =
2(k2 − u3 ) + 6ku(1 − u)
,
(u2 + k)3
it follow that f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show
that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
We have
g(u) =
u
,
2
u +k
h(x, y) =
k−xy
n2 − 4(n − 1)x y
≥
.
(x 2 + k)( y 2 + k) 4(n − 1)(x 2 + k)( y 2 + k)
38
Vasile Cîrtoaje
We only need to show that n2 ≥ 4(n − 1)x y. Indeed, this follows by the AM-GM inequality, as follows:
Æ
2
n2 = [x + (n − 1) y]2 ≥ 2 (n − 1)x y
= 4(n − 1)x y.
The equality holds for a1 = a2 = · · · = an = 1, and also for a1 = n/2 and a2 = a3 =
· · · = an = n/(2n − 2) (or any cyclic permutation).
P 1.21. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n,
then
1 − an
1 − a1
1 − a2
+
+ ··· +
≤ 0.
2
2
(n − 2a1 )
(n − 2a2 )
(n − 2an )2
(Vasile Cîrtoaje, 2012)
Solution. For n = 2, the inequality is an identity. Consider further n ≥ 3 and write the
inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
From
f 0 (u) =
s=
a1 + a2 + · · · + an
= 1,
n
u−1
, u ∈ I = [0, n] \ {n/2}.
(n − 2u)2
2u + n − 4
,
(n − 2u)3
f 00 (u) =
8(u + n − 3)
,
(n − 2u)4
it follow that f is convex on [0, 1]. By HCF Theorem, Remark 1 and Remark 4, it suffices
to show that h(x, y) ≥ 0 for x, y ∈ I such that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
We have
g(u) =
and
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
1
(n − 2u)2
4(n − x − y)
4(n − 2) y
=
≥ 0.
2
2
(n − 2x) (n − 2 y)
(n − 2x)2 (n − 2 y)2
In accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1, and also
for a1 = n and a2 = a3 = · · · = an = 0 (or any cyclic permutation).
Half Convex Function Method
39
P 1.22. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
n
, then
k ≥1+ p
n−1
a12 − 1
(a1 + k)2
+
a22 − 1
(a2 + k)2
+ ··· +
an2 − 1
(an + k)2
≥ 0.
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
s=
a1 + a2 + · · · + an
= 1,
n
u2 − 1
, u ∈ [0, n].
(u + k)2
For u ∈ [0, 1], we have
f 00 (u) =
2(k2 − 3 − 2ku) 2(k2 − 2k − 3) 2(k + 1)(k − 3)
≥
=
≥ 0.
(u + k)4
(u + k)4
(u + k)4
Thus, f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have
g(u) =
and
h(x, y) =
f (u) − f (1)
u+1
=
u−1
(u + k)2
g(x) − g( y) (k − 1)2 − (1 + x)(1 + y)
=
.
x−y
(x + k)2 ( y + k)2
Since
(k − 1)2 ≥
n2
,
n−1
we need to show that
n2 ≥ (n − 1)(1 + x)(1 + y).
Indeed,
n2 − (n − 1)(1 + x)(1 + y) = n2 − (1 + x)(2n − 1 − x) = (x − n + 1)2 ≥ 0.
Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1. If k =
n
1
1+ p
, then the equality holds also for a1 = n − 1 and a2 = a3 = · · · = an =
n−1
n−1
(or any cyclic permutation).
40
Vasile Cîrtoaje
P 1.23. Let as
1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
2n − 1
0< k ≤1+
, then
n−1
a12 − 1
(a1 + k)2
+
a22 − 1
(a2 + k)2
+ ··· +
an2 − 1
(an + k)2
≤ 0.
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
s=
a1 + a2 + · · · + an
= 1,
n
1 − u2
, u ∈ [0, n].
(u + k)2
For u ≥ 1, we have
f 00 (u) =
2(2ku − k2 + 3) 2(2k − k2 + 3) 2(1 + k)(3 − k)
≥
=
> 0.
(u + k)4
(u + k)4
(u + k)4
Thus, f is convex on [1, n]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. We have
g(u) =
and
h(x, y) =
f (u) − f (1)
−u − 1
=
u−1
(u + k)2
2k − k2 + x + y
g(x) − g( y) 2k − k2 + x + y + x y
=
≥
.
x−y
(x + k)2 ( y + k)2
(x + k)2 ( y + k)2
Since
x+y≥
x + (n − 1) y
n
=
,
n−1
n−1
we get
2k − k2 + x + y ≥ 2k − k2 +
n
2n − 1
= −(k − 1)2 +
≥ 0,
n−1
n−1
hence h(x, y) ≥ 0.sThus, the proof is completed. The equality holds for a1 = a2 = · · · =
2n − 1
an = 1. If k = 1 +
, then the equality holds also for a1 = 0 and a2 = a3 = · · · =
n−1
n
an =
(or any cyclic permutation).
n−1
Half Convex Function Method
41
P 1.24. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
(n − 1)(2n − 1)
k≥
, then
n2
1
1
n
1
+
+ ··· +
≥
.
3
3
3
1 + kan
1+k
1 + ka1 1 + ka2
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
For u ∈ [1, n], we have
f 00 (u) =
s=
a1 + a2 + · · · + an
= 1,
n
1
, u ∈ [0, n].
1 + ku3
6ku(2ku3 − 1) 6ku(2k − 1)
≥
> 0.
(1 + ku3 )3
(1 + ku3 )3
Thus, f is convex on [1, n]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
We have
g(u) =
and
g(u) =
f (u) − f (1)
.
u−1
−k(u2 + u + 1)
(1 + k)(1 + ku3 )
x 2 y 2 + x y(x + y − 1) + (x + y)2 − (x + y + 1)/k
1
h(x,
y)
=
.
k2
(1 + k)(1 + k x 3 )(1 + k y 3 )
Since
x+y≥
x + (n − 1) y
n
=
> 1,
n−1
n−1
it suffices to show that k(x + y)2 ≥ x + y + 1. From x + y ≥
k(x + y) ≥
n
, we get
n−1
2n − 1
n
and
n
2n − 1
− 1 = 1.
n−1
n
(n − 1)(2n − 1)
The equality holds for a1 = a2 = · · · = an = 1. If k =
, then the equality
n2
n
holds also for a1 = 0 and a2 = a3 = · · · = an =
(or any cyclic permutation).
n−1
k(x + y)2 − x − y ≥ (x + y)[k(x + y) − 1] ≥
42
Vasile Cîrtoaje
P 1.25. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
n−1
, then
0<k≤ 2
n − 2n + 2
n
1
1
1
≤
+
+ ··· +
.
3
3
3
1 + kan
1+k
1 + ka1 1 + ka2
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
For u ∈ [0, 1], we have
f 00 (u) =
s=
a1 + a2 + · · · + an
= 1,
n
−1
, u ∈ [0, n].
1 + ku3
6ku(1 − 2ku3 ) 6ku(1 − 2k)
≥
> 0.
(1 + ku3 )3
(1 + ku3 )3
Thus, f is convex on [0, 1]. By HCF Theorem and Remark 1, it suffices to show that
h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
We have
g(u) =
g(u) =
f (u) − f (1)
.
u−1
k(u2 + u + 1)
(1 + k)(1 + ku3 )
and
(x + y + 1)/k − x 2 y 2 − x y(x + y − 1) − (x + y)2
1
h(x,
y)
=
.
k2
(1 + k)(1 + k x 3 )(1 + k y 3 )
It suffices to show that
(n2 − 2n + 2)(x + y + 1)
− x 2 y 2 − x y(x + y − 1) − (x + y)2 ≥ 0,
n−1
which is equivalent to
[2 + n y − (n − 1) y 2 ][1 − (n − 1) y]2 ≥ 0.
This is true since
2 + n y − (n − 1) y 2 = 2 + y[n − (n − 1) y] = 2 + x y > 0.
In accord with Remark 3, the equality holds for a1 = a2 = · · · = an = 1. If k =
n−1
1
, then the equality holds also for a1 = n − 1 and a2 = a3 = · · · = an =
2
n − 2n + 2
n−1
(or any cyclic permutation).
Half Convex Function Method
43
P 1.26. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
n2
k≥
, then
n−1
v
v
v
t a1
t a2
t an
n
+
+ ··· +
≤p
.
k − a1
k − a2
k − an
k−1
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) = −
s
s=
a1 + a2 + · · · + an
= 1,
n
u
, u ∈ [0, n].
k−u
For u ∈ [0, 1], we have
f 00 (u) =
k(k − 4u)
4u3/2 (k − u)5/2
≥
k(k − 4)
4u3/2 (k − u)5/2
≥ 0.
Thus, f is convex on [0, 1]. By HCF Theorem, it suffices to prove that f (x) + (n −
1) f ( y) ≥ n f (1) for all x, y ≥ 0 such that x + (n − 1) y = n. We write the inequality as
v
v
t (k − 1)x
t (k − 1) y
+ (n − 1)
≤ n,
k−x
k− y
v
v
t (k − 1) y
t
(n − 1)k(1 − y)
1+
≤ 1 + (n − 1) 1 −
.
(n − 1) y + k − n
k− y
Let
z=
v
t (k − 1) y
k− y
,
which yields
y=
1− y =
(k − 1)(1 − z 2 )
,
z2 + k − 1
kz 2
,
z2 + k − 1
(n − 1) y + k − n =
(k − 1)(nz 2 + k − n)
,
z2 + k − 1
hence
k(1 − y)
k(1 − z 2 )
1
=
=
2
(n − 1) y + k − n
k − n(1 − z ) 1/(1 − z 2 ) − n/k
1
n(1 − z 2 )
≤
=
.
1/(1 − z 2 ) − (n − 1)/n (n − 1)z 2 + 1
44
Vasile Cîrtoaje
Therefore, it suffices to show that
v
t
n(n − 1)(1 − z 2 )
1+
≤ 1 + (n − 1)(1 − z).
(n − 1)z 2 + 1
By squaring, we get the obvious inequality
(z − 1)2 [(n − 1)z − 1]2 ≥ 0.
So, we only need to show that 1 + (n − 1)(1 − z) ≥ 0, that is, z ≤ n/(n − 1). Since
y=
and
we have
n− x
n
≤
n−1
n−1
1
n−1
1
≤ 2
= 2
,
k−1
n /(n − 1) − 1
n −n+1
v
v
v
t
t k−1
t
k−1
n
z=
≤
=
k/ y − 1
k(n − 1)/n − 1
n − 1 − 1/(k − 1)
p
v
t
n
n
n2 − n + 1
≤
=
<
.
2
n − 1 − (n − 1)/(n − n + 1)
n−1
n−1
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1. If k =
n2
n(n − 1)2
, then the equality holds also for a1 = 2
and a2 = a3 = · · · = an =
n−1
n − 2n + 2
n
(or any cyclic permutation).
2
(n − 1)(n − 2n + 2)
P 1.27. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n,
then
2
2
2
n−a1 + n−a2 + · · · + n−an ≥ 1.
(Vasile Cîrtoaje, 2006)
Solution. Let k = ln n. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
s=
a1 + a2 + · · · + an
= 1,
n
2
f (u) = n−u , u ∈ [0, n].
For u ≥ 1, we have
2
2
2
f 00 (u) = 2kn−u (2ku2 − 1) ≥ 2kn−u (2k − 1) ≥ 2kn−u (2 ln 2 − 1) > 0;
Half Convex Function Method
45
therefore, f is convex on [1, n]. By HCF Theorem and Remark 5, it suffices to show
that f (x) + (n − 1) f ( y) ≥ n f (1) for 0 ≤ x ≤ 1 ≤ y and x + (n − 1) y = n. The desired
inequality is equivalent to g(x) ≥ 0, where
2
2
g(x) = n−x + (n − 1)n− y ,
y=
n− x
, 0 ≤ x ≤ 1.
n−1
Since y 0 = −1/(n − 1), we get
2
2
2
2
g 0 (x) = −2x kn−x − 2(n − 1)k y y 0 n− y = 2k( y n− y − x n−x ).
The derivative g 0 (x) has the same sign as g1 (x), where
2
2
g1 (x) = ln( y n− y ) − ln(x n−x ) = ln y − ln x + k(x 2 − y 2 ).
From
g10 (x) =
y0 1
−1
2k + 2(n − 2)k x
− + 2k(x − y y 0 ) = n
+
,
y
x
x(n − x)
(n − 1)2
we see that g10 (x) has for 0 < x ≤ 1 the same sign as
h(x) =
−(n − 1)2
+ x(n − x)(1 + nx − 2x).
2k
Since
h0 (x) = n + 2(n2 − 2n − 1)x − 3(n − 2)x 2 ≥ nx + 2(n2 − 2n − 1)x − 3(n − 2)x
= 2(n − 1)(n − 2)x ≥ 0,
h is strictly increasing on [0, 1]. From
−(n − 1)2
h(0) =
< 0,
2k
1
h(1) = (n − 1) 1 −
> 0,
2k
2
it follows that there is x 1 ∈ (0, 1) such that h(x) < 0 for x ∈ [0, x 1 ), h(x 1 ) = 0 and
h(x) > 0 for x ∈ (x 1 , 1]. Therefore, g1 is strictly decreasing on (0, x 1 ] and strictly
increasing on [x 1 , 1]. Since lim x→0 g1 (x) = ∞ and g1 (1) = 0, there is x 2 ∈ (0, x 1 ) such
that g1 (x) > 0 for x ∈ (0, x 2 ), g1 (x 2 ) = 0 and g1 (x) < 0 for x ∈ (x 2 , 1). Consequently,
g is strictly increasing on [0, x 2 ] and strictly decreasing on [x 2 , 1]. From g(0) > 0 and
g(1) = 0, it follows that g(x) ≥ 0 for x ∈ [0, 1]. The proof is completed. The equality
holds for a1 = a2 = · · · = an = 1.
P 1.28. If a, b, c, d, e are nonnegative real numbers such that a + b + c + d + e = 5, then
then
(a2 + 1)(b2 + 1)(c 2 + 1)(d 2 + 1)(e2 + 1) ≥ (a + 1)(b + 1)(c + 1)(d + 1)(e + 1).
46
Vasile Cîrtoaje
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) + f (e) ≥ n f (s),
s=
a+b+c+d+e
= 1,
5
where
f (u) = ln(u2 + 1) − ln(u + 1),
u ∈ [0, 5].
For u ∈ [0, 1], we have
f 00 (u) =
2(1 − u2 )
1
(u2 − u4 ) + 4u(1 − u2 ) + u2 + 3
+
=
> 0.
(u2 + 1)2 (u + 1)2
(u2 + 1)2 (u + 1)2
Therefore, f is convex on [0, 1]. By HCF Theorem and Remark 2, we only need to show
that H(x, y) ≥ 0 for x, y ≥ 0 such that x + 4 y = 5, where
H(x, y) =
2(1 − x y)
f 0 (x) − f 0 ( y)
1
= 2
+
;
x−y
(x + 1)( y 2 + 1) (x + 1)( y + 1)
that is,
(x 2 + 1)( y 2 + 1)H(x, y) = 2(1 − x y) +
Since
x2 + 1
x +1
≥
,
x +1
2
(x 2 + 1)( y 2 + 1)
.
(x + 1)( y + 1)
y2 + 1
y +1
≥
,
y +1
2
it suffices to prove that
2(1 − x y) +
(x + 1)( y + 1)
≥ 0,
4
which is equivalent to x + y + 9 − 7x y ≥ 0. Indeed,
x + y + 9 − 7x y = 28x 2 − 38x + 14 = 28(x − 19/28)2 + 31/28 > 0.
The proof is completed. The equality holds for a = b = c = d = e = 1.
P 1.29. If a1 , a2 , · · · , a10 are nonnegative real numbers such that a1 + a2 + · · · + a10 = 10,
then
2
(1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a10 + a10
) ≥ 1.
(Vasile Cîrtoaje, 2006)
Half Convex Function Method
47
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (a10 ) ≥ 10 f (s),
s=
a1 + a2 + · · · + a10
= 1,
10
where
f (u) = ln(1 − u + u2 ),
From
f 00 (u) =
u ∈ [0, 10].
1 + 2u(1 − u)
,
(1 − u + u2 )2
it follows that f 00 (u) > 0 for u ∈ [0, 1], hence f is convex on [0, 1]. According to HCF
Theorem, we only need to show that f (x) + 9 f ( y) ≥ 10 f (1) for all x, y ≥ 0 such that
x + 9 y = 10. Using Remark 2, it suffices to prove that H(x, y) ≥ 0, where
H(x, y) =
Since
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
1 + x + y − 2x y
,
(1 − x + x 2 )(1 − y + y 2 )
we need to show that
1 + x + y − 2x y ≥ 0.
Indeed,
7
1 + x + y − 2x y = 18 y − 28 y + 11 = 18 y −
9
2
2
+
1
> 0.
9
The proof is completed. The equality holds for a1 = a2 = · · · = a10 = 1.
P 1.30. If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 + a2 + · · · + an = 1,
then
p
p
p
1
1
1
1 n
p
n− p
.
p − a2 · · · p − an ≥
p − a1
a1
a2
an
n
(Vasile Cîrtoaje, 2006)
Solution. Apply HCF Theorem to the function
p
1
1
f (u) = ln p − u = ln(1 − u) − ln u,
2
u
u ∈ (0, 1),
for s = 1/n. From
f 0 (u) =
−1
1
−
,
1 − u 2u
f 00 (u) =
1 − 2u − u2
,
2u2 (1 − u)2
48
Vasile Cîrtoaje
p
it follows that f is convex on (0, 2 − 1]. Since
s=
1 1 p
≤ < 2 − 1,
n 3
f is also convex on (0, s].
First Solution. By HCF Theorem, it suffices to show that f (x) + (n − 1) f ( y) ≥ n f (1/n)
for all x, y > 0 such that x + (n − 1) y = 1; that is, to show that
p
1
p − x
x
1
p
p − y
y
n−1
≥
p
1
n− p
n
n
.
Write this inequality as
nn/2 (1 − y)n−1 ≥ (n − 1)n−1 x 1/2 y (n−3)/2 .
By squaring, this inequality becomes as follows
nn (1 − y)2n−2 ≥ (n − 1)2n−2 x y n−3 ,
(2 − 2 y)2n−2 ≥
1
n · + x + (n − 3) y
n
2n−2
(2n − 2)2n−2
x y n−3 ,
nn
≥ [n + 1 + (n − 3)]n+1+(n−3) ·
1
· x · y n−3 .
nn
Clearly, the last inequality follows from the AM-GM inequality. The proof is completed.
The equality holds for a1 = a2 = · · · = an = 1/n.
Second Solution. By HCF Theorem and Remark 2, it suffices to prove that H(x, y) ≥ 0
for x, y > 0 such that x + (n − 1) y = 1, where
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
We have
(n − 1)(1 + y) − 2
1− x − y − xy
=
2x y(1 − x)(1 − y)
2x(1 − x)(1 − y)
2(1 + y) − 2
y
≥
=
> 0.
2x(1 − x)(1 − y)
x(1 − x)(1 − y)
H(x, y) =
Remark 1. We may write the inequality in P 1.30 in the form
Y
n n
Y
p
p
1 n
1
−
1
·
(1
+
a
)
≥
n
−
.
p
p
i
ai
n
i=1
i=1
Half Convex Function Method
49
On the other hand, by the AM-GM inequality and the Cauchy-Schwarz inequality, we
have
v
!n
n
u X
n
n
n
Y
p
1 n
1 Xp
t1
ai
= 1+ p
ai ≤ 1 +
(1 + ai ) ≤ 1 +
.
n i=1
n i=1
n
i=1
Thus, the following statement follows.
• If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 + a2 + · · · + an = 1,
then
p
1
1
1
n
p − 1 · · · p − 1 ≥ ( n − 1) ,
p −1
a1
a2
an
with equality for a1 = a2 = · · · = an = 1/n.
Remark 2. By squaring, the inequality in P 1.30 becomes
n
Y
(1 − ai )2
(n − 1)2n
.
≥
ai
nn
i=1
1+ x
is convex on (0, 1), by Jensen’s
1− x
inequality,
a1 + a2 + · · · + an n
n
1+
Y
1 + ai
n+1 n
n
≥
.
=
a1 + a2 + · · · + an
1 − ai
n−1
i=1
1−
n
Multiplying these inequality yields the following result (Kee-Wai Lau, 2000).
On the other hand, since the function f (x) = ln
• If a1 , a2 , . . . , an (n ≥ 3) are positive real numbers such that a1 + a2 + · · · + an = 1,
then
1
1
1 n
1
− a1
− a2 · · ·
− an ≥ n −
,
a1
a2
an
n
with equality for a1 = a2 = · · · = an = 1/n.
P 1.31. Let a1 , a2 , . . . , an be positive real numbers such that a1 + a2 + · · · + an = n. If
2
p
2 n−1
0< k ≤ 1+
,
n
then
1
ka1 +
a1
1
1
ka2 +
· · · kan +
≥ (k + 1)n .
a2
an
(Vasile Cîrtoaje, 2006)
50
Vasile Cîrtoaje
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
s=
1
f (u) = ln ku +
,
u
We have
f 0 (u) =
ku2 − 1
,
u(ku2 + 1)
f 00 (u) =
a1 + a2 + · · · + an
= 1,
n
u ∈ (0, n).
1 + 4ku2 − k2 u4
.
u2 (ku2 + 1)2
For u ∈ (0, 1], we have f 00 (u) > 0, since
1 + 4ku2 − k2 u4 > ku2 (4 − ku2 ) ≥ ku2 (4 − k) ≥ 0.
Therefore, f is convex on (0, 1]. By HCF Theorem and Remark 2, it suffices to prove
that H(x, y) ≥ 0 for x, y > 0 such that x + (n − 1) y = n, where
H(x, y) =
Since
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
1 + k(x + y)2 − k2 x 6 y 2
k[(x + y)2 − k x 6 y 2 ]
>
,
x y(k x 2 + 1)(k y 2 + 1)
x y(k x 2 + 1)(k y 2 + 1)
it suffices to show that
x+y≥
p
k x y.
Indeed, by the Cauchy-Schwarz inequality, we have
p
(x + y)[(n − 1) y + x] ≥ ( n − 1 + 1)2 x y,
hence
p
1 p
( n − 1 + 1)2 x y ≥ k x y.
n
The equality holds for a1 = a2 = · · · = an = 1.
x+y≥
P 1.32. If a, b, c, d are nonzero real numbers such that
a, b, c, d ≥
then
−1
,
2
a + b + c + d = 4,
1
1
1
1
1 1 1 1
3 2 + 2 + 2 + 2 + + + + ≥ 16.
a
b
c
d
a b c d
Half Convex Function Method
51
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s), s =
where
a+b+c+d
= 1,
4
3
1
−1 11
f (u) = 2 + , u ∈ I =
,
\ {0}.
u
u
2 2
Clearly, f is convex for u ∈ I, u ≥ s. Therefore, by HCF Theorem and Remark 4, it
suffices to prove that f (x) + 3 f ( y) ≥ 4 f (1) for all x, y ∈ I such that x + 3 y = 4.
According to Remark 1, we only need to show that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
Indeed, we have
3
4
g(u) = − − 2 ,
u u
h(x, y) =
4x y + 3x + 3 y
2(1 + 2x)(6 − x)
=
≥ 0.
2
2
x y
3x 2 y 2
The proof is completed. In accord with Remark 3, the equality holds for a = b = c =
−1
3
d = 1, and also for a =
and b = c = d = (or any cyclic permutation).
2
2
P 1.33. If a1 , a2 , . . . , an are nonnegative real numbers such that a12 + a22 + · · · + an2 = n,
then
s
n
3
3
3
a1 + a2 + · · · + an − n +
(a1 + a2 + · · · + an − n) ≥ 0.
n−1
(Vasile Cirtoaje, 2007)
Solution. Replacing each ai by
p
ai , we have to prove that
a1 + a2 + · · · + an
= 1,
n
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
p
p
f (u) = u u + k u,
n
,
n−1
where
For u ≥ 1, we have
f 00 (u) =
k=
s
u ∈ [0, n].
3u − k
3−k
p ≥ p > 0.
4u u
4u u
52
Vasile Cîrtoaje
Therefore, f is convex for u ≥ 1. According to HCF Theorem and Remark 1, it suffices
to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n. Since
g(u) =
and
f (u) − f (1)
u+k
=1+ p
u−1
u+1
p
p
p
x + y+ xy−k
g(x) − g( y)
h(x, y) =
= p
,
p
p
p
x−y
( x + y)( x + 1)( y + 1)
we need to show that
p
x+
p
y+
This is true if
p
x+y≥
p
x y ≥ k.
s
n
.
n−1
Indeed, we have
x
n
+y=
.
n−1
n−1
The proof is completed. In accord with Remark 3,
sthe equality holds for a1 = a2 = · · · =
n
an = 1, and also for a1 = 0 and a2 = · · · = an =
(or any cyclic permutation).
n−1
x+y≥
P 1.34. If a, b, c, d, e are nonnegative real numbers such that a2 + b2 + c 2 + d 2 + e2 = 5,
then
1
1
1
1
1
+
+
+
+
≤ 1.
7 − 2a 7 − 2b 7 − 2c 7 − 2d 7 − 2e
(Vasile Cirtoaje, 2010)
p
p
p
p
p
Solution. Replacing a, b, c, d, e by a, b, c, d, e, we have to prove that
f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s),
where
1
f (u) = p
,
2 u−7
For u ∈ [0, 1], we have
s=
a+b+c+d+e
= 1,
5
u ∈ [0, 5].
p
7−6 u
f (u) = p
> 0.
p
2u u(7 − 2 u)3
00
Therefore, f is convex for u ∈ [0, 1]. According to HCF Theorem and Remark 1, it
suffices to show that h(x, y) ≥ 0 for x, y ≥ 0 such that x + 4 y = 5. Since
g(u) =
f (u) − f (1)
−2
=
p
p
u−1
5(7 − 2 u)(1 + u)
Half Convex Function Method
53
and
p
p
2(5 − 2 x − 2 y)
g(x) − g( y)
= p
h(x, y) =
p
p
p
p
p ,
x−y
( x + y)(1 + x)(1 + y)(7 − 2 x)(7 − 2 y)
we need to show that
p
5
.
2
Indeed, by the Cauchy-Schwarz inequality, we have
p
1
25
p 2
( x + y) ≤ 1 +
(x + 4 y) =
.
4
4
x+
p
y≤
The proof is completed. The equality holds for a = b = c = d = e = 1, and also for
1
a = 2 and b = c = d = e = (or any cyclic permutation).
2
Remark Similarly, we can prove the following generalization.
• Let a1 , a2 , . . . , an be nonnegative real numbers such that a12 + a22 + · · · + an2 = n. If
n
k ≥1+ p
, then
n−1
1
1
1
n
+
+ ··· +
≤
,
k − a1 k − a2
k − an
k−1
with equality for a1 = a2 = · · · = an = 1. If k = 1 + p
for a1 =
p
n − 1 and a2 = · · · = an = p
P 1.35. If 0 ≤ a1 , a2 , . . . , an <
then
1
n−1
n
n−1
, then the equality holds also
.
k such that a12 +a22 +· · ·+an2
= n, where 1 < k ≤ 1+
s
n
,
n−1
1
1
1
n
+
+ ··· +
≥
.
k − a1 k − a2
k − an
k−1
(Vasile Cirtoaje, 2010)
Solution. Replacing a1 , a2 , . . . , an by
p
a1 ,
p
a2 , . . . ,
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
1
p ,
k− u
p
s=
an , we have to prove that
a1 + a2 + · · · + an
= 1,
n
u ∈ [0, k2 ).
54
Vasile Cîrtoaje
From
p
3 u−k
f (u) = p
,
p
4u u(k − u)3
00
it follows that f is convex for u ≥ 1, since
p
3 u−k ≥3−k ≥2−
s
n
> 0.
n−1
According to HCF Theorem and Remark 1, it suffices to show that h(x, y) ≥ 0 for all
0 ≤ x, y < k2 such that x + (n − 1) y = n. Since
g(u) =
f (u) − f (1)
1
=
p
p
u−1
(k − 1)(k − u)(1 + u)
and
p
p
x + y +1−k
g(x) − g( y)
h(x, y) =
=
p
p
p
p
p
p ,
x−y
(k − 1)( x + y)(1 + x)(1 + y)(k − x)(k − y)
we need to show that
p
x+
p
y ≥ k − 1.
Indeed,
p
x+
p
y≥
p
x+y≥
s
x
+y=
n−1
s
n
≥ k − 1.
n−1
The proof is completed. In accord with Remark 3,
sthe equality holds for a1 = a2 = · · · =
n
an = 1, and also for a1 = 0 and a2 = · · · = an =
(or any cyclic permutation).
n−1
P 1.36. If a, b, c are nonnegative real numbers, no two of which are zero, then
v
v
v
t
48a t
48b t
48c
1+
+ 1+
+ 1+
≥ 15.
b+c
c+a
a+b
(Vasile Cirtoaje, 2005)
Solution. Due to homogeneity, we may assume that a + b + c = 1. Thus, we need to
show that
a+b+c
1
f (a) + f (b) + f (c) ≥ 3 f (s), s =
= ,
3
3
where
v
t 1 + 47u
f (u) =
, u ∈ [0, 1).
1−u
Half Convex Function Method
55
From
f 00 (u) = p
48(47u − 11)
(1 − u)5 (1 + 47u)3
,
it follows that f is convex on [1/3, 1). By HCF Theorem, it suffices to show that f (x) +
2 f ( y) ≥ 3 f (1/3) for x, y ≥ 0 such that x + 2 y = 1; that is,
v
v
t 1 + 47x
t 49 − 47x
+2
≥ 15.
1− x
1+ x
Setting
t=
v
t 49 − 47x
1+ x
, 1 < t ≤ 7,
the inequality turns into
v
t 1175 − 23t 2
t2 − 1
≥ 15 − 2t.
By squaring, this inequality becomes
350 − 15t − 61t 2 + 15t 3 − t 4 ≥ 0,
(5 − t)2 (2 + t)(7 − t) ≥ 0.
The last inequality is clearly true. From x + 2 y = 1 and f (x) + 2 f ( y) = 3 f (1/3), we
get x = y = 1/3 and x = 0, y = 1/2. Therefore, in accordance with Remark 3, the
equality in the case a + b + c = 1 holds for a = b = c = 1/3, and also for a = 0 and
b = c = 1/2 (or any cyclic permutation). For the original inequality, the equality holds
for a = b = c, and also for a = 0 and b = c (or any cyclic permutation).
P 1.37. If a, b, c are nonnegative real numbers, then
v
v
v
t
t
t
3a2
3b2
3c 2
+
+
≤ 1.
7a2 + 5(b + c)2
7b2 + 5(c + a)2
7c 2 + 5(a + b)2
(Vasile Cirtoaje, 2008)
Solution. Due to homogeneity, we may assume that a + b + c = 3. Thus, we need to
show that
a+b+c
f (a) + f (b) + f (c) ≥ 3 f (s), s =
= 1,
3
where
v
t
3u2
−u
=p
, u ∈ [0, 3].
f (u) = −
2
2
7u + 5(3 − u)
4u2 − 10u + 15
56
Vasile Cîrtoaje
From
f 00 (u) =
10(u − 1)(15 − 4u)
5(−8u2 + 41u − 30) 5(−8u2 + 38u − 30)
≥
=
,
2
5/2
2
5/2
(4u − 10u + 15)
(4u − 10u + 15)
(4u2 − 10u + 15)5/2
it follows that f is convex on [1, 3]. By HCF Theorem, it suffices to prove the original
homogeneous inequality for b = c = 1; that is
v
v
t 3a2
t
3
+
2
≤ 1.
7a2 + 20
5a2 + 10a + 12
By squaring two times, the inequality becomes
Æ
a(5a3 + 10a2 + 16a + 50) ≥ 3a (7a2 + 20)(5a2 + 10a + 12),
a2 (5a6 + 20a5 − 11a4 + 38a3 − 80a2 − 40a + 68) ≥ 0,
a2 (a − 1)2 (5a4 + 30a3 + 44a2 + 96a + 68) ≥ 0.
The last inequality is clearly true. The equality holds for a = b = c, and also for a = 0
and b = c (or any cyclic permutation).
P 1.38. If a, b, c are nonnegative real numbers, then
v
v
v
t
t
t
a2
b2
c2
+
+
≥ 1.
a2 + 2(b + c)2
b2 + 2(c + a)2
c 2 + 2(a + b)2
(Vasile Cirtoaje, 2008)
Solution. Due to homogeneity, we may assume that a + b + c = 3. Thus, we need to
show that
a+b+c
f (a) + f (b) + f (c) ≥ 3 f (s), s =
= 1,
3
where
v
t
3u2
u
=p
, u ∈ [0, 3].
f (u) =
2
2
u + 2(3 − u)
u2 − 4u + 6
From
f 00 (u) =
2(2u2 − 11u + 12)
2(−11u + 12)
≥ 2
,
2
5/2
(u − 4u + 6)
(u − 4u + 6)5/2
it follows that f is convex on [0, 1]. By HCF Theorem, it suffices to prove the original
homogeneous inequality for b = c = 1; that is
p
2
+p
≥ 1.
2a2 + 4a + 3
a2 + 8
a
Half Convex Function Method
57
By squaring, the inequality becomes
Æ
a (a2 + 8)(2a2 + 4a + 3) ≥ 3a2 + 8a − 2.
For the nontrivial case 3a2 + 8a − 2 > 0, squaring both sides, we get
a6 + 2a5 + 5a4 − 8a3 − 14a2 + 16a − 2 ≥ 0,
(a − 1)2 [a4 + 4a3 + 9a2 + 4a + (3a2 + 8a − 2)] ≥ 0.
The last inequality is clearly true. The equality holds for a = b = c.
P 1.39. Let a, b, c be nonnegative real numbers, no two of which are zero. If
k0 =
k ≥ k0 ,
then
2a
b+c
k
+
ln 3
− 1 ≈ 0.585,
ln 2
2b
c+a
k
+
2c
a+b
k
≥ 3.
(Vasile Cirtoaje, 2005)
Solution. For k = 1, the inequality is just the well known Nesbitt’s inequality
2b
2c
2a
+
+
≥ 3,
b+c c+a a+b
while for k ≥ 1, the inequality follows from Jensens’s inequality applied to the convex
function f (u) = uk :
2a
b+c
k
2b
+
c+a
k
2c
+
a+b
k
≥3
2a
b+c
+
2b
c+a
+
2c
a+b
k
3
≥ 3.
Consider now that k0 ≤ k < 1. Due to homogeneity, we may assume that a + b + c = 1.
Thus, we need to show that
f (a) + f (b) + f (c) ≥ 3 f (s),
where
2u
f (u) =
1−u
k
,
From
f 00 (u) =
4k
(1 − u)4
2u
1−u
s=
a+b+c
1
= ,
3
3
u ∈ [0, 1).
k−2
(2u + k − 1),
58
Vasile Cîrtoaje
it follows that f is convex on [1/3, 1) (since u ≥ 1/3 involves 2u + k − 1 ≥ 2/3 + k − 1 =
k −1/3 > 0). By HCF Theorem, it suffices to prove the original homogeneous inequality
for b = c = 1; that is, to show that h(a) ≥ 3 for all a ≥ 0, where
k
2
k
h(a) = a + 2
.
a+1
For a > 0, the derivative
0
h (a) = ka
k−1
2
−k
a+1
k+1
has the same sign as
g(a) = (k − 1) ln a − (k + 1) ln
From
g 0 (a) =
2
.
a+1
2ka + k − 1
,
a(a + 1)
it follows that g 0 (a) = 0 for a0 = (1 − k)/(2k) < 1, g 0 (a) < 0 for a ∈ (0, a0 ) and
g 0 (a) > 0 for a ∈ (a0 , ∞). Then, g is strictly decreasing on (0, a0 ] and strictly increasing
on (a0 , ∞). Since lima→0 g(a) = ∞ and g(1) = 0, there exists a1 ∈ (0, a0 ) such that
g(a1 ) = 0, g(a) > 0 for a ∈ (0, a1 ) ∪ (1, ∞) and g(a) < 0 for a ∈ (a1 , 1); hence, h(a) is
strictly increasing on [0, a1 ] ∪ [1, ∞) and strictly decreasing on [a1 , 1]. Consequently,
h(a) ≥ min{h(0), h(1)}.
Since h(0) = 2k+1 ≥ 3 and h(1) = 3, we get h(a) ≥ 3. The proof is completed. The
equality holds for a = b = c. If k = k0 , then the equality holds also for a = 0 and b = c
(or any cyclic permutation).
Remark. For k = 2/3, we can give the following solution (based on the AM-GM inequality):
X 2a 2/3 X
2a
=
p
3
b+c
2a · (b + c) · (b + c)
X
6a
≥
= 3.
2a + (b + c) + (b + c)
p
P 1.40. If a, b, c ∈ [1, 7 + 4 3], then
v
v
v
t 2b
t 2a
t 2c
+
+
≥ 3.
b+c
c+a
a+b
(Vasile Cirtoaje, 2007)
Half Convex Function Method
59
Solution. Denoting
s=
a+b+c
,
3
we need to show that
f (a) + f (b) + f (c) ≥ 3 f (s),
where
v
t 2u
f (u) =
, 1 ≤ u < 3s.
3s − u
For u ≥ s, we have
3s − u
f (u) = 3s
2u
00
3/2
4u − 3s
> 0.
(3s − u)4
Therefore, f is convex for u ≥ s. By HCF Theorem, it suffices to prove the original
inequality for b = c; that is,
v
s
t 2b
a
+2
≥ 3.
b
a+b
v
tb
p
Putting t =
, the condition a, b ∈ [1, 7 + 4 3] involves
a
p
p
2 − 3 ≤ t ≤ 2 + 3.
We need to show that
v
t 2t 2
1
2
≥3− .
t2 + 1
t
This is true if
8t 2
1 2
≥ 3−
.
t2 + 1
t
This is equivalent to the obvious inequality
(t − 1)2 (t 2 − 4t + 1) ≥ 0.
The proof is completed. In accord
p with Remark 3, the equality holds for a = b = c, and
also for a = 1 and b = c = 7 + 4 3 (or any cyclic permutation).
P 1.41. Let a, b, c be nonnegative real numbers such that a + b + c = 3. If
0 < k ≤ k0 ,
k0 =
ln 2
≈ 1.71,
ln 3 − ln 2
then
a k (b + c) + b k (c + a) + c k (a + b) ≤ 6.
60
Vasile Cîrtoaje
Solution. For 0 < k ≤ 1, the inequality follows from Jensens’s inequality applied to the
convex function f (u) = uk :
(b + c)a + (c + a)b + (a + b)c k
(b + c)a k + (c + a)b k + (a + b)c k ≤ 2(a + b + c)
2(a + b + c)
a b + bc + ca k
a + b + c 2k
=6
≤6
= 6.
3
3
Consider now that 1 < k ≤ k0 , and write the inequality as
f (a) + f (b) + f (c) ≥ 3 f (s), s =
a+b+c
= 1,
3
where
f (u) = uk (u − 3), u ∈ [0, 3].
For u ≥ 1, we have
f 00 (u) = kuk−2 [(k + 1)u − 3k + 3] ≥ kuk−2 [(k + 1) − 3k + 3] = 2k(2 − k)uk−2 > 0;
therefore, f is convex for u ≥ 1. By HCF Theorem, it suffices to consider the case where
two of a, b, c are equal. Write the desired inequality in the homogeneous form
a + b + c k+1
k
k
k
a (b + c) + b (c + a) + c (a + b) ≤ 6
.
3
Since this inequality is trivial for b = c = 0, we may consider that b = c = 1. So, we
need to show that g(a) ≥ 0 for a ≥ 0, where
a + 2 k+1
g(a) = 3
− a k − a − 1.
3
We have
a+2
g (a) = (k + 1)
3
0
k
− ka
k−1
− 1,
1 00
k+1
g (a) =
k
3
a+2
3
k−1
−
k−1
.
a2−k
Since g 00 is strictly increasing, lima→0 g(a) = −∞ and g 00 (1) = 2k(2 − k)/3 > 0, there
exists a1 ∈ (0, 1) such that g 00 (a1 ) = 0, g 00 (a) < 0 for a ∈ (0, a1 ), and g 00 (a) > 0 for
a > 1. Therefore, g 0 is strictly decreasing on [0, a1 ] and strictly increasing on [a1 , ∞).
Since
g 0 (0) = (k + 1)(2/3)k − 1 ≥ (k + 1)(2/3)k0 − 1 =
k+1
k−1
−1=
> 0,
2
2
g 0 (1) = 0,
there exists a2 ∈ (0, a1 ) such that g 0 (a2 ) = 0, g 0 (a) > 0 for a ∈ [0, a2 ) ∪ (1, ∞), and
g 0 (a) < 0 for a ∈ (a2 , 1). Thus, g is strictly increasing on [0, a2 ] ∪ [1, ∞) and strictly
decreasing on [a2 , 1]. Consequently,
g(a) ≥ min{g(0), g(1)},
Half Convex Function Method
61
and from
g(0) = 3(2/3)k+1 − 1 ≥ 3(2/3)k0 +1 − 1 = 1 − 1 = 0,
g(1) = 0,
we get g(a) ≥ 0. This completes the proof. The equality holds for a = b = c = 1. If k =
k0 , then the equality holds also for a = 0 and b = c = 3/2 (or any cyclic permutation).
Remark 1. Using the Cauchy-Schwarz inequality, we get
X
(a + b + c)2
9
3
a
P
≥
=P
≥ .
k
k
k
k
k
b +c
a(b + c )
a (b + c) 2
Thus, the following statement holds.
• Let a, b, c be nonnegative real numbers such that a + b + c = 3. If
0 < k ≤ k0 ,
then
k0 =
ln 2
≈ 1.71,
ln 3 − ln 2
b
c
3
a
+
+
≥ ,
bk + c k c k + ak ak + bk
2
with equality for a = b = c = 1. If k = k0 , then the equality holds also for a = 0 and
b = c = 3/2 (or any cyclic permutation).
Remark 2. Also, the following statement holds.
• Let a, b, c be nonnegative real numbers such that a + b + c = 3. If
k ≥ k1 ,
k1 =
ln 9 − ln 8
≈ 0.2905,
ln 3 − ln 2
then
ak
bk
ck
3
+
+
≥ ,
b+c c+a a+b
2
with equality for a = b = c = 1. If k = k1 , then the equality holds also for a = 0 and
b = c = 3/2 (or any cyclic permutation).
For k1 ≤ k ≤ 2, the inequality can be proved using the Cauchy-Schwarz inequality, as
follows:
X ak
(a + b + c)2
9
3
≥P
=P
= .
2−k
2−k
b+c
a (b + c)
a (b + c) 2
62
Vasile Cîrtoaje
For k > 2, the inequality can be deduced from the Cauchy-Schwarz inequality and
Bernoulli’s inequality, as follows:
P k/2 2
P k/2 2
X ak
a
a
≥ P
=
,
b+c
6
(b + c)
X
X
k
k/2
a ≥
1 + (a − 1) = 3.
2
P 1.42. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
n2
, then
k≥
4(n − 1)
(a12 + k)(a22 + k) · · · (an2 + k) ≥ (1 + k)n .
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = ln(u2 + k),
u ∈ [0, n].
We have
f 0 (u) =
2u
,
u2 + k
f 00 (u) =
2(k − u2 )
.
(u2 + k)2
For u ∈ [0, 1], we have f 00 (u) > 0, since
k − u2 ≥ k − 1 ≥
n2
− 1 ≥ 0.
4(n − 1)
Therefore, f is convex on [0, 1]. By HCF Theorem and Remark 2, it suffices to prove
that H(x, y) ≥ 0 for x, y ≥ 0 such that x + (n − 1) y = n, where
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
Since
2(k − x y)
2[n2 − 4(n − 1)x y]
≥
,
(x 2 + k)( y 2 + k) 4(n − 1)(x 2 + k)( y 2 + k)
we only need to show that
n2 ≥ 4(n − 1)x y.
H(x, y) =
Indeed, we have
n2 = [x + (n − 1) y]2 ≥ 4(n − 1)x y.
The equality holds for a1 = a2 = · · · = an = 1.
Half Convex Function Method
63
P 1.43. Let a, b, c be nonnegative real numbers such a + b + c = 3. If k ≥ k0 , where
p
p
p
6−2
= (2 + 2)(2 + 3) ≈ 12.74 ,
k0 = p
p
6− 2−1
then
p
a+
p
b+
p
c−3≥ k
v
ta + b
2
+
v
tb+c
2
+
s
c+a
−3 .
2
(Vasile Cirtoaje, 2008)
Solution. By the Cauchy-Schwarz inequality
(1 + 1 + 1)[(a + b) + (b + c) + (c + a)] ≥
p
a+b+
p
b+c+
p
c+a
2
,
we get
v
ta + b
2
+
v
tb+c
2
+
s
c+a
≤ 3.
2
Therefore, it suffices to consider the case k = k0 . Write the inequality as
f (a) + f (b) + f (c) ≥ 3 f (s),
where
s=
v
t3 − u
p
f (u) = u − k0
,
2
a+b+c
= 1,
3
u ∈ [0, 3).
For u ≥ 1, we have
00
4 f (u) = −u
−3/2
k0
+
4
3−u
2
−3/2
≥ −1 +
k0
> 0.
4
Therefore, f is convex on [1, 3]. By HCF Theorem and Remark 5, it suffices to consider
only the case a ≤ 1 ≤ b = c. Write the original inequality in the homogeneous form
v
v
v
v
ta + b t b + c sc + a
ta + b + c
ta + b + c
p
p
p
a+ b+ c−3
≥ k0
+
+
−3
.
3
2
2
2
3
Due to homogeneity, we may assume that b = c = 1. Moreover, it is convenient to use
p
the substitution x = a. Thus, we need to show that g(x) ≥ 0 for x ∈ [0, 1], where
v
v
t x2 + 1
t x2 + 2
g(x) = x + 2 − k0 + 3(k0 − 1)
− 2k0
.
3
2
We have
0
g (x) = 1 + (k0 − 1)x
v
t
v
t 2
3
−
k
x
,
0
x2 + 2
x2 + 1
64
Vasile Cîrtoaje
k0
g 00 (x) =
2
2
2
x +1
3/2
x2 + 1
m· 2
x +2
3/2
−1 ,
where
v
u
t
1 2
3
m= 6 1−
≈ 1.72.
k0
Clearly, g 00 (x) has the same sign as h(x), where
x2 + 1
− 1.
x2 + 2
h(x) = m ·
Since
h(0) =
m
− 1 < 0,
2
h(1) =
2m
− 1 > 0,
3
there is x 1 ∈ (0, 1) such that h(x) < 0 for x ∈ [0, x 1 ), h(x 1 ) = 0 and h(x) > 0 for
x ∈ (x 1 , 1]. Therefore, g 0 is strictly decreasing on [0, x 1 ] and strictly increasing on
[x 1 , 1]. Since g 0 (0) = 1 and g 0 (1) = 0, there is x 2 ∈ (0, x 1 ) such that g 0 (x) > 0 for
x ∈ [0, x 2 ), g 0 (x 2 ) = 0 and g 0 (x) < 0 for x ∈ (x 2 , 1). Thus, g(x) is strictly
p increasing
p on
[0, x 2 ] and strictly decreasing on [x 2 , 1]. From g(0) = 2 − k0 + (k0 − 1) 6 − k0 2 = 0
and g(1) = 0, it follows that g(x) ≥ 0 for x ∈ [0, 1]. This completes the proof. The
equality holds for a = b = c. If k = k0 , then the equality holds also for a = 0 and
b = c = 3/2 (or any cyclic permutation).
P 1.44. Let a, b, c be nonnegative real numbers such a + b + c = 3. If k ≤ k1 , where
p
p
p
k1 = ( 3 − 1)( 3 + 2) ≈ 2.303 ,
then
p
a+
p
b+
p
c−3≤ k
v
ta + b
2
+
v
tb+c
2
+
s
c+a
−3 .
2
(Vasile Cirtoaje, 2008)
Solution. By the Cauchy-Schwarz inequality
(1 + 1 + 1)[(a + b) + (b + c) + (c + a)] ≥
p
a+b+
p
we get
v
ta + b
2
+
v
tb+c
2
+
s
c+a
≤ 3.
2
b+c+
p
c+a
2
,
Half Convex Function Method
65
Therefore, it suffices to consider the case k = k1 . Write the inequality as
f (a) + f (b) + f (c) ≥ 3 f (s),
where
s=
v
t3 − u
p
f (u) = − u + k1
,
2
a+b+c
= 1,
3
u ∈ [0, 3).
For 0 ≤ u ≤ 1, we have
4 f 00 (u) = u−3/2 −
k1
4
−3/2
3−u
2
≥1−
k1
> 0.
4
Therefore, f is convex on [0, 1]. By HCF Theorem and Remark 5, it suffices to consider
only the case a ≥ 1 ≥ b = c. Write the original inequality in the homogeneous form
p
a+
p
b+
p
c−3
v
ta + b + c
3
≤ k1
v
ta + b
2
+
v
tb+c
2
+
s
v
ta + b + c
c+a
−3
.
2
3
Due to homogeneity, we may assume that b = c = 1. Moreover, it is convenient to use
p
the substitution x = a. Thus, we need to show that g(x) ≤ 0 for x ≥ 1, where
v
v
t x2 + 1
t x2 + 2
− 2k1
.
g(x) = x + 2 − k1 + 3(k1 − 1)
3
2
We have
v
t
v
t 2
3
g (x) = 1 + (k1 − 1)x
−
k
x
,
1
x2 + 2
x2 + 1
3/2
3/2
2
k
2
x
+
1
1
g 00 (x) =
m· 2
−1 ,
2 x2 + 1
x +2
0
where
v
u
t
1 2
3
m= 6 1−
≈ 1.2431 .
k1
Clearly, g 00 (x) has the same sign as h(x), where
h(x) = m ·
x2 + 1
− 1.
x2 + 2
Since h is strictly increasing on [1, ∞) and
h(1) =
2m
− 1 < 0,
3
lim h1 (x) = m − 1 > 0,
x→∞
66
Vasile Cîrtoaje
there is x 1 ∈ (1, ∞) such that h(x) < 0 for x ∈ [1, x 1 ), h(x 1 ) = 0 and h(x) > 0
for x ∈ (x 1 , ∞). Therefore, g 0 is strictly decreasing on [1, x 1 ] and strictly increasing
on [x 1 , ∞). Since g 0 (1) = 0 and lim x→∞ g 0 (x) = 0, it follows that g 0 (x) < 0 for
x ∈ (1, ∞). Thus, g(x) is strictly decreasing on [1, ∞), hence g(x) ≤ g(1) = 0. This
completes the proof. The equality holds for a = b = c = 1. If k = k0 , then the equality
holds also for a = 0 and b = c = 3/2 (or any cyclic permutation).
P 1.45. If a, b, c are positive real numbers such that a bc = 1, then
a2 + b2 + c 2 − 3 ≥ 18(a + b + c − a b − bc − ca).
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
we need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
s=
x + y +z
= 0,
3
where
f (u) = e2u − 1 − 18(eu − e−u ), u ∈ R.
For u ≤ 0, we have
f 00 (u) = 4e2u + 18(e−u − eu ) > 0,
hence f is convex on (−∞, 0]. By HCF Theorem, it suffices to prove the original inequality for b = c := t and a = 1/t 2 , where t > 0. Since
a2 + b2 + c 2 − 3 =
and
a + b + c − a b − bc − ca =
1
(t 2 − 1)2 (2t 2 + 1)
2
+
2t
−
3
=
t4
t4
−(t 4 − 2t 3 + 2t − 1) −(t − 1)3 (t + 1)
=
,
t2
t2
we get
a2 + b2 + c 2 − 3 − 18(a + b + c − a b − bc − ca) =
(t − 1)2 (2t − 1)2 (t + 1)(5t + 1)
≥ 0.
t4
The equality holds for a = b = c = 1, and also for a = 4 and b = c = 1/2 (or any cyclic
permutation).
Half Convex Function Method
67
P 1.46. If a, b, c are positive real numbers such that a bc = 1, then
p
p
p
a2 − a + 1 + b2 − b + 1 + c 2 − c + 1 ≥ a + b + c.
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
we need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
where
f (u) =
p
s=
x + y +z
= 0,
3
e2u − eu + 1 − eu , u ∈ R.
We claim that f is convex for u ≥ 0. Since
e−u f 00 (u) =
4e3u − 6e2u + 9eu − 2
− 1,
4(e2u − eu + 1)3/2
we need to show that
(4x 3 − 6x 2 + 9x − 2)2 ≥ 16(x 2 − x + 1)3 ,
where x = eu ≥ 1. Indeed,
(4x 3 − 6x 2 + 9x − 2)2 − 16(x 2 − x + 1)3 = 12x 3 (x − 1) + 9x 2 + 12(x − 1) > 0.
By HCF Theorem, it suffices to prove the original inequality for b = c := t and a = 1/t 2 ,
where t > 0; that is,
p
p
t4 − t2 + 1
1
+
2
t 2 − t + 1 ≥ 2 + 2t,
t2
t
p
t2 − 1
t4 − t2 + 1 + 1
Since
p
it suffices to show that
+p
t2 − 1
t4 − t2 + 1
2(1 − t)
t2 − t + 1 + t
≥
≥ 0.
t2 − 1
,
t2 + 1
t2 − 1
2(1 − t)
+p
≥ 0,
2
t +1
t2 − t + 1 + t
which is equivalent to
t +1
2
(t − 1) 2
−p
≥ 0,
t +1
t2 − t + 1 + t
68
Vasile Cîrtoaje
p
(t − 1) (t + 1) t 2 − t + 1 − t 2 + t − 2 ≥ 0,
(t − 1)2 (3t 2 − 2t + 3)
≥ 0.
p
(t + 1) t 2 − t + 1 + t 2 − t + 2
Clearly, the last inequality is true. The equality holds for a = b = c = 1.
P 1.47. If a, b, c, d ≥
1
p such that a bcd = 1, then
1+ 6
1
1
1
1
4
+
+
+
≤ .
a+2 b+2 c+2 d +2 3
(Vasile Cirtoaje, 2005)
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
d = ew,
we need to show that
f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s),
where
f (u) =
For u ≤ 0, we have
f 00 (u) =
s=
x + y +z+w
= 0,
4
−1
, u ∈ R.
+2
eu
eu (2 − eu )
> 0,
(eu + 2)3
hence f is convex on (−∞, 0]. By HCF Theorem, it suffices to prove the original in1
equality for b = c = d := t and a = 1/t 3 , where t ≥
p ; that is,
1+ 6
t3
3
4
+
≤ ,
3
2t + 1 t + 2 3
which is equivalent to the obvious inequality
(t − 1)2 (5t 2 + 2t − 1) ≥ 0.
In accord with Remark 3, the equality holds for a = b = c = d = 1, and also for
p
1
a = 19 + 9 6 and b = c = d =
p (or any cyclic permutation).
1+ 6
Half Convex Function Method
69
P 1.48. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
p
1
1
1
2
2
2
.
−
− ··· −
a1 + a2 + · · · + an − n ≥ 6 3 a1 + a2 + · · · + an −
a1 a2
an
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
s=
x1 + x2 + · · · + x n
= 0,
n
p
f (u) = e2u − 1 − 6 3 (eu − e−u ), u ∈ R.
For u ≤ 0, we have
p
f 00 (u) = 4e2u + 6 3(e−u − eu ) > 0,
hence f is convex on (−∞, 0]. By HCF Theorem and Remark 2, it suffices to show that
H(x, y) ≥ 0 for x, y ∈ R such that x + (n − 1) y = 0, where
H(x, y) =
From
f 0 (x) − f 0 ( y)
.
x−y
p
f 0 (u) = 2e2u − 6 3 (eu + e−u ),
we get
H(x, y) =
p
p
2(e x − e y ) x
e + e y − 3 3 + 3 3 e−x e− y .
x−y
Since (e x − e y )/(x − y) > 0, we need to prove that
p
p
e x + e y + 3 3 e−x e− y ≥ 3 3.
Indeed, by the AM-GM inequality, we have
Æ
p
p
p
3
e x + e y + 3 3 e−x e− y ≥ 3 e x · e y · 3 3 e−x e− y = 3 3.
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
P 1.49. If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that a1 a2 · · · an = 1, then
(n − 1)(a12 + a22 + · · · + an2 ) + n(n + 3) ≥ (2n + 2)(a1 + a2 + · · · + an ).
70
Vasile Cîrtoaje
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
s=
x1 + x2 + · · · + x n
= 0,
n
where
f (u) = (n − 1)e2u − (2n + 2)eu , u ∈ R.
For u ≥ 0, we have
f 00 (u) = 4(n − 1)e2u − (2n + 2)eu = 2eu [2(n − 1)eu − n − 1]
≥ 2eu [2(n − 1) − n − 1] = 2(n − 3)eu > 0.
Therefore, f is convex on [0, ∞). By HCF Theorem and Remark 2, it suffices to show
that H(x, y) ≥ 0 for x, y ∈ R such that x + (n − 1) y = 0, where
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
From
f 0 (u) = 2(n − 1)e2u − (2n + 2)eu ,
we get
H(x, y) =
2(e x − e y )
[(n − 1)(e x + e y ) − (n + 1)] .
x−y
Since (e x − e y )/(x − y) > 0, we need to prove that (n − 1)(e x + e y ) ≥ n + 1. Using the
AM-GM inequality, we have
Æ
n
(n − 1)(e x + e y ) = (n − 1)e x + e y + e y + · · · + e y ≥ n (n − 1)e x · e y · e y · · · e y
Æ
p
n
n
= n (n − 1)e x+(n−1) y = n n − 1.
Thus, it suffices to show that
which is equivalent to
p
n
n n − 1 ≥ n + 1,
1 n
n−1≥ 1+
.
n
This is true for n ≥ 4, since
1 n
n−1≥3> 1+
.
n
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
Remark. From the proof above, it follows that the following sharper inequality holds
in the same conditions (Gabriel Dospinescu and Calin Popa):
p
n
2n n − 1
2
2
2
a1 + a2 + · · · + an − n ≥
(a1 + a2 + · · · + an − n).
n−1
Half Convex Function Method
71
P 1.50. Let a, b, c, d be positive real numbers such that a bcd = 1. If p and q are nonnegative real numbers such that p + q = 3, then then
1
1
1
1
+
+
+
≥ 1.
1 + pa + qa3 1 + p b + q b3 1 + pc + qc 3 1 + pd + qd 3
(Vasile Cirtoaje, 2007)
Solution. Using the substitutions
a = ex ,
b = ey,
c = ez ,
d = ew,
we need to show that
f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s),
where
f (u) =
1
1+
peu
+ qe3u
s=
x + y +z+w
= 0,
4
, u ∈ R.
We will show that f 00 (u) > 0 for u ≥ 0, hence f is convex on [0, ∞). Since
f 00 (u) =
th(t)
,
(1 + pt + qt 3 )3
where
h(t) = 9q2 t 5 + 2pqt 3 − 9qt 2 + p2 t − p,
t = eu , t ≥ 1,
we need to show that h(t) > 0 for t ≥ 1. Indeed, we have
h(t) ≥ 9q2 t 3 + 2pqt 3 − 9qt 2 + p2 t − pt = t[(9q2 + 2pq)t 2 − 9qt + p2 − p]
and
(9q2 + 2pq)t 2 − 9qt + p2 − p ≥ (9q2 + 2pq)(2t − 1) − 9qt + p2 − p
= q(18q + 4p − 9)t − 9q2 − 2pq + p2 − p ≥ q(18q + 4p − 9) − 9q2 − 2pq + p2 − p
= p2 + 2pq + 9q2 − p − 9q = p2 + 2pq + 9q2 −
=
(p + 9q)(p + q)
3
2(p − q)2 + 16q2
> 0.
3
By HCF Theorem, it suffices to prove the original inequality for b = c = d = t and
a = 1/t 3 , where t > 0; that is,
3
t9
+
≥ 1,
9
6
t + pt + q 1 + pt + qt 3
72
Vasile Cîrtoaje
pt 6 + q
3
≥
,
1 + pt + qt 3
t 9 + pt 6 + q
(3 − pq)t 9 − p2 t 7 + 2pt 6 − q2 t 3 − pqt + 2q ≥ 0,
[(p + q)2 − 3pq]t 9 − 3p2 t 7 + 2p(p + q)t 6 − 3q2 t 3 − 3pqt + 2q(p + q) ≥ 0,
Ap2 + Bq2 ≥ C pq,
where
A = t 9 − 3t 7 + 2t 6 = t 6 (t − 1)2 (t + 2) ≥ 0,
B = t 9 − 3t 3 + 2 = (t 3 − 1)2 (t 3 + 2) ≥ 0,
C = t 9 − 2t 6 + 3t − 2.
Since A ≥ 0 and B ≥ 0, it suffices to consider that C ≥ 0 and to show that 4AB ≥ C 2 .
From
t 3 − 3t + 2 = (t − 1)2 (t + 2) ≥ 0,
we get 3t − 2 ≤ t 3 . Therefore
C ≤ t 9 − 2t 6 + t 3 = t 3 (t 3 − 1)2 ,
hence
4AB − C 2 ≥ 4AB − t 6 (t 3 − 1)4 = t 6 (t − 1)2 (t 3 − 1)2 [4(t + 2)(t 3 + 2) − (t 2 + t + 1)2 ]
= t 6 (t − 1)2 (t 3 − 1)2 (3t 4 + 6t 3 − 3t 2 + 6t + 15) ≥ 0.
The proof is completed. The inequality holds for a = b = c = d = 1.
Remark. Similarly, we can prove the following generalization.
• Let a, b, c, d be positive real numbers such that a bcd = 1. If p and q are nonnegative
real numbers such that p + q ≥ 3, then then
1
1
1
1
4
+
+
+
≥
.
3
3
3
3
1 + pa + qa
1 + pb + qb
1 + pc + qc
1 + pd + qd
1+p+q
P 1.51. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 . . . an = 1. If
p, q ≥ 0 such that p + q ≥ n − 1, then
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
1
n
≥
.
2
1 + pan + qan
1+p+q
(Vasile Cirtoaje, 2007)
Half Convex Function Method
73
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
f (u) =
1
1+
peu
+ qe2u
s=
x1 + x2 + · · · + x n
= 0,
n
, u ∈ R.
For u ≥ 0, we have
eu [4q2 e3u + 3pqe2u + (p2 − 4q)eu − p]
(1 + peu + qe2u )3
2u
2
e [4q + 3pq + (p2 − 4q) − p]
≥
(1 + peu + qe2u )3
e2u [(p + 2q)(p + q − 2) + 2q2 + p]
=
> 0,
(1 + peu + qe2u )3
f 00 (u) =
therefore f (u) is convex for u ≥ 0. By HCF Theorem, it suffices to prove the original
inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where t > 0. Write this inequality as
t 2n−2
n−1
n
+
≥
.
2n−2
n−1
2
t
+ pt
+ q 1 + pt + qt
1+p+q
Applying the Cauchy-Schwarz inequality, it suffices to prove that
n
(t n−1 + n − 1)2
≥
,
(t 2n−2 + pt n−1 + q) + (n − 1)(1 + pt + qt 2 ) 1 + p + q
which is equivalent to
pB + qC ≥ A,
where
A = (n − 1)(t n−1 − 1)2 ≥ 0,
A
+ nE, E = t n−1 + n − 2 − (n − 1)t,
n−1
A
C = (t n−1 − 1)2 + nF =
+ nF, F = 2t n−1 + n − 3 − (n − 1)t 2 .
n−1
By the AM-GM inequality applied to n − 1 positive numbers, we have E ≥ 0 and F ≥ 0
for n ≥ 3. Since A ≥ 0 and p + q ≥ n − 1, we have
B = (t n−1 − 1)2 + nE =
pB + qC − A ≥ pB + qC −
(p + q)A
= n(pE + qF ) ≥ 0.
n−1
The equality holds for a1 = a2 = · · · = an = 1.
74
Vasile Cîrtoaje
Remark 1. For p = 2k and q = k2 , we get the following result.
• Let a1 , a2 , · · · , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If
p
k ≥ n − 1, then
1
1
n
1
+
+ ··· +
≥
,
(1 + ka1 )2 (1 + ka2 )2
(1 + kan )2
(1 + k)2
with equality for a1 = a2 = · · · = an = 1.
In addition, for n = 4 and k = 1, we get the known inequality (Vasile Cirtoaje, 1999):
1
1
1
1
+
+
+
≥ 1,
2
2
2
(1 + a)
(1 + b)
(1 + c)
(1 + d)2
where a, b, c, d > 0 such that a bcd = 1.
Remark 2. For p + q = n − 1, we get the beautiful inequality
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
1
≥ 1,
1 + pan + qan2
n ≥ 3,
which is a generalization of the following inequalities:
1
1
1
+
+ ··· +
≥ 1,
1 + (n − 1)a1 1 + (n − 1)a2
1 + (n − 1)an
1
1
1
+
+
·
·
·
+
≥ 1,
p
p
p
[1 + ( n − 1)a1 ]2 [1 + ( n − 1)a1 ]2
[1 + ( n − 1)a1 ]2
1
2 + (n − 1)(a1 + a12 )
+
1
2 + (n − 1)(a2 + a22 )
+ ··· +
1
1
≥ .
2
2 + (n − 1)(an + an ) 2
Remark 3. Similarly, we can prove the following statement:
• Let a1 , a2 , . . . , an (n ≥ 4) be positive real numbers such that a1 a2 . . . an = 1. If
p, q, r ≥ 0 such that p + q + r ≥ n − 1, then
n
X
i=1
1
1+
pai + qai2
+
r ai3
≥
n
.
1+p+q+r
For n = 4 and p + q + r = 3, we get the beautiful inequality
4
X
1
i=1
1 + pai + qai2 + r ai3
≥ 1.
Half Convex Function Method
75
If p = q = r = 1, then we get the known inequality (Vasile Cirtoaje, 1999):
4
X
1
i=1
1 + ai + ai2 + ai3
≥ 1.
Since 2ai2 ≤ ai + ai3 , the best inequality with respect to q if for q = 0; that is,
4
X
1
i=1
1 + pai + r ai3
p + r = 3.
≥ 1,
For p = 1 and p = 2, we get the following strong inequalities:
4
X
1
i=1
1 + ai + 2ai3
4
X
1
i=1
1 + 2ai + ai3
≥ 1,
≥ 1.
Actually, the following generalization holds.
• Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1, and let k1 , k2 , . . . , km ≥
0 such that k1 + k2 + · · · + km ≥ n − 1. If m ≤ n − 1, then
n
X
1
1 + k1 ai + k2 ai2
i=1
+ · · · + km aim
≥
n
1 + k1 + k2 + · · · + km
.
(*)
For m = n − 1 and k1 = k2 = · · · = km = 1, (*) turns into the known beautiful inequality
n
X
1
1 + ai + ai2
i=1
+ · · · + ain−1
≥ 1.
Since
(m − 1)aik ≤ (m − k)ai + (k − 1)aim ,
k = 2, 3, . . . , m − 1
(by the AM-GM inequality), the best inequality (*) with respect to k2 , · · · , km−1 is for
k2 , · · · , km−1 = 0; that is,
n
X
i=1
n
1
,
m ≥
1 + k1 ai + km ai
1 + k1 + km
k1 + km ≥ n − 1, 1 ≤ m ≤ n − 1.
If k1 + km = n − 1 and m = n − 1, then
n
X
1
i=1
1 + k1 ai + kn−1 ain−1
≥ 1.
76
Vasile Cîrtoaje
For k1 = 1 and k1 = n − 2, we get the following strong inequalities:
n
X
1
i=1
1 + ai + (n − 2)ain−1
n
X
1
i=1
1 + (n − 2)ai + ain−1
≥ 1,
≥ 1.
P 1.52. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If k ≥ n2 − 1,
then
1
1
1
n
.
+p
+ ··· + p
≥p
p
1+k
1 + kan
1 + ka1
1 + ka2
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
s=
x1 + x2 + · · · + x n
= 0,
n
1
, u ∈ R.
f (u) = p
1 + keu
For u ≥ 0, we have
f 00 (u) =
keu (keu − 2)
keu (k − 2)
≥
> 0.
4(1 + keu )5/2
4(1 + keu )5/2
Therefore, f is convex on [0, ∞). By HCF Theorem and Remark 5, it suffices to prove
the original inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where t ≥ 1. Write this
inequality as h(t) ≥ 0, where
v
t t n−1
n−1
n
h(t) =
+p
−p
.
n−1
t
+k
1 + kt
1+k
The derivative
h0 (t) =
(n − 1)kt (n−3)/2
(n − 1)k
−
n−1
3/2
2(t
+ k)
2(kt + 1)3/2
has the same sign as
h1 (t) = t n/3−1 (kt + 1) − t n−1 − k.
Denoting m = n/3, m ≥ 2/3, we see that
h1 (t) = kt m + t m−1 − t 3m−1 − k = k(t m − 1) − t m−1 (t 2m − 1) = (t m − 1)h2 (t),
Half Convex Function Method
77
where
h2 (t) = k − t m−1 − t 2m−1 .
For t > 1, we have
h02 (t) = t m−2 [−m + 1 − (2m − 1)t m ] < t m−2 [−m + 1 − (2m − 1)]
= −(3m − 2)t m−2 ≤ 0,
hence h2 (t) is strictly decreasing for t ≥ 1. Since h2 (1) = k − 2 > 0 and lim t→∞ h2 (t) =
−∞, there exists t 1 > 1 such that h2 (t 1 ) = 0, h2 (t) > 0 for t ∈ (1, t 1 ), and h2 (t) < 0 for
t ∈ (t 1 , ∞). Since h2 , h1 and h0 has the same sign for t > 1, h(t) is strictly increasing for
t ∈ [1, t 1 ] and strictly decreasing for t ∈ [t 1 , ∞); this yields h(t) ≥ min{h(1), h(∞)}.
n
≥ 0, it follows that h(t) ≥ 0 for all t ≥ 1. The
From h(1) = 0 and h(∞) = 1 − p
1+k
proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
Remark. The following generalization holds (Vasile Cirtoaje, 2005):
• Let a1 , a2 , · · · , an be positive real numbers such that a1 a2 · · · an = 1. If k and m are
positive numbers such that
m ≤ n − 1,
k ≥ n1/m − 1,
then
1
1
1
n
+
+ ··· +
≥
,
m
m
m
(1 + ka1 )
(1 + ka2 )
(1 + kan )
(1 + k)m
with equality for a1 = a2 = · · · = an = 1.
For 0 < m ≤ n − 1 and k = n1/m − 1, we get the beautiful inequality
1
1
1
+
+ ··· +
≥ 1.
m
m
(1 + ka1 )
(1 + ka2 )
(1 + kan )m
P 1.53. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then
1
1 + a1 + · · · + a1n−1
+
1
1 + a2 + · · · + a2n−1
+ ··· +
1
≥ 1.
1 + an + · · · + ann−1
(Vasile Cirtoaje, 2007)
78
Vasile Cîrtoaje
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
f (u) =
s=
x1 + x2 + · · · + x n
= 0,
n
1
, u ∈ R.
1 + eu + · · · + e(n−1)u
We will show that f is convex for u ≥ 0. Setting t = eu , t ≥ 1, the necessary and
sufficient condition f 00 (u) ≥ 0 for u ≥ 0 is equivalent to
2A2 ≥ B(1 + C),
where
A = t + 2t 2 + · · · + (n − 1)t n−1 ,
B = t + 4t 2 + · · · + (n − 1)2 t n−1 ,
C = t + t 2 + · · · + t n−1 .
We will prove this inequality by induction on n. For n = 2, the inequality becomes
t(t − 1) ≥ 0, which is clearly true for t ≥ 1. Assume now that the inequality is true for
n and prove it for n + 1, n ≥ 2. So, we need to show that 2A2 ≥ B(1 + C) involves
2(A + nt n )2 ≥ (B + n2 t n )(1 + C + t n ),
which is equivalent to
2A2 − B(1 + C) + t n [n2 (t n − 1) + D] ≥ 0,
where
2
D = 2nA − B − n C =
n−1
X
bi t i ,
bi = 3n2 − (2n − i)2 .
i=1
Since 2A − B(1 + C) ≥ 0, it suffices to show that D ≥ 0. Since
2
b1 < b2 < · · · < bn−1 ,
t ≤ t 2 ≤ · · · ≤ t n−1 ,
we may apply Chebyshev’s inequality to get
D≥
1
(b1 + b2 + · · · + bn−1 )(t + t 2 + · · · + t n−1 ).
n
Thus, it suffices to show that b1 + b2 + · · · + bn−1 ≥ 0. Indeed,
b1 + b2 + · · · + bn−1 =
n−1
X
n(n − 1)(4n + 1)
[3n2 − (2n − i)2 ] =
> 0.
6
i=1
Half Convex Function Method
79
By HCF Theorem and Remark 5, it suffices to prove the original inequality for a2 = · · · =
an := t and a1 = 1/t n−1 , where t ≥ 1. Setting k = n − 1, k ≥ 1, we need to show that
2
k
tk
+
≥ 1.
2
1 + t + · · · + tk
1 + tk + · · · + tk
For the nontrivial case t > 1, this inequality is equivalent to the following sequence of
inequalities:
k
1 + t k + · · · + t (k−1)k
,
≥
1 + t + · · · + tk
1 + t k + · · · + t k2
2
k(t − 1)
tk − 1
tk − 1
≥
·
,
t k+1 − 1
t k − 1 t (k+1)k − 1
2
k(t − 1)
tk − 1
≥
,
t k+1 − 1
t (k+1)k − 1
2
t k(k+1) − 1
tk − 1
≥
,
t k+1 − 1
t −1
k 1 + t k+1 + t 2(k+1) + · · · + t (k−1)(k+1) ≥ 1 + t + t 2 + · · · + t (k−1)(k+1) ,
k 1 · 1 + t · t k + · · · + t k−1 · t (k−1)k ≥ 1 + t + · · · + t k−1 1 + t k + · · · + t (k−1)k .
k
Since 1 < t < · · · < t k−1 and 1 < t k < · · · < t (k−1)k , the last inequality follows from
Chebyshev’s inequality. This completes the proof. The equality holds for a1 = a2 = · · · =
an = 1.
P 1.54. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If p, q ≥ 0
1
such that 0 < p + q ≤
, then
n−1
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
1
n
≤
.
1 + pan + qan2
1+p+q
(Vasile Cirtoaje, 2007)
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
f (u) =
s=
x1 + x2 + · · · + x n
= 0,
n
−1
, u ∈ R.
1 + peu + qe2u
80
Vasile Cîrtoaje
For u ≤ 0, we have
eu [−4q2 e3u − 3pqe2u + (4q − p2 )eu + p]
(1 + peu + qe2u )3
e2u [−4q2 − 3pq + (4q − p2 ) + p]
≥
(1 + peu + qe2u )3
e2u [(p + 4q)(1 − p − q) + 2pq]
=
≥ 0,
(1 + peu + qe2u )3
f 00 (u) =
therefore f (u) is convex for u ≤ 0. By HCF Theorem, it suffices to prove the original
inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where t > 0. Write this inequality as
n−1
n
t 2n−2
+
≤
,
t 2n−2 + pt n−1 + q 1 + pt + qt 2
1+p+q
p2 A + q2 B + pqC ≤ pD + qE,
where
A = t n−1 (t n − nt + n − 1),
B = t 2n − nt 2 + n − 1,
C = t 2n−1 + t 2n − nt n+1 + (n − 1)t n−1 − nt + n − 1,
D = t n−1 [(n − 1)t n − nt n−1 + 1],
E = (n − 1)t 2n − nt 2n−2 + 1.
Applying the AM-GM inequality to n positive numbers yields D ≥ 0 and E ≥ 0. Then,
since p + q ≤ 1/(n − 1) involves pD + qE ≥ (n − 1)(p + q)(pD + qE), it suffices to show
that
p2 A + q2 B + pqC ≤ (n − 1)(p + q)(pD + qE).
Write this inequality as
p2 A1 + q2 B1 + pqC1 ≥ 0,
where
A1 = (n − 1)D − A = nt n [(n − 2)t n−1 − (n − 1)t n−2 + 1],
B1 = (n − 1)E − B = nt 2 [(n − 2)t 2n−2 − (n − 1)t 2n−4 + 1],
C1 = (n − 1)(D + E) − C = n y[(n − 2)(t 2n−1 + t 2n−2 ) − 2(n − 1)t 2n−3 + t n + 1].
Applying the AM-GM inequality to n−1 nonnegative numbers yields A1 ≥ 0 and B1 ≥ 0.
So, it suffices to show that C1 ≥ 0. Indeed, we have
(n − 2)(t 2n−1 + t 2n−2 ) − 2(n − 1)t 2n−3 + t n + 1 = A2 + B2 + C2 ,
where
A2 = (n − 2)t 2n−1 − (n − 1)t 2n−3 + t ≥ 0,
B2 = (n − 2)t 2n−2 − (n − 1)t 2n−3 + t n−1 ≥ 0,
Half Convex Function Method
81
C2 = t n − t n−1 − t + 1 = (t − 1)(t n−1 − 1) ≥ 0.
The inequalities A2 ≥ 0 and B2 ≥ 0 follows by the AM-GM inequality applied to n − 1
nonnegative numbers. This completes the proof. The equality holds for a1 = a2 = · · · =
an = 1.
Remark 1. For p + q =
1
, we get the inequality
n−1
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
1
≤ n − 1,
1 + pan + qan2
which is a generalization of the following inequalities:
1
1
1
+
+ ··· +
≤ 1,
n − 1 + a1 n − 1 + a2
n − 1 + an
1
2n − 2 + a1 + a12
+
1
2n − 2 + a2 + a22
+ ··· +
1
1
≤ .
2
2n − 2 + an + an
2
Remark 2. For p = 2k and q = k2 , we get the following result:
• Let a1 , a2 , · · · , an be positive real numbers such that a1 a2 · · · an = 1. If
0<k≤
then
s
n
− 1,
n−1
1
1
1
n
+
+ ··· +
≤
,
2
2
2
(1 + ka1 )
(1 + ka2 )
(1 + kan )
(1 + k)2
with equality for a1 = a2 = · · · = an = 1.
P 1.55. Let a1 , a2 , . . . , an (n ≥ 3) be positive real numbers such that a1 a2 . . . an = 1. If
0<k≤
then
1
p
1 + ka1
+p
1
1 + ka2
n 2
− 1,
n−1
+ ··· + p
1
n
≤p
.
1+k
1 + kan
82
Vasile Cîrtoaje
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
s=
x1 + x2 + · · · + x n
= 0,
n
−1
, u ∈ R.
f (u) = p
1 + keu
For u ≤ 0, we have
f 00 (u) =
keu (2 − keu )
keu (2 − k)
≥
> 0.
4(1 + keu )5/2
4(1 + keu )5/2
Therefore, f is convex on (−∞, 0]. By HCF Theorem and Remark 5, it suffices to prove
the original inequality for a2 = · · · = an := t and a1 = 1/t n−1 , where 0 < t ≤ 1. Write
this inequality as h(t) ≤ 0, where
h(t) =
v
t
The derivative
h0 (t) =
t n−1
n−1
n
+p
−p
.
t n−1 + k
1 + kt
1+k
(n − 1)kt (n−3)/2
(n − 1)k
−
n−1
3/2
2(t
+ k)
2(kt + 1)3/2
has the same sign as
h1 (t) = t n/3−1 (kt + 1) − t n−1 − k.
Denoting m = n/3, m ≥ 1, we see that
h1 (t) = kt m + t m−1 − t 3m−1 − k = −k(1 − t m ) + t m−1 (1 − t 2m ) = (1 − t m )h2 (t),
where
h2 (t) = t m−1 + t 2m−1 − k
is strictly increasing for t ∈ (0, 1]. There are two possible cases: h2 (0) ≥ 0 and h2 (0) <
0.
Case 1: h2 (0) ≥ 0. This case is possible only for m = 1 (n = 3) and k ≤ 1, when
h2 (t) = t + 1 − k > 0 for t ∈ (0, 1]. Also, we have h1 (t) > 0 and h0 (t) > 0 for t ∈ (0, 1).
Therefore, h is strictly increasing on [0, 1], hence h(t) ≤ h(1) = 0.
Case 2: h2 (0) < 0. This case is possible for either m = 1 (n = 3) and 1 < k ≤ 5/4, or
m > 1 (n ≥ 4). Since h2 (1) = 2 − k > 0, there exists t 1 ∈ (0, 1) such that h2 (t 1 ) = 0,
h2 (t) < 0 for t ∈ (0, t 1 ), and h2 (t) > 0 for t ∈ (t 1 , 1). Since h0 has the same sign as h2 on
(0, 1), it follows that h is strictly decreasing on [0, t 1 ], and strictly increasing on [t 1 , 1].
n
Therefore, h(t) ≤ min{h(0), h(1)}. Since h(0) = n − 1 − p
≤ 0 and h(1) = 0,
1+k
Half Convex Function Method
83
we have h(t) ≤ 0 for all t ∈ (0, 1]. This completes the proof. The equality holds for
a1 = a2 = · · · = an = 1.
Remark. The following generalization holds (Vasile Cirtoaje, 2005):
• Let a1 , a2 , · · · , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If k and
m are positive numbers such that
1
m≥
,
n−1
then
n 1/m
k≤
− 1,
n−1
1
1
1
n
+
+ ··· +
≤
,
m
m
m
(1 + ka1 )
(1 + ka2 )
(1 + kan )
(1 + k)m
with equality for a1 = a2 = · · · = an = 1.
For n ≥ 3, m ≥
n 1/m
1
and k =
− 1, we get the beautiful inequality
n−1
n−1
1
1
1
+
+ ··· +
≤ n − 1.
m
m
(1 + ka1 )
(1 + ka2 )
(1 + kan )m
P 1.56. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then
1
1
1
n−1
n−1
n−1
.
a1 + a2 + · · · + an + n(n − 2) ≥ (n − 1)
+
+ ··· +
a1 a2
an
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
s=
x1 + x2 + · · · + x n
= 0,
n
f (u) = e(n−1)u − (n − 1)e−u , u ∈ R.
For u ≥ 0, we have
f 00 (u) = (n − 1)2 e(n−1)u − (n − 1)e−u = (n − 1)e−u [(n − 1)e nu − 1] ≥ 0;
therefore, f (u) is convex on [0, ∞). By HCF Theorem and Remark 2, it suffices to show
that H(x, y) ≥ 0 for x, y ∈ R such that x + (n − 1) y = 0, where
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
84
Vasile Cîrtoaje
From
f 0 (u) = (n − 1)[e(n−1)u + e−u ],
we get
(n − 1)(e x − e y ) (n−2)x
e
+ e(n−3)x+ y + · · · + e x+(n−3) y + e(n−2) y − e−x− y
x−y
(n − 1)(e x − e y ) (n−2)x
=
e
+ e(n−3)x+ y + · · · + e x+(n−3) y ) .
x−y
H(x, y) =
Since (e x − e y )/(x − y) > 0, we have H(x, y) > 0. The equality holds for a1 = a2 =
· · · = an = 1.
P 1.57. Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 . . . an = 1. If k ≥ n, then
1
1
1
k
k
k
a1 + a2 + · · · + an + kn ≥ (k + 1)
+
+ ··· +
.
a1 a2
an
(Vasile Cirtoaje, 2006)
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
s=
x1 + x2 + · · · + x n
= 0,
n
where
f (u) = e ku − (k + 1)e−u , u ∈ R.
For u ≥ 0, we have
f 00 (u) = k2 e ku − (k + 1)e−u = e−u k2 e(k+1)u − k − 1 ≥ e−u (k2 − k − 1) > 0;
therefore, f is convex on [0, ∞). By HCF Theorem and Remark 5, it suffices to to prove
the original inequality for a2 = · · · = an := b ≥ 1 and a1 := a ≤ 1, a b n−1 = 1; that is
a k + (n − 1)b k −
k + 1 (k + 1)(n − 1)
−
+ kn ≥ 0.
a
b
By the weighted AM-GM inequality, we have
1
a k + (kn − k − 1) ≥ [1 + (kn − k − 1)](a k ) 1+(kn−k−1) =
Then, we still have to show that
1
1
(n − 1) b k −
− (k + 1)
− 1 ≥ 0,
b
a
k(n − 1)
.
b
Half Convex Function Method
85
which is equivalent to h(b) ≥ 0 for b ≥ 1, where
h(b) = (n − 1)(b k+1 − 1) − (k + 1)(b n − b).
Since
h0 (b)
= (n − 1)b k − nb n−1 + 1 ≥ (n − 1)b n − nb n−1 + 1
k+1
= nb n−1 (b − 1) − (b n − 1)
= (b − 1) (b n−1 − b n−2 ) + (b n−1 − b n−3 ) + · · · + (b n−1 − 1) ≥ 0,
h is increasing on [1, ∞), hence h(b) ≥ h(1) = 0. The proof is completed. The equality
holds for a1 = a2 = · · · = an = 1.
P 1.58. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 . . . an = 1, then
1 a2
1 an
1 a1
+ 1−
+ ··· + 1 −
≤ n − 1.
1−
n
n
n
(Vasile Cîrtoaje, 2006)
Solution. Let
k=
n
,
n−1
k > 1,
and
m = ln k, 0 < m ≤ ln 2 < 1.
Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
s=
x1 + x2 + · · · + x n
= 0,
n
u
f (u) = −k−e , u ∈ R.
From
u
f 00 (u) = meu k−e (1 − meu ),
it follows that f 00 (u) > 0 for u ≤ 0, since
1 − meu ≥ 1 − m ≥ 1 − ln 2 > 0.
Therefore, f is convex on (−∞, 0]. By HCF Theorem and Remark 5, it suffices to prove
the original inequality for a2 = · · · = an := t and a1 = t −n+1 , where 0 < t ≤ 1. Write
this inequality as
h(t) ≤ n − 1,
86
Vasile Cîrtoaje
where
h(t) = k−t
−n+1
We have
h0 (t) = (n − 1)mt −n k−t
h01 (t) = k−t
−n+1
−t
+ (n − 1)k−t ,
−n+1
t ∈ (0, 1].
h1 (t), h1 (t) = 1 − t n k−t
−n+1
−t
,
h2 (t), h2 (t) = [m(n − 1 + t n ) − nt n−1 ,
h02 (t) = nt n−2 (mt − n + 1) ≤ nt n−2 (m − n + 1) ≤ nt n−2 (m − 1) < 0,
hence h2 is strictly decreasing on [0, 1]. Since h2 (0) = (n − 1)m > 0 and h2 (1) =
n(m − 1) < 0, there is t 1 ∈ (0, 1) such that h2 (t 1 ) = 0, h2 (t) > 0 for t ∈ [0, t 1 ) and
h2 (t) < 0 for t ∈ (t 1 , 1]. Therefore, h1 is strictly increasing on (0, t 1 ] and is strictly
decreasing on [t 1 , 1]. Since lim t→0 h1 (u) = −∞ and h1 (1) = 0, there is t 2 ∈ (0, t 1 ) such
that h1 (t 2 ) = 0, h1 (t) < 0 for t ∈ (0, t 2 ) and h1 (t) > 0 for t ∈ (t 2 , 1). Thus, h is strictly
decreasing on (0, t 2 ] and is strictly increasing on [t 2 , 1]. Since lim t→0 h(u) = n − 1 and
h(1) = n − 1, we have h(t) ≤ n − 1 for all t ∈ (0, 1]. This completes the proof. The
equality holds for a1 = a2 = · · · = an = 1.
P 1.59. If a, b, c are positive real numbers such that a bc = 1, then
1
1
1
+
+
≤ 1.
p
p
p
1 + 1 + 3a 1 + 1 + 3b 1 + 1 + 3c
(Vasile Cîrtoaje, 2008)
Solution. Using the substitution
a = e−x ,
b = e− y ,
c = e−z ,
we need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
where
f (u) =
s=
x + y +z
= 0,
3
p
−3
= eu − e2u + 3eu , u ∈ R.
p
1 + 1 + 3e−u
For u ≥ 0, which involves t = eu ≥ 1, we have
4t 2 + 18t + 9
00
f (u) = t 1 −
>0
p
4(t + 3) t(t + 3)
since
16t(t + 3)3 − (4t 2 + 18t + 9)2 = 9(4t 2 + 12t − 9) > 0.
Half Convex Function Method
87
Therefore, f is half convex for u ≥ 0. By HCF Theorem, it suffices to prove that f (x) +
2 f ( y) ≥ 3 f (0), where x, y ∈ R such that x + 2 y = 0. Substituting a = e x and b = e y ,
we need to show that
p
p
a − a2 + 3a + 2(b − b2 + 3b ) ≥ −3,
where a, b > 0 such that a b2 = 1. Write this inequality as
2b3 + 3b2 + 1 ≥
p
3b2 + 1 + 2b2
p
b2 + 3b.
Squaring and dividing by b2 , the inequality becomes
9b2 + 4b + 3 ≥ 4
Æ
(b2 + 3b)(3b2 + 1).
Since
2
Æ
(b2 + 3b)(3b2 + 1) ≤ (b2 + 3b) + (3b2 + 1) = 4b2 + 3b + 1,
we get
9b2 + 4b + 3 − 4
Æ
(b2 + 3b)(3b2 + 1) ≥ 9b2 + 4b + 3 − 2(4b2 + 3b + 1) = (b − 1)2 ≥ 0.
The equality holds for a = b = c = 1.
Remark. Similarly, we can prove the following generalization.
• Let a1 , a2 , . . . , an be positive real numbers such that a1 a2 · · · an = 1. If
0<k≤
4n
,
(n − 1)2
then
1
1+
p
1 + ka1 1 +
1
p
1 + ka2
+ ··· +
1
1+
p
1 + kan
≤
n
.
p
1+ 1+k
P 1.60. If a1 , a2 , . . . , an are positive real numbers such that a1 a2 · · · an = 1, then
1
1+
p
1 + 4n(n − 1)a1 1 +
1
p
1 + 4n(n − 1)a2
+ ··· +
1
1+
p
1 + 4n(n − 1)an
≥
1
.
2
(Vasile Cîrtoaje, 2008)
88
Vasile Cîrtoaje
Solution. Let k = 4n(n − 1), k ≥ 8. Using the substitutions ai = e−x i for i = 1, 2, . . . , n,
we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
s=
x1 + x2 + · · · + x n
= 0,
n
where
p
k
= e2u + keu − eu , u ∈ R.
p
1 + 1 + ke−u
For u ≤ 0, which involves t = eu ∈ (0, 1], we have
4t 2 + 6kt + k2
00
−1 >0
f (u) = t
p
4(t + k) t(t + k)
f (u) =
since
(4t 2 + 6kt + k2 )2 − 16t(t + k)3 = k2 (k2 − 4kt − 4t 2 ) ≥ k2 (k2 − 4k − 4) > 0.
Therefore, f is convex on (−∞, 0]. By HCF Theorem, it suffices to prove that f (x) +
(n − 1) f ( y) ≥ n f (0), where x, y ∈ R such that x + (n − 1) y = 0. Substituting a = e x
and b = e y , we need to show that
p
p
p
a2 + ka − a + (n − 1)( b2 + k b − b) ≥ n( 1 + k − 1),
where a, b > 0 such that a b n−1 = 1. Write this inequality as
Æ
Æ
4n(n − 1)b n−1 + 1 + (n − 1) 4n(n − 1)b2n−1 + b2n ≥ (n − 1)b n + 2n(n − 1)b n−1 + 1.
By Minkowski’s inequality, we have
Æ
Æ
4n(n − 1)b n−1 + 1 + (n − 1) 4n(n − 1)b2n−1 + b2n ≥
Æ
≥ 4n(n − 1)b n−1 [1 + (n − 1)b n/2 ]2 + [1 + (n − 1)b n ]2 .
Thus, it suffices to show that
4n(n − 1)b n−1 [1 + (n − 1)b n/2 ]2 + [1 + (n − 1)b n ]2 ≥ [(n − 1)b n + 2n(n − 1)b n−1 + 1]2 ,
which is equivalent to
4n(n − 1)2 b
3n−2
2
n
2 + (n − 2)b 2 − nb
n−2
2
≥ 0.
This inequality follows immediately by the AM-GM inequality. Thus, the proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
Half Convex Function Method
89
P 1.61. If a, b, c are positive real numbers such that a bc = 1, then
b6
c6
a6
+
+
≥ 1.
1 + 2a5 1 + 2b5 1 + 2c 5
(Vasile Cîrtoaje, 2008)
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
we need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
s=
x + y +z
= 0,
3
where
e6u
, u ∈ R.
1 + 2e5u
For u ≤ 0, which involves w = eu ∈ (0, 1], we have
f (u) =
f 00 (u) =
2w6 (2 − w5 )(9 − 2w5 )
> 0.
(1 + 2w5 )3
Therefore, f is convex for u ≤ 0. By HCF Theorem, it suffices to prove the original
inequality for b = c := t and a = 1/t 2 , where t > 0; that is,
1
t 2 (t 10 + 2)
+
2t 6
≥ 1.
1 + 2t 5
Since
1 + 2t 5 ≤ 1 + t 4 + t 6 ,
it suffices to show that
1
2x 3
+
≥ 1,
x(x 5 + 2) 1 + x 2 + x 3
x=
p
t.
This inequality can be written as follows:
x 3 (x 6 − x 5 − x 3 + 2x − 1) + (x − 1)2 ≥ 0,
x 3 (x − 1)2 (x 4 + x 3 + x 2 − 1) + (x − 1)2 ≥ 0,
(x − 1)2 [x 7 + x 5 + (x 6 − x 3 + 1)] ≥ 0.
Clearly, the last inequality is true. The equality holds for a = b = c = 1.
90
Vasile Cîrtoaje
P 1.62. If a, b, c are positive real numbers such that a bc = 1, then
p
p
p
25a2 + 144 + 25b2 + 144 + 25c 2 + 144 ≤ 5(a + b + c) + 24.
(Vasile Cîrtoaje, 2008)
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
we need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
where
f (u) = 5eu −
p
s=
x + y +z
= 0,
3
25e2u + 144, u ∈ R.
We will show that f is convex for u ≤ 0. From
5w(25w2 + 288)
00
f (u) = 5w 1 −
,
(25w2 + 144)3/2
w = eu ∈ (0, 1],
we need to show that
(25w2 + 144)3 ≥ 25w2 (25w2 + 288)2 .
Setting 25w2 = 144z, z ∈ (0, 25/144], we have
(25w2 +144)3 −25w2 (25w2 +288)2 = 1443 (z+1)3 −1443 z(z+2)2 = 1443 (1−z−z 2 ) > 0.
By HCF Theorem, it suffices to prove the original inequality for a = t 2 and b = c = 1/t,
where t > 0; that is,
p
p
5t 3 + 24t + 10 ≥ 25t 6 + 144t 2 + 2 25 + 144t 2 .
Squaring and dividing by 4t give
60t 3 + 25t 2 − 36t + 120 ≥
Æ
(25t 4 + 144)(144t 2 + 25).
Squaring again and dividing by 120, the inequality becomes
25t 5 − 36t 4 + 105t 3 − 112t 2 − 72t + 90 ≥ 0,
(t − 1)2 (25t 3 + 14t 2 + 108t + 90) ≥ 0.
Since the last inequality is true, the proof is completed. The equality holds for a = b =
c = 1.
Half Convex Function Method
91
P 1.63. If a, b, c are positive real numbers such that a bc = 1, then
p
p
p
16a2 + 9 + 16b2 + 9 + 16c 2 + 9 ≥ 4(a + b + c) + 3.
(Vasile Cîrtoaje, 2008)
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
we need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
where
f (u) =
p
s=
x + y +z
= 0,
3
16e2u + 9 − 4eu , u ∈ R.
We will show that f is convex for u ≥ 0. From
4w(16w2 + 18)
00
f (u) = 4w
−1 ,
(16w2 + 9)3/2
w = eu ≥ 1,
we need to show that
16w2 (16w2 + 18)2 ≥ (16w2 + 9)3 .
Setting 16w2 = 9z, z ≥ 16/9, we have
Indeed,
16w2 (16w2 + 18)2 − (16w2 + 9)3 = 729z(z + 2)2 − 729(z + 1)3 = 729(z 2 + z − 1) > 0.
By HCF Theorem, it suffices to prove the original inequality for a = t 2 and b = c = 1/t,
where t > 0; that is,
p
p
16a6 + 9a2 + 2 16 + 9a2 ≥ 4t 3 + 3t + 8.
Squaring and dividing by 4a give
Æ
(16a4 + 9)(9a2 + 16) ≥ 6a3 + 16a2 − 9a + 12.
Squaring again and dividing by 12a, the inequality becomes
9a5 − 16a4 + 9a3 + 12a2 − 32a + 18 ≥ 0,
(a − 1)2 (9a3 + 2a2 + 4a + 18) ≥ 0.
Since the last inequality is true, the proof is completed. The equality holds for a = b =
c = 1.
92
Vasile Cîrtoaje
P 1.64. If a, b, c, d are real numbers such that a + b + c + d = 4, then
a2
b
c
d
a
+ 2
+ 2
+ 2
≤ 1.
−a+4 b − b+4 c −c+4 d −d +4
(Sqing, 2015)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
where
f (u) =
From
f 0 (u) =
−u
,
u2 − u + 4
u2 − 4
,
(u2 − u + 4)2
f 00 (u) =
s=
a+b+c+d
= 1,
4
u ∈ R.
2(−u3 + 12u − 4)
,
(u2 − u + 4)3
it follows that f is increasing on (−∞, −2] ∪ [2, ∞), decreasing on [−2, 2] and convex
on [1, 2]. Define the function
f (u), u ≤ 2
f0 (u) =
.
f (2), u > 2
Since f0 (u) ≤ f (u) for u ∈ R and f0 (1) = f (1), it suffices to show that
f0 (a) + f0 (b) + f0 (c) + f0 (d) ≥ 4 f0 (s),
s=
a+b+c+d
= 1.
4
It is easy to check that f0 is convex on [1, ∞). Therefore, by HCF Theorem and Remark
5, we only need to show that f0 (x)+3 f0 ( y) ≥ 4 f0 (1) for all x, y ∈ R such that x ≤ 1 ≤ y
and x + 3 y = 4. There are two cases to consider: y ≤ 2 and y > 2.
Case 1: y ≤ 2. We need to show that f (x) + 3 f ( y) ≥ 3 f (1) for all x, y ∈ R such that
x + 3 y = 4. According to Remark 1, this is true if h(x, y) ≥ 0 for x + 3 y = 4. We have
g(u) =
h(x, y) =
f (u) − f (1)
u−4
=
,
2
u−1
4(u − u + 4)
g(x) − g( y)
4(x + y) − x y
3( y − 2)2 + 4
=
=
> 0.
x−y
4(x 2 − x + 4)( y 2 − y + 4) 4(x 2 − x + 4)( y 2 − y + 4)
Case 2: y > 2. From y > 2 and x + 3 y = 4, we get x < −2 and
f0 (x) + 3 f0 ( y) − 4 f0 (1) = f (x) + 3 f (2) − 4 f (1) =
The equality holds for a = b = c = d = 1.
x2
−x
> 0.
− x +4
Half Convex Function Method
93
P 1.65. If a, b, c are nonnegative real numbers such that a + b + c = 12, then
(a2 + 10)(b2 + 10)(c 2 + 10) ≥ 13310.
(Vasile Cîrtoaje, 2006)
Solution. Let
f (u) = ln(u2 + 10),
From
f 00 (u) =
u ∈ [0, ∞).
2(10 − u2 )
,
(u2 + 10)2
p
p
it follows that f is convex on [0, 10] and concave on [ 10, ∞). According to LCRCFTheorem, the sum f (a) + f (b) + f (c) is minimum when two of a, b, c are equal. Therefore, it suffices to prove the original inequality for b = c. So, we need to show that
(a2 + 10)(b2 + 10)2 ≥ 13310
for a + 2b = 12. Write this inequality as follows:
[(12 − 2b)2 + 10](b2 + 10)2 ≥ 13310,
(2b2 − 24b + 77)(b4 + 20b2 + 100) ≥ 6655,
2b6 − 24b5 + 117b4 − 480b3 + 1740b2 − 2400b + 1045 ≥ 0,
(b − 1)2 (2b4 − 20b3 + 75b2 − 310b + 1045) ≥ 0,
(b − 1)2 [2b2 (b − 5)2 + 5(5b2 − 62b + 209)] ≥ 0.
The last inequality is true since
31
5b − 62b + 209 = 5 b −
5
2
2
+
84
> 0.
5
The proof is completed. The equality holds for a = 10 and b = c = 1 (or any cyclic
permutation).
Remark. Similarly, we can prove the following generalization.
• If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = 2n(n − 1),
then
(a12 + k)(a22 + k) · · · (an2 + k) ≥ k(k + 1)n ,
k = (n − 1)(2n − 1).
The equality holds for a1 = k and a2 = · · · = an = 1 (or any cyclic permutation).
94
Vasile Cîrtoaje
P 1.66. Let a, b, c be nonnegative real numbers. If
k0 ≤ k ≤ 3,
then
k0 =
ln 2
≈ 1.71,
ln 3 − ln 2
a+b+c
a (b + c) + b (c + a) + c (a + b) ≤ 2
2
k
k
k
k+1
.
Solution. Due to homogeneity, we may assume that a + b + c = 2. Write the inequality
as
f (a) + f (b) + f (c) ≤ 2,
where
f (u) = uk (2 − u), u ∈ [0, ∞).
From
f 00 (u) = kuk−2 [2k − 2 − (k + 1)u],
2k − 2
2k − 2
and concave on
, 2 . By
it follows that therefore, f is convex on 0,
k+1
k+1
LCRCF-Theorem, the sum f (a) + f (b) + f (c) is maximum when a = 0 or 0 < a ≤ b = c.
Case 1: a = 0. We need to show that
bc(b k−1 + c k−1 ) ≤ 2
for b + c = 2. Since 0 < (k − 1)/2 ≤ 1, Bernoulli’s inequality gives
b k−1 + c k−1 = (b2 )(k−1)/2 + (c 2 )(k−1)/2 ≤ 1 +
=3−k+
k−1 2
k−1 2
(b − 1) + 1 +
(b − 1)
2
2
k−1 2
(b + c 2 ).
2
Thus, it suffices to show that
(3 − k)bc +
k−1
bc(b2 + c 2 ) ≤ 2.
2
Since
bc ≤
b+c
2
2
= 1,
we only need to show that
bc(b2 + c 2 ) ≤ 2.
Indeed, we have
8[2 − bc(b2 + c 2 )] = (b + c)4 − 8bc(b2 + c 2 ) = (b − c)4 ≥ 0.
Half Convex Function Method
95
Case 2: 0 < a ≤ b = c. We only need to prove the original homogeneous inequality for
b = c = 1 and 0 < a ≤ 1; that is,
a k+1
− a k − a − 1 ≥ 0.
1+
2
a k+1
Since 1 +
is increasing and a k is decreasing when k increases, it suffices to prove
2
that g(a) ≥ 0 for 0 < a ≤ 1, where
a k0 +1
g(a) = 1 +
− a k0 − a − 1.
2
We have
k0 + 1 a k0
− k0 a k0 −1 − 1,
1+
2
2
k0 + 1 1 00
a k0 k0 − 1
− 2−k .
g (a) =
1+
k0
4
2
a 0
g 0 (a) =
Since g 00 is increasing on (0, 1], lim x→0 g 00 (a) = −∞ and
k0 + 1 3 k0
k0 + 1
3 − k0
1 00
g (1) =
− k0 + 1 =
− k0 + 1 =
> 0,
k0
4
2
2
2
there exists a1 ∈ (0, 1) such that g 00 (a1 ) = 0, g 00 (a) < 0 for a ∈ (0, a1 ), and g 00 (a) > 0
for a ∈ (a1 , 1]. Therefore, g 0 is strictly decreasing on [0, a1 ] and strictly increasing on
k0 − 1
k0 + 1 [a1 , 1]. Since g 0 (0) =
> 0 and g 0 (1) =
(3/2)k0 − 2 = 0, there exists
2
2
a2 ∈ (0, a1 ) such that g 0 (a2 ) = 0, g 0 (a) > 0 for a ∈ [0, a2 ), and g 0 (a) < 0 for a ∈ (a2 , 1).
Thus, g is strictly increasing on [0, a2 ], and strictly decreasing on [a2 , 1]. Consequently,
g(a) ≥ min{g(0), g(1)},
and from
g(0) = 0,
g(1) = (3/2)k0 +1 − 3 = 0,
we get g(a) ≥ 0.
This completes the proof. The equality holds for a = 0 and b = c (or any cyclic
permutation). If k = k0 , then the equality holds also for a = b = c.
P 1.67. If a1 , a2 , . . . , an are positive real numbers such that a1 + a2 + · · · + an = n, then
1
1
1
(n + 1)2
+
+ ··· +
≥ 4(n + 2)(a12 + a22 + · · · + an2 ) + n(n2 − 3n − 6).
a1 a2
an
(Vasile Cîrtoaje, 2006)
96
Vasile Cîrtoaje
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n(n2 − 3n − 6),
where
f (u) =
(n + 1)2
− 4(n + 2)u2 ,
u
u ∈ (0, ∞).
From
2(n + 1)2
− 8(n + 2),
u3
it follows that f is strictly convex on (0,c] and strictly concave on [c, ∞), where
v
t
2
3 (n + 1)
c=
.
4(n + 2)
f 00 (u) =
According to LCRCF-Theorem and Remark 6, it suffices to consider the case
a1 = a2 = · · · = an−1 = x, 0 < x ≤ 1,
an = n − (n − 1)x,
when the inequality becomes as follows:
1
2 n−1
(n + 1)
+
≥ 4(n + 2)[(n − 1)x 2 + an2 ) + n(n2 − 3n − 6),
x
an
n(n − 1)(2x − 1)2 [(n + 2)(n − 1)x 2 − (n + 2)(2n − 1)x + (n + 1)2 ] ≥ 0.
The last inequality is true since
2n − 1
(n+2)(n−1)x −(n+2)(2n−1)x +(n+1) = (n+2)(n−1) x −
2n − 2
2
2
The equality holds for a1 = a2 = · · · = an−1 =
tation).
2
+
3(n − 2)
≥ 0.
4(n − 1)
n+1
1
and an =
(or any cyclic permu2
2
Chapter 2
Half Convex Function Method for
Ordered Variables
2.1
Theoretical Basis
Half Convex Function Theorem for Ordered Variables (Vasile Cirtoaje, 2007). Let
f (u) be a function defined on a real interval I and convex on Iu≥s / Iu≤s , where s ∈ I. The
inequality
a + a + ··· + a 1
2
n
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
n
holds for all a1 , a2 , . . . , an ∈ I such that a1 + a2 + · · · + an = ns and at least n − m of
a1 , a2 , . . . , an are smaller/greater than or equal to s if and only if
f (x) + m f ( y) ≥ (1 + m) f (s)
for all x, y ∈ I such that x + m y = (1 + m)s.
Proof (for the right convexity). The necessity is obvious. Without loss of generality,
assume that
a1 ≤ a2 ≤ · · · ≤ an , a1 ≤ · · · ≤ an−m ≤ s.
Since at least n − m of a1 , a2 , . . . , an are smaller than or equal to s , there is an integer
k ∈ {n − m, . . . , n − 1} such that
a1 ≤ · · · ≤ ak ≤ s ≤ ak+1 ≤ · · · ≤ an .
Since f is convex on Iu≥s , we can apply Jensen’s inequality to get
f (ak+1 ) + · · · + f (an ) ≥ (n − k) f (z),
97
z=
ak+1 + · · · + an
.
n−k
98
Vasile Cîrtoaje
Thus, it suffices to show that
f (a1 ) + · · · + f (ak ) + (n − k) f (z) ≥ n f (s).
Let b1 , . . . , bk ∈ I defined by
ai + mbi = (1 + m)s,
i = 1, . . . , k.
We claim that
z ≥ b1 ≥ · · · ≥ bk ≥ s.
Indeed, we have
b1 ≥ · · · ≥ bk ,
bk − s =
s − ak
≥ 0,
m
mb1 = (1 + m)s − a1 = (m − n + 1)s + (a2 + · · · + ak ) + (ak+1 + · · · + an )
≤ (m − n + 1)s + (k − 1)s + ak+1 + · · · + an = (m − n + k)s + (ak+1 + · · · + an )
= (m − n + k)s + (n − k)z = (m − n + k)(s − z) + mz ≤ mz.
By hypothesis, we have
f (a1 ) + m f (b1 ) ≥ (1 + m) f (s),
···
f (ak ) + m f (bk ) ≥ (1 + m) f (s),
hence
f (a1 ) + · · · + f (ak ) + m[ f (b1 ) + · · · + f (bk )] ≥ k(1 + m) f (s).
Consequently, it suffices to show that
k(1 + m) f (s) − m[ f (b1 ) + · · · + f (bk )] + (n − k) f (z) ≥ n f (s),
which is equivalent to
p f (z) + (k − p) f (s) ≥ f (b1 ) + · · · + f (bk ),
p=
n−k
≤ 1.
m
By Jensen’s inequality, we have
p f (z) + (1 − p) f (s) ≥ f (w),
w = pz + (1 − p)s ≥ s.
Thus, we only need to show that
f (w) + (k − 1) f (s) ≥ f (b1 ) + · · · + f (bk ).
HCF Method for Ordered Variables
99
Since the decreasingly ordered vector A~k = (w, s, . . . , s) majorizes the decreasingly ordered vector B~k = (b1 , b2 , . . . , bk ), this inequality follows from Karamata’s inequality for
convex functions.
The proof for the left convexity is similar.
Remark 1. Let us denote
g(u) =
f (u) − f (s)
,
u−s
h(x, y) =
g(x) − g( y)
.
x−y
In many applications, it is useful to replace the hypothesis
f (x) + m f ( y) ≥ (1 + m) f (s)
in Half Convex Function Theorem for Ordered Variables (HCF-OV Theorem) by the
equivalent condition
h(x, y) ≥ 0 for all x, y ∈ I such that x + m y = (1 + m)s.
This equivalence is true since
f (x) + m f ( y) − (1 + m) f (s) = [ f (x) − f (s)] + m[ f ( y) − f (s)]
= (x − s)g(x) + m( y − s)g( y)
m
=
(x − y)[g(x) − g( y)]
1+m
m
(x − y)2 h(x, y).
=
1+m
Remark 2. Assume that f is differentiable on I, and let
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
Then, the desired inequality of Jensen’s type in HCF-OV Theorem holds true by replacing
the hypothesis
f (x) + m f ( y) ≥ (1 + m) f (s)
with the more restrictive condition
H(x, y) ≥ 0 for all x, y ∈ I such that x + m y = (1 + m)s.
To prove this claim, we will show that the new condition implies f (x) + m f ( y) ≥
(1 + m) f (s) for all x, y ∈ I such that x + m y = (1 + m)s. Write this inequality as
f1 (x) ≥ (1 + m) f (s), where
(1 + m)s − x
f1 (x) = f (x) + m f ( y) = f (x) + m f
.
m
100
Vasile Cîrtoaje
From
f10 (x) = f 0 (x) − f 0
(1 + m)s − x
m
= f 0 (x) − f 0 ( y) =
1+m
(x − s)H(x, y),
m
it follows that f1 is decreasing for x ≤ s and increasing for x ≥ s; therefore,
f1 (x) ≥ f1 (s) = (1 + m) f (s).
Remark 3. HCF-OV Theorem is also valid in the case when I = [a, b] \ {u0 } or I =
(a, b) \ {u0 }, where a, b, u0 are real numbers such that a < u0 < b. Clearly, two cases
are possible:
(1) u0 < s - when f is right convex, i.e. convex on Iu≥s ;
(2) u0 > s - when f is left convex, i.e. convex on Iu≤s .
Remark 4. In HCF-OV Theorem for the right convexity (when HCF-OV Theorem is called
RHCF-OV Theorem), it suffices to consider
x ≤ s ≤ y.
This claim is true because in the proof of HCF-OV Theorem, the hypothesis
f (x) + m f ( y) ≥ (1 + m) f (s)
is used to get the inequalities
f (ai ) + m f (bi ) ≥ (1 + m) f (s),
i = 1, 2, · · · , k,
where ai ≤ s ≤ bi .
Similarly, in HCF-OV Theorem for the left convexity (when HCF-OV Theorem is called
LHCF-OV Theorem), it suffices to consider
x ≥ s ≥ y.
Remark 5. HCF-OV Theorem is a generalization of HCF-Theorem, because the last theorem can be obtained from the first theorem for m = n − 1.
Remark 6. If
a1 ≥ · · · ≥ am ≥ s ≥ am+1 ≥ · · · ≥ an ,
then at least n − m of a1 , a2 , . . . , an are smaller than or equal to s. Also, if
a1 ≥ · · · ≥ an−m ≥ s ≥ an−m+1 ≥ · · · ≥ an ,
then at least n − m of a1 , a2 , . . . , an are greater than or equal to s. Thus, from HCF-OV
Theorem and Remark 4, we get the following corollaries.
HCF Method for Ordered Variables
101
RHCF-OV Corollary. Let f (u) be a function defined on a real interval I and convex on
Iu≥s , where s ∈ I. The inequality
a + a + ··· + a 1
2
n
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
n
holds for all a1 , a2 , . . . , an ∈ I satisfying
a1 ≥ · · · ≥ am ≥ s ≥ am+1 ≥ · · · ≥ an ,
a1 + a2 + · · · + an = ns,
if and only if
f (x) + m f ( y) ≥ (1 + m) f (s)
for all x, y ∈ I such that x ≤ s ≤ y and x + m y = (1 + m)s.
LHCF-OV Corollary. Let f (u) be a function defined on a real interval I and convex on Iu≤s ,
where s ∈ I. The inequality
a + a + ··· + a 1
2
n
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
n
holds for all a1 , a2 , . . . , an ∈ I satisfying
a1 ≤ · · · ≤ am ≤ s ≤ am+1 ≤ · · · ≤ an ,
a1 + a2 + · · · + an = ns,
if and only if
f (x) + m f ( y) ≥ (1 + m) f (s)
for all x, y ∈ I such that x ≥ s ≥ y and x + m y = (1 + m)s.
Remark 7. The inequality in RHCF-OV Corollary becomes an equality for
a1 = a2 = · · · = an = s
and also for
a1 = · · · = am = y,
am+1 = · · · = an−1 = s,
an = x (x < y),
where x, y ∈ I satisfy the equations
x + m y = (1 + m)s,
f (x) + m f ( y) = (1 + m) f (s).
For x 6= y, these equations are equivalent to
x + m y = (1 + m)s, h(x, y) = 0.
Remark 8. The inequality in LHCF-OV Corollary becomes an equality for
a1 = a2 = · · · = an = s
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Vasile Cîrtoaje
and also for
a1 = x,
a2 = · · · = an−m = s,
an−m+1 = · · · = an = y (x > y),
where x, y ∈ I satisfy the equations
x + m y = (1 + m)s,
f (x) + m f ( y) = (1 + m) f (s).
For x 6= y, these equations are equivalent to
x + m y = (1 + m)s, h(x, y) = 0.
HCF Method for Ordered Variables
2.2
103
Applications
2.1. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n.
(a) If at least n − m of a1 , a2 , . . . , an are smaller than or equal to 1, then
m(a13 + a23 + · · · + an3 − n) ≥ (2m + 1)(a12 + a22 + · · · + an2 − n);
(b) If at least n − m of a1 , a2 , . . . , an are greater than or equal to 1, then
a13 + a23 + · · · + an3 − n ≤ (m + 2)(a12 + a22 + · · · + an2 − n).
2.2. If a, b, c, d are real numbers such that
a + b + c + d = 4,
a ≥ b ≥ 1 ≥ c ≥ d,
then
(3a2 − 2)(a − 1)2 + (3b2 − 2)(b − 1)2 + (3c 2 − 2)(c − 1)2 + (3d 2 − 2)(d − 1)2 ≥ 0.
2.3. If a1 , a2 , . . . , a2n ≥
−2n − 1
such that
n−1
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
a1 + a2 + · · · + a2n = 2n,
then
3
≥ 2n.
a13 + a23 + · · · + a2n
2.4. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n.
(a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then
a14 + a24 + · · · + an4 − n ≥ 6(a12 + a22 + · · · + an2 − n);
(b) If a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , then
a14 + a24 + · · · + an4 − n ≥
14 2
(a + a22 + · · · + an2 − n);
3 1
(c) If a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then
a14 + a24 + · · · + an4 − n ≥
2(n2 − 3n + 3) 2
(a1 + a22 + · · · + an2 − n).
n2 − 5n + 7
104
Vasile Cîrtoaje
2.5. Let a, b, c, d, e be nonnegative real numbers such that a + b + c + d + e = 5.
(a) If a ≥ b ≥ 1 ≥ c ≥ d ≥ e, then
21(a2 + b2 + c 2 + d 2 + e2 ) ≥ a4 + b4 + c 4 + d 4 + e4 + 100;
(b) If a ≥ b ≥ c ≥ 1 ≥ d ≥ e, then
13(a2 + b2 + c 2 + d 2 + e2 ) ≥ a4 + b4 + c 4 + d 4 + e4 + 60.
2.6. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n.
(a) If a1 ≥ · · · ≥ an−1 ≥ 1 ≥ an , then
7(a13 + a23 + · · · + an3 ) ≥ 3(a14 + a24 + · · · + an4 ) + 4n;
(b) If a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then
13(a13 + a23 + · · · + an3 ) ≥ 4(a14 + a24 + · · · + an4 ) + 9n.
2.7. If a1 , a2 , . . . , an are positive real numbers such that
a1 + a2 + · · · + an = n,
a1 ≥ · · · ≥ an−m ≥ 1 ≥ an−m+1 ≥ · · · ≥ an ,
then
(m + 1)2
1
1
1
+
+ ··· +
− n ≥ 4m(a12 + a22 + · · · + an2 − n).
a1 a2
an
2.8. If a1 , a2 , . . . , an are positive real numbers such that
1
1
1
+
+ ··· +
= n,
a1 a2
an
a1 ≤ · · · ≤ an−m ≤ 1 ≤ an−m+1 ≤ · · · ≤ an ,
then
a12
+ a22
+ · · · + an2
p
m
(a1 + a2 + · · · + an − n).
−n≥2 1+
1+m
2.9. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n.
(a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then
1
a12
+2
+
1
a22
+2
+ ··· +
an2
1
n
≥ ;
+2 3
(b) If a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , then
1
2a12
+3
+
1
2a22
+3
+ ··· +
1
n
≥ .
+3 5
2an2
HCF Method for Ordered Variables
105
2.10. If a1 , a2 , . . . , a2n are nonnegative real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
a1 + a2 + · · · + a2n = 2n,
then
1
na12
+ n2
+n+1
+
1
na22
+ n2
+n+1
+ ··· +
1
2
na2n
+ n2
+n+1
≤
2n
.
(n + 1)2
2.11. If a, b, c, d, e, f are nonnegative real numbers such that
a ≥ b ≥ c ≥ 1 ≥ d ≥ e ≥ f,
then
a + b + c + d + e + f = 6,
3f + 4
3a + 4
3b + 4
3c + 4
3d + 4
3e + 4
+ 2
+ 2
+ 2
+ 2
+
≤ 6.
2
3a + 4 3b + 4 3c + 4 3d + 4 3e + 4 3 f 2 + 4
2.12. If a, b, c, d, e, f are nonnegative real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d ≥ e ≥ f,
a + b + c + d + e + f = 6,
then
f2−1
b2 − 1
c2 − 1
d2 − 1
e2 − 1
a2 − 1
+
+
+
+
+
≥ 0.
(2a + 7)2 (2b + 7)2 (2c + 7)2 (2d + 7)2 (2e + 7)2 (2 f + 7)2
2.13. If a, b, c, d, e, f are nonnegative real numbers such that
a ≥ b ≥ c ≥ d ≥ 1 ≥ e ≥ f,
a + b + c + d + e + f = 6,
then
f2−1
b2 − 1
c2 − 1
d2 − 1
e2 − 1
a2 − 1
+
+
+
+
+
≤ 0.
(2a + 5)2 (2b + 5)2 (2c + 5)2 (2d + 5)2 (2e + 5)2 (2 f + 5)2
2.14. If a, b, c, d are nonnegative real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
then
a + b + c + d = 4,
1
1
1
1
4
+ 3
+ 3
+ 3
≤ .
+ 5 2b + 5 2c + 5 2d + 5 7
2a3
106
Vasile Cîrtoaje
2.15. If a1 , a2 , . . . , a8 are nonnegative real numbers such that
a1 ≥ a2 ≥ a3 ≥ a4 ≥ 1 ≥ a5 ≥ a6 ≥ a7 ≥ a8 ,
a1 + a2 + · · · + a8 = 8,
then
(a12 + 1)(a22 + 1) · · · (a82 + 1) ≥ (a1 + 1)(a2 + 1) · · · (a8 + 1).
2.16. If a, b, c, d are positive real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
then
a bcd = 1,
1 1 1 1
a + b + c + d − 4 ≥ 18 a + b + c + d − − − −
.
a b c d
2
2
2
2
2.17. If a, b, c, d are positive real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
a bcd = 1,
then
p
a2 − a + 1 +
p
b2 − b + 1 +
p
c2 − c + 1 +
p
d 2 − d + 1 ≥ a + b + c + d.
2.18. If a, b, c, d are positive real numbers such that
a ≥ 1 ≥ b ≥ c ≥ d,
then
a3
a bcd = 1,
1
1
1
1
2
+ 3
+ 3
+ 3
≥ .
+ 3a + 2 b + 3b + 2 c + 3c + 2 d + 3d + 2 3
2.19. Let a1 , a2 , . . . , an be positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
If k ≥ 1, then
a1 a2 · · · an = 1.
1
1
1
n
+
+ ··· +
≥
.
1 + ka1 1 + ka2
1 + kan
1+k
HCF Method for Ordered Variables
107
2.20. If a1 , a2 , . . . , a9 are positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ a9 ,
then
a1 a2 · · · a9 = 1,
1
1
1
+
+ ··· +
≥ 1.
(a1 + 2)2 (a2 + 2)2
(a9 + 2)2
2.21. If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that
a1 ≥ a2 ≥ a3 ≥ 1 ≥ a4 ≥ · · · ≥ an ,
then
a1 a2 · · · an = 1,
1
1
1
n
+
+ ··· +
≥ .
(a1 + 1)2 (a2 + 1)2
(an + 1)2
4
2.22. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,
then
a1 a2 · · · an = 1,
1
1
1
n
+
+ ··· +
≤
.
2
2
2
(a1 + 3)
(a2 + 3)
(an + 3)
16
2.23. Let a1 , a2 , . . . , an be positive real numbers such that
a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,
a1 a2 · · · an = 1.
If p, q ≥ 0 such that p + q ≤ 1, then
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
1
n
≤
.
2
1 + pan + qan
1+p+q
2.24. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≤ a2 ≤ 1 ≤ a3 ≤ · · · ≤ an ,
then
a1 a2 · · · an = 1,
1
1
1
n
+
+ ··· +
≤
.
2
2
2
(a1 + 5)
(a2 + 5)
(an + 5)
36
108
Vasile Cîrtoaje
2.25. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
then
1
p
1 + 3a1
+p
1
1 + 3a2
a1 a2 · · · an = 1,
+ ··· + p
1
1 + 3an
≥
n
.
2
2.26. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,
then
1
p
1 + 2a1
+p
1
1 + 2a2
a1 a2 · · · an = 1,
+ ··· + p
1
n
≤p .
3
1 + 2an
HCF Method for Ordered Variables
2.3
109
Solutions
P 2.1. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n.
(a) If at least n − m of a1 , a2 , . . . , an are smaller than or equal to 1, then
m(a13 + a23 + · · · + an3 − n) ≥ (2m + 1)(a12 + a22 + · · · + an2 − n);
(b) If at least n − m of a1 , a2 , . . . , an are greater than or equal to 1, then
a13 + a23 + · · · + an3 − n ≤ (m + 2)(a12 + a22 + · · · + an2 − n).
(Vasile Cirtoaje, 2007)
Solution. (a) Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = mu3 − (2m + 1)u2 ,
u ∈ [0, n].
From f 00 (u) = 2(3mu − 2m − 1), it follows that f is convex on [1, n]. By HCF-OV
Theorem (or RHCF-OV Corollary) and Remark 1, it suffices to show that h(x, y) ≥ 0 for
all x, y ≥ 0 such that x + m y = m + 1. We have
g(u) =
f (u) − f (1)
= m(u2 + u + 1) − (2m + 1)(u + 1) = mu2 − (m + 1)u − m − 1,
u−1
h(x, y) =
g(x) − g( y)
= m(x + y) − m − 1 = (m − 1)x ≥ 0.
x−y
From x + m y = m + 1 and h(x, y) = 0, we get x = 0, y = (m + 1)/m. Therefore, in
accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for
a1 = · · · = am =
m+1
,
m
am+1 = · · · = an−1 = 1,
an = 0
(or any thereof permutation).
(b) Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = (m + 2)u2 − u3 ,
u ∈ [0, n].
110
Vasile Cîrtoaje
From f 00 (u) = 2(m + 2 − 3u), it follows that f is convex on [0, 1]. By HCF-OV Theorem
(or LRHCF-OV Corollary) and Remark 1, it suffices to show that h(x, y) ≥ 0 for all
x, y ≥ 0 such that x + m y = m + 1. We have
g(u) =
f (u) − f (1)
= (m + 2)(u + 1) − (u2 + u + 1) = −u2 + (m + 1)u + m + 1,
u−1
h(x, y) =
g(x) − g( y)
= −(x + y) + m + 1 = (m − 1) y ≥ 0.
x−y
From x +m y = m+1 and h(x, y) = 0, we get x = m+1, y = 0. Therefore, in accordance
with Remark 8, the equality holds for a1 = a2 = · · · = an = 1, and also for
a1 = m + 1,
a2 = · · · = an−m = 1,
an−m+1 = · · · = an = 0
(or any thereof permutation).
Remark 1. For m = n − 1, we get the following results:
• If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
(n − 1)(a13 + a23 + · · · + an3 − n) ≥ (2n − 1)(a12 + a22 + · · · + an2 − n),
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 0 and a2 = a3 = · · · = an =
n
(or any cyclic permutation).
n−1
• If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n, then
a13 + a23 + · · · + an3 − n ≤ (n + 1)(a12 + a22 + · · · + an2 − n),
with equality for a1 = a2 = · · · = an = 1, and also for a1 = n and a2 = a3 = · · · = an = 0
(or any cyclic permutation).
Remark 2. For m = 1, we get the following results:
• If a1 , a2 , . . . , an are nonnegative real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
a1 + a2 + · · · + an = n,
then
a13 + a23 + · · · + an3 + 2n ≥ 3(a12 + a22 + · · · + an2 ),
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 2, a2 = · · · = an−1 = 1,
an = 0.
• If a1 , a2 , . . . , an are nonnegative real numbers such that
a1 ≥ · · · ≥ an−1 ≥ 1 ≥ an ,
a1 + a2 + · · · + an = n,
HCF Method for Ordered Variables
111
then
a13 + a23 + · · · + an3 + 2n ≤ 3(a12 + a22 + · · · + an2 ),
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 2, a2 = · · · = an−1 = 1,
an = 0.
Remark 3. Replacing n with 2n and choosing m = n, we get the following results:
• If a1 , a2 , . . . , a2n are nonnegative real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
a1 + a2 + · · · + a2n = 2n,
then
2
3
− 2n),
− 2n) ≥ (2n + 1)(a12 + a22 + · · · + a2n
n(a13 + a23 + · · · + a2n
with equality for a1 = a2 = · · · = a2n = 1, and also for
a1 = · · · = an =
n+1
,
n
an+2 = · · · = a2n−1 = 1,
a2n = 0.
• If a1 , a2 , . . . , a2n are nonnegative real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
a1 + a2 + · · · + a2n = 2n,
then
2
3
− 2n),
− 2n ≤ (n + 2)(a12 + a22 + · · · + a2n
a13 + a23 + · · · + a2n
with equality for a1 = a2 = · · · = a2n = 1, and also for
a1 = n + 1,
a2 = · · · = an = 1,
an+1 = · · · = a2n = 0.
P 2.2. If a, b, c, d are real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
a + b + c + d = 4,
then
(3a2 − 2)(a − 1)2 + (3b2 − 2)(b − 1)2 + (3c 2 − 2)(c − 1)2 + (3d 2 − 2)(d − 1)2 ≥ 0.
(Vasile Cirtoaje, 2007)
112
Vasile Cîrtoaje
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
s=
a+b+c+d
= 1,
4
where
f (u) = (3u2 − 2)(u − 1)2 ,
u ∈ R.
From
f 00 (u) = 2(18u2 − 18u + 1),
it follows that f 00 (u) > 0 for u ≥ 1, hence f is convex for u ≥ 1. Therefore, we may
apply HCF-OV Theorem or RHCF-OV Corollary for n = 4 and m = 2. By these, it suffices
to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3. Using Remark
1, we only need to show that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
We have
g(u) = 3(u3 + u2 + u + 1) − 6(u2 + u + 1) + u + 1 = 3u3 − 3u2 − 2u − 2,
h(x, y) = 3(x 2 + x y + y 2 ) − 3(x + y) − 2 = (3 y − 4)2 ≥ 0.
From x + 2 y = 3 and h(x, y) = 0, we get x = 1/3, y = 4/3. Therefore, in accordance
with Remark 7, the equality holds for a = b = c = d = 1, and also for a = b = 4/3,
c = 1 and d = 1/3.
Remark. Similarly, we can prove the following generalization:
• Let a1 , a2 , . . . , a2n be real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
If k =
n2
a1 + a2 + · · · + a2n = 2n.
n
, then
−n+1
2
(a12 − k)(a1 − 1)2 + (a22 − k)(a2 − 1)2 + · · · + (a2n
− k)(a2n − 1)2 ≥ 0,
with equality for a1 = a2 = · · · = a2n = 1, and also for
a1 = · · · = an =
n2
,
n2 − n + 1
an+1 = · · · = a2n−1 = 1,
a2n =
n2
1
.
−n+1
HCF Method for Ordered Variables
P 2.3. If a1 , a2 , . . . , a2n ≥
113
−2n − 1
such that
n−1
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
a1 + a2 + · · · + a2n = 2n,
then
3
a13 + a23 + · · · + a2n
≥ 2n.
(Vasile Cirtoaje, 2007)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (a2n ) ≥ 2n f (s),
where
f (u) = u3 ,
u≥
s=
a1 + a2 + · · · + a2n
= 1,
2n
−2n − 1
.
n−1
From f 00 (u) = 6u, it follows that f is convex for u ≥ 1. Therefore, we may apply HCF-OV
Theorem or RHCF-OV Corollary for 2n numbers and m = n. By Remark 1, it suffices to
−2n − 1
show that h(x, y) ≥ 0 for all x, y ≥
such that x + n y = n + 1. We have
n−1
g(u) =
h(x, y) =
f (u) − f (1)
= u2 + u + 1,
u−1
g(x) − g( y)
(n − 1)x + 2n + 1
= x + y +1=
≥ 0.
x−y
n−1
n+2
−2n − 1
and y =
. Therefore,
n−1
n−1
in accordance with Remark 7, the equality holds for a1 = a2 = · · · = a2n = 1, and also
for
n+2
−2n − 1
a1 = · · · = an =
, an+1 = · · · = a2n−1 = 1, a2n =
.
n−1
n−1
From x + n y = n + 1 and h(x, y) = 0, we get x =
P 2.4. Let a1 , a2 , . . . , an be real numbers such that a1 + a2 + · · · + an = n.
(a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then
a14 + a24 + · · · + an4 − n ≥ 6(a12 + a22 + · · · + an2 − n);
(b) If a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , then
a14 + a24 + · · · + an4 − n ≥
14 2
(a + a22 + · · · + an2 − n);
3 1
114
Vasile Cîrtoaje
(c) If a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then
a14 + a24 + · · · + an4 − n ≥
2(n2 − 3n + 3) 2
(a1 + a22 + · · · + an2 − n).
n2 − 5n + 7
(Vasile Cirtoaje, 2009)
Solution. Consider the inequality
a14 + a24 + · · · + an4 − n ≥ k(a12 + a22 + · · · + an2 − n),
k ≤ 6,
and write it as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = u4 − ku2 ,
u ≥ 0.
From f 00 (u) = 2(6u2 − k), it follows that f is convex for u ≥ 1. Therefore, we may apply
HCF-OV Theorem or RHCF-OV Corollary for m = 1, m = 2 and m = n − 2, respectively.
By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x +m y = m+1.
We have
f (u) − f (1)
g(u) =
= u3 + u2 + u + 1 − k(u + 1),
u−1
h(x, y) =
g(x) − g( y)
= x 2 + x y + y 2 + x + y − k + 1.
x−y
(a) We have k = 6, m = 1 and x + y = 2, which involve
h(x, y) = 1 − x y =
1
(x − y)2 ≥ 0.
4
From x + y = 2 and h(x, y) = 0, we get x = y = 1. Therefore, in accordance with
Remark 7, the equality holds for a1 = a2 = · · · = an = 1.
(b) We have k = 14/3, m = 2 and x + 2 y = 3, which involve
h(x, y) =
1
(3 y − 5)2 ≥ 0.
3
From x + 2 y = 3 and h(x, y) = 0, we get x = −1/3 and y = 5/3. Therefore, in
accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for
a1 = a2 =
5
,
3
a3 = · · · = an−1 = 1,
an =
−1
.
3
HCF Method for Ordered Variables
(c) We have k =
115
2(n2 − 3n + 3)
, m = n − 2 and x + (n − 2) y = n − 1, which involve
n2 − 5n + 7
h(x, y) =
n2
1
[(n2 − 5n + 7) y − n2 + 3n − 1]2 ≥ 0.
− 5n + 7
−n2 + 5n − 4
n2 − 3n + 1
and
.
n2 − 5n + 7
n2 − 5n + 7
Therefore, in accordance with Remark 7, the equality holds for a1 = a2 = · · · = an = 1,
and also for
From x + (n − 2) y = n − 1 and h(x, y) = 0, we get x =
a1 = · · · = an−2 =
n2 − 3n + 1
,
n2 − 5n + 7
an−1 = 1,
an =
−n2 + 5n − 4
.
n2 − 5n + 7
P 2.5. Let a, b, c, d, e be nonnegative real numbers such that a + b + c + d + e = 5.
(a) If a ≥ b ≥ 1 ≥ c ≥ d ≥ e, then
21(a2 + b2 + c 2 + d 2 + e2 ) ≥ a4 + b4 + c 4 + d 4 + e4 + 100;
(b) If a ≥ b ≥ c ≥ 1 ≥ d ≥ e, then
13(a2 + b2 + c 2 + d 2 + e2 ) ≥ a4 + b4 + c 4 + d 4 + e4 + 60.
(Vasile Cirtoaje, 2009)
Solution. Consider the inequality
k(a2 + b2 + c 2 + d 2 + e2 − 5) ≥ a4 + b4 + c 4 + d 4 + e4 − 5,
k ≥ 6,
and write it as
f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s),
s=
a+b+c+d+e
= 1,
5
where
f (u) = ku2 − u4 ,
u ≥ 0.
From f 00 (u) = 2(k − 6u2 ), it follows that f is convex on [0, 1]. Therefore, we may apply
HCF-OV Theorem or LHCF-OV Corollary for m = 3 and m = 2, respectively. By Remark
1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + m y = m + 1. We
have
f (u) − f (1)
g(u) =
= k(u + 1) − (u3 + u2 + u + 1),
u−1
116
Vasile Cîrtoaje
h(x, y) =
g(x) − g( y)
= k − (x 2 + x y + y 2 + x + y + 1).
x−y
(a) We have k = 21, m = 3 and x + 3 y = 4, which involve y ≤ 4/3 and
h(x, y) = 20 − (x 2 + x y + y 2 + x + y) = y(22 − 7 y) ≥ 0.
From x + 3 y = 4 and h(x, y) = 0, we get x = 4 and y = 0. Therefore, in accordance
with Remark 8, the equality holds for a = b = c = d = e = 1 and also for a = 4, b = 1,
c = d = e = 0.
(b) We have k = 13, m = 2 and x + 2 y = 3, which involve y ≤ 3/2 and
h(x, y) = 12 − (x 2 + x y + y 2 + x + y) = y(10 − 3 y) ≥ 0.
From x + 2 y = 3 and h(x, y) = 0, we get x = 3 and y = 0. Therefore, in accordance
with Remark 8, the equality holds for a = b = c = d = e = 1 and also for a = 3,
b = c = 1, d = e = 0.
P 2.6. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n.
(a) If a1 ≥ · · · ≥ an−1 ≥ 1 ≥ an , then
7(a13 + a23 + · · · + an3 ) ≥ 3(a14 + a24 + · · · + an4 ) + 4n;
(b) If a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an , then
13(a13 + a23 + · · · + an3 ) ≥ 4(a14 + a24 + · · · + an4 ) + 9n.
(Vasile Cirtoaje, 2009)
Solution. Consider the inequality
k(a13 + a23 + · · · + an3 − n) ≥ a14 + a24 + · · · + an4 − n,
k > 2,
and write it as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = ku3 − u4 ,
u ≥ 0.
From f 00 (u) = 6u(k − 2u2 ), it follows that f is convex on [0, 1]. Therefore, we may
apply HCF-OV Theorem or LHCF-OV Corollary for m = 1 and m = 2, respectively. By
HCF Method for Ordered Variables
117
Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0 such that x + m y = m + 1.
We have
f (u) − f (1)
g(u) =
= k(u2 + u + 1) − (u3 + u2 + u + 1),
u−1
g(x) − g( y)
h(x, y) =
= −(x 2 + x y + y 2 ) + (k − 1)(x + y + 1).
x−y
(a) We have k = 7/3, m = 1 and x + y = 2, which involve
h(x, y) = x y ≥ 0.
From x + y = 2 and h(x, y) = 0, we get x = 2 and y = 0. Therefore, in accordance
with Remark 8, the equality holds for a1 = a2 = · · · = an = 1, and also for
a1 = 2,
a2 = · · · = an−1 = 1,
an = 0.
(b) We have k = 13/4, m = 2 and x + 2 y = 3, which involve
h(x, y) = 3 y(9 − 4 y) = 3 y(3 + 2x) ≥ 0.
From x + 2 y = 3 and h(x, y) = 0, we get x = 3 and y = 0. Therefore, in accordance
with Remark 8, the equality holds for a1 = a2 = · · · = an = 1, and also for
a1 = 3,
a2 = · · · = an−2 = 1,
an−1 = an = 0.
P 2.7. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≥ · · · ≥ an−m ≥ 1 ≥ an−m+1 ≥ · · · ≥ an ,
then
(m + 1)2
a1 + a2 + · · · + an = n,
1
1
1
+
+ ··· +
− n ≥ 4m(a12 + a22 + · · · + an2 − n).
a1 a2
an
(Vasile Cirtoaje, 2007)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
s=
a1 + a2 + · · · + an
= 1,
n
(m + 1)2
− 4mu2 , u ∈ (0, n).
u
118
Vasile Cîrtoaje
For u ∈ (0, 1], we have
f 00 (u) =
2(m + 1)2
− 8m ≥ 2(m + 1)2 − 8m = 2(m − 1)2 ≥ 0.
u3
Since f is convex on (0, 1], we may apply HCF-OV Theorem or LHCF-OV Corollary. By
Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y > 0 such that x + m y = m + 1.
We have
f (u) − f (1) −(m + 1)2
g(u) =
=
− 4m(u + 1),
u−1
u
h(x, y) =
[m + 1 − 2m y]2
(m + 1)2
− 4m =
≥ 0.
xy
xy
1+m
1+m
and y =
. Therefore, in
2
2m
accordance with Remark 8, the equality holds for a1 = a2 = · · · = an = 1, and also for
From x + m y = m + 1 and h(x, y) = 0, we get x =
a1 =
1+m
,
2
a2 = a3 = · · · = an−m = 1,
an−m+1 = · · · = an =
1+m
.
2m
Remark 1. For m = 1, we get the following elegant statement:
• If a1 , a2 , . . . , an are positive real numbers such that
a1 ≥ · · · ≥ an−1 ≥ 1 ≥ an ,
then
a1 + a2 + · · · + an = n,
1
1
1
+
+ ··· +
≥ a12 + a22 + · · · + an2 ,
a1 a2
an
with equality for a1 = a2 = · · · = an = 1
Remark 2. Replacing n with 2n and choosing m = n, we get the statement:
• If a1 , a2 , . . . , a2n are positive real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
then
(n + 1)2
a1 + a2 + · · · + a2n = 2n,
1
1
1
2
+
+ ··· +
− 2n ≥ 4n(a12 + a22 + · · · + a2n
− 2n),
a1 a2
a2n
with equality for a1 = a2 = · · · = a2n = 1, and also for
a1 =
1+n
,
2
a2 = a3 = · · · = an = 1,
an+1 = · · · = a2n =
1+n
.
2n
HCF Method for Ordered Variables
119
P 2.8. If a1 , a2 , . . . , an are positive real numbers such that
1
1
1
+
+ ··· +
= n,
a1 a2
an
a1 ≤ · · · ≤ an−m ≤ 1 ≤ an−m+1 ≤ · · · ≤ an ,
then
a12
+ a22
p
m
−n≥2 1+
(a1 + a2 + · · · + an − n).
1+m
+ · · · + an2
(Vasile Cirtoaje, 2007)
Solution. Replacing each ai by 1/ai , we need to prove that
a1 ≥ · · · ≥ an−m ≥ 1 ≥ an−m+1 ≥ · · · ≥ an ,
a1 + a2 + · · · + an = n
involves
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
s=
a1 + a2 + · · · + an
= 1,
n
p
1
1
m
, u ∈ (0, n).
f (u) = 2 − 2 1 +
u
1+m u
For u ∈ (0, 1], we have
00
f (u) =
6−4 1+
p
m
1+m u
u4
≥
6−4 1+
p
m
m+1
u4
p
2( m − 1)2
=
≥ 0.
(1 + m)u4
Thus, f is convex on (0, 1]. By HCF-OV Theorem (or LHCF-OV Corollary) and Remark
1, it suffices to show that h(x, y) ≥ 0 for x, y > 0 such that x + m y = 1 + m, where
h(x, y) =
We have
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
p
2 m 1
−1
g(u) = 2 + 1 +
u
1+m u
and
h(x, y) =
We only need to show that
1
xy
p
1 1
2 m
+ −1−
.
x
y
1+m
p
1 1
2 m
+ ≥1+
.
x
y
1+m
Indeed, using the Cauchy-Schwarz inequality, we get
p
p
p
(1 + m)2
(1 + m)2
2 m
1 1
+ ≥
=
=1+
.
x
y
x + my
1+m
1+m
120
Vasile Cîrtoaje
1+m
1+m
p and y =
p . In
1+ m
m+ m
accordance with Remark 8, we have f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (1) for a1 = a2 =
· · · = an = 1, and also for
From x + m y = 1 + m and h(x, y) = 0, we get x =
a1 =
1+m
p ,
1+ m
a2 = a3 = · · · = an−m = 1,
an−m+1 = · · · = an =
1+m
p .
m+ m
Therefore, the original inequality becomes an equality for a1 = a2 = · · · = an = 1, and
also for
p
p
m+ m
1+ m
, a2 = a3 = · · · = an−m = 1, an−m+1 = · · · = an =
.
a1 =
1+m
1+m
Remark. Replacing n with 2n and choosing m = n, we get the statement:
• If a1 , a2 , . . . , a2n are positive real numbers such that
a1 ≤ · · · ≤ an ≤ 1 ≤ an+1 ≤ · · · ≤ a2n ,
then
a12
+ a22
1
1
1
+
+ ··· +
= 2n,
a1 a2
a2n
p
n
− 2n ≥ 2 1 +
(a1 + a2 + · · · + a2n − 2n).
1+n
2
+ · · · + a2n
with equality for a1 = a2 = · · · = a2n = 1, and also for
a1 =
p
1+ n
,
1+n
a2 = a3 = · · · = an = 1,
an+1 = · · · = a2n =
p
n+ n
.
1+n
P 2.9. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n.
(a) If a1 ≥ 1 ≥ a2 ≥ · · · ≥ an , then
1
a12
+2
+
1
a22
+2
+ ··· +
1
n
≥ ;
an2 + 2 3
(b) If a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an , then
1
2a12
+3
+
1
2a22
+3
+ ··· +
1
n
≥
.
2an2 + 3 5
(Vasile Cirtoaje, 2007)
HCF Method for Ordered Variables
121
Solution. Consider the inequality
1
a12
+k
+
1
a22
+k
+ ··· +
an2
1
n
≥
,
+k
1+k
k ∈ [0, 3];
and write it as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
1
,
u2 + k
s=
a1 + a2 + · · · + an
= 1,
n
u ∈ [0, n].
For u ∈ [1, n], we have
f 00 (u) =
2(3u2 − k)
2(3 − k)
≥ 2
≥ 0,
2
3
(u + k)
(u + k)3
hence f is convex on [1, n]. Therefore, we may apply HCF-OV Theorem or RHCF-OV
Corollary for m = 1 and m = 2, respectively. By Remark 1, it suffices to show that
h(x, y) ≥ 0 for all x, y ≥ 0 such that x + m y = m + 1. Since
g(u) =
h(x, y) =
f (u) − f (1)
−u − 1
=
,
u−1
(1 + k)(u2 + k)
xy+x + y−k
g(x) − g( y)
=
,
x−y
(1 + k)(x 2 + k)( y 2 + k)
we only need to show that
x y + x + y − k ≥ 0.
(a) We have k = 2, m = 1 and x + y = 2, which involve
x y + x + y − k = x y ≥ 0.
From x + y = 2 and h(x, y) = 0, we get x = 0 and y = 2. Therefore, in accordance
with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for
a1 = 2,
a2 = · · · = an−1 = 1,
an = 0.
(b) We have k = 3/2, m = 2 and x + 2 y = 3, which involve
xy+x + y−k=
x(1 + 2 y)
x(4 − x)
=
≥ 0.
2
2
From x + 2 y = 3 and h(x, y) = 0, we get x = 0 and y = 3/2. Therefore, in accordance
with Remark 7, the equality holds for a1 = a2 = · · · = an = 1, and also for
a1 = a2 = 3/2,
a3 = · · · = an−1 = 1,
an = 0.
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Vasile Cîrtoaje
P 2.10. If a1 , a2 , . . . , a2n are nonnegative real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
a1 + a2 + · · · + a2n = 2n,
then
1
na12
+ n2
+n+1
+
1
na22
+ n2
+n+1
+ ··· +
1
2
na2n
+ n2
+n+1
≤
2n
.
(n + 1)2
(Vasile Cirtoaje, 2007)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (a2n ) ≥ 2n f (s),
where
f (u) =
−1
,
nu2 + n2 + n + 1
s=
a1 + a2 + · · · + a2n
= 1,
2n
u ∈ [0, 2n].
For u ∈ [0, 1], we have
f 00 (u) =
2nu(n2 + n + 1 − 3nu2 ) 2nu(n2 + n + 1 − 3n)
≥
≥ 0,
(nu2 + n2 + n + 1)3
(nu2 + n2 + n + 1)3
hence f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV
Corollary for m = n. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0
such that x + n y = n + 1. Since
g(u) =
h(x, y) =
f (u) − f (1)
n(u + 1)
=
,
2
u−1
(n + 1) (nu2 + n2 + n + 1)
g(x) − g( y)
n(n2 + n + 1 − nx − n y − nx y)
=
,
x−y
(n + 1)2 (nx 2 + n2 + n + 1)(n y 2 + n2 + n + 1)
we only need to show that
n2 + n + 1 − n(x + y + x y) ≥ 0.
Indeed,
n2 + n + 1 − n(x + y + x y) = (n y − 1)2 ≥ 0.
From x + n y = n + 1 and h(x, y) = 0, we get x = n and y = 1/n. Therefore, in
accordance with Remark 7, the equality holds for a1 = a2 = · · · = a2n = 1, and also for
a1 = n,
a2 = · · · = an = 1,
an+1 = · · · = an = 1/n.
HCF Method for Ordered Variables
123
P 2.11. If a, b, c, d, e, f are nonnegative real numbers such that
a ≥ b ≥ c ≥ 1 ≥ d ≥ e ≥ f,
then
a + b + c + d + e + f = 6,
3f + 4
3b + 4
3c + 4
3d + 4
3e + 4
3a + 4
+ 2
+ 2
+ 2
+ 2
+
≤ 6.
2
3a + 4 3b + 4 3c + 4 3d + 4 3e + 4 3 f 2 + 4
(Vasile Cirtoaje, 2009)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) + f (e) + f ( f ) ≥ 6 f (s),
where
f (u) =
−3u − 4
,
3u2 + 4
s=
a+b+c+d+e+ f
= 1,
6
u ∈ [0, 6].
For u ∈ [0, 1], we have
f 00 (u) =
6(16 − 9u3 ) + 216u(1 − u)
≥ 0,
(3u2 + 4)3
hence f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV
Corollary for m = 3. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0
such that x + 3 y = 4. We have
g(u) =
h(x, y) =
f (u) − f (1)
3u
= 2
,
u−1
3u + 4
g(x) − g( y)
3(4 − 3x y)
3(x − 2)2
=
=
≥ 0.
x−y
(3x 2 + 4)(3 y 2 + 4) (3x 2 + 4)(3 y 2 + 4)
From x + 3 y = 4 and h(x, y) = 0, we get x = 2 and y = 2/3. Therefore, in accordance
with Remark 8, the equality holds only for a = b = c = d = e = f = 1, and also for
a = 2, b = c = 1, d = e = f = 2/3.
P 2.12. If a, b, c, d, e, f are nonnegative real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d ≥ e ≥ f,
a + b + c + d + e + f = 6,
then
f2−1
b2 − 1
c2 − 1
d2 − 1
e2 − 1
a2 − 1
+
+
+
+
+
≥ 0.
(2a + 7)2 (2b + 7)2 (2c + 7)2 (2d + 7)2 (2e + 7)2 (2 f + 7)2
(Vasile Cirtoaje, 2009)
124
Vasile Cîrtoaje
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) + f (e) + f ( f ) ≥ 6 f (s),
where
f (u) =
u2 − 1
,
(2u + 7)2
For u ∈ [0, 1], we have
f 00 (u) =
s=
a+b+c+d+e+ f
= 1,
6
u ∈ [0, 6].
2(37 − 28u)
≥ 0,
(2u + 7)4
hence f is convex on [0, 1]. Therefore, we may apply HCF-OV Theorem or LHCF-OV
Corollary for m = 4. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0
such that x + 4 y = 5. We have
g(u) =
h(x, y) =
f (u) − f (1)
u+1
=
,
u−1
(2u + 7)2
g(x) − g( y) 21 − 4x − 4 y − 4x y
(x − 4)2
=
=
≥ 0.
x−y
(2x + 7)2 (2 y + 7)2
(2x + 7)2 (2 y + 7)2
From x + 4 y = 5 and h(x, y) = 0, we get x = 4 and y = 1/4. Therefore, in accordance
with Remark 8, the equality holds only for a = b = c = d = e = f = 1, and also for
a = 4, b = 1, c = d = e = f = 1/4.
P 2.13. If a, b, c, d, e, f are nonnegative real numbers such that
a ≥ b ≥ c ≥ d ≥ 1 ≥ e ≥ f,
a + b + c + d + e + f = 6,
then
f2−1
a2 − 1
b2 − 1
c2 − 1
d2 − 1
e2 − 1
+
+
+
+
+
≤ 0.
(2a + 5)2 (2b + 5)2 (2c + 5)2 (2d + 5)2 (2e + 5)2 (2 f + 5)2
(Vasile Cirtoaje, 2009)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) + f (e) + f ( f ) ≥ 6 f (s),
where
f (u) =
1 − u2
,
(2u + 5)2
s=
u ∈ [0, 6].
a+b+c+d+e+ f
= 1,
6
HCF Method for Ordered Variables
125
For u ≥ 1, we have
f 00 (u) =
2(20u − 13)
≥ 0,
(2u + 5)4
hence f is convex on [1, 6]. Therefore, we may apply HCF-OV Theorem or RHCF-OV
Corollary for m = 4. By Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ≥ 0
such that x + 4 y = 5. We have
g(u) =
h(x, y) =
f (u) − f (1)
−u − 1
=
,
u−1
(2u + 5)2
g(x) − g( y)
4x y + 4x + 4 y − 5
x(8 − x)
=
=
≥ 0.
2
2
x−y
(2x + 5) (2 y + 5)
(2x + 5)2 (2 y + 5)2
From x + 4 y = 5 and h(x, y) = 0, we get x = 0 and y = 5/4. Therefore, in accordance
with Remark 7, the equality holds only for a = b = c = d = e = f = 1, and also for
a = b = c = d = 5/4, e = 1, f = 0.
P 2.14. If a, b, c, d are nonnegative real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
then
a + b + c + d = 4,
1
1
1
4
1
+ 3
+ 3
+ 3
≤ .
+ 5 2b + 5 2c + 5 2d + 5 7
(Vasile Cirtoaje, 2009)
2a3
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
where
f (u) =
−1
,
2u3 + 5
From
f 00 (u) =
s=
a+b+c+d
= 1,
4
u ∈ [0, 4].
12u(5 − 4u3 )
,
(2u3 + 5)3
it follows that f 00 (u) ≥ 0 for u ∈ [0, 1], hence f is convex on [0, 1]. Therefore, we may
apply HCF-OV Theorem or LHCF-OV Corollary for n = 4 and m = 2. Thus, it suffices to
show that f (x) + 2 f ( y) ≥ 3 f (1) for all x, y ≥ 0 such that x + 2 y = 3. Using Remark 1,
we only need to show that h(x, y) ≥ 0. We have
g(u) =
f (u) − f (1) 2(u2 + u + 1)
=
,
u−1
7(2u3 + 5)
126
Vasile Cîrtoaje
h(x, y) =
g(x) − g( y)
2E
=
,
3
x−y
7(2x + 5)(2 y 3 + 5)
where
E = −2x 2 y 2 − 2x y(x + y) − 2(x 2 + x y + y 2 ) + 5(x + y) + 5.
Since
E = (1 − 2 y)2 (2 + 3 y − 2 y 2 ) = (1 − 2 y)2 (2 + x y) ≥ 0,
the proof is completed. From x + 2 y = 3 and h(x, y) = 0, we get x = 2, y = 1/2.
Therefore, in accordance with Remark 8, the equality holds for a = b = c = d = 1, and
also for a = 2, b = 1, c = d = 1/2.
Remark. Similarly, we can prove the following generalization:
• If a1 , a2 , . . . , a2n are nonnegative real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
a1 + a2 + · · · + a2n = 2n.
then
1
a13 + n +
1
n
+
1
a23 + n +
1
n
+ ··· +
1
3
a2n
+n+
1
n
≥
2n2
,
n2 + n + 1
with equality for a1 = a2 = · · · = a2n = 1, and also for
a1 = n,
a2 = · · · = an = 1,
an+1 = · · · = a2n = 1/n.
P 2.15. If a1 , a2 , . . . , a8 are nonnegative real numbers such that
a1 ≥ a2 ≥ a3 ≥ a4 ≥ 1 ≥ a5 ≥ a6 ≥ a7 ≥ a8 ,
a1 + a2 + · · · + a8 = 8,
then
(a12 + 1)(a22 + 1) · · · (a82 + 1) ≥ (a1 + 1)(a2 + 1) · · · (a8 + 1).
(Vasile Cirtoaje, 2008)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (a8 ) ≥ 8 f (s),
s=
a1 + a2 + · · · + a8
= 1,
8
where
f (u) = ln(u2 + 1) − ln(u + 1),
u ∈ [0, 8].
HCF Method for Ordered Variables
127
For u ∈ [0, 1], we have
f 00 (u) =
1
(u2 − u4 ) + 4u(1 − u2 ) + u2 + 3
2(1 − u2 )
+
=
> 0.
(u2 + 1)2 (u + 1)2
(u2 + 1)2 (u + 1)2
Therefore, f is convex on [0, 1]. According to HCF-OV Theorem or LHCF-OV Corollary
applied for n = 8 and m = 4, it suffices to show that f (x) + 4 f ( y) ≥ 5 f (1) for all
x, y ≥ 0 such that x + 4 y = 5. Using Remark 2, we only need to show that H(x, y) ≥ 0
for x, y ≥ 0 such that x + 4 y = 5, where
H(x, y) =
f 0 (x) − f 0 ( y)
2(1 − x y)
1
= 2
+
.
2
x−y
(x + 1)( y + 1) (x + 1)( y + 1)
Since 2(x 2 + 1) ≥ (x + 1)2 and 2( y 2 + 1) ≥ ( y + 1)2 , it suffices to prove that 8(1 − x y) +
(x + 1)( y + 1) ≥ 0. Indeed,
8(1 − x y) + (x + 1)( y + 1) = 28x 2 − 38x + 14 = 28(x − 19/28)2 + 31/28 > 0.
The proof is completed. The equality holds for a1 = a2 = · · · = a8 .
P 2.16. If a, b, c, d are positive real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
then
a bcd = 1,
1 1 1 1
.
a + b + c + d − 4 ≥ 18 a + b + c + d − − − −
a b c d
2
2
2
2
(Vasile Cirtoaje, 2008)
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
d = ew,
we need to show that
f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s),
where
x ≥ y ≥ 0 ≥ z ≥ w,
s=
x + y +z+w
= 0,
4
f (u) = e2u − 1 − 18(eu − e−u ), u ∈ R.
For u ≤ 0, we have
f 00 (u) = 4e2u + 18(e−u − eu ) > 0,
128
Vasile Cîrtoaje
hence f is convex on (−∞, 0]. By HCF-OV Theorem or LHCF-OV Corollary applied for
n = 4 and m = 2, it suffices to show that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y such
that x + 2 y = 0; that is, to show that
1 2
2
2
a + 2b − 3 − 18 a + 2b − −
≥0
a b
for all a, b > 0 such that a b2 = 1. This inequality is equivalent to
(b2 − 1)2 (2b2 + 1) 18(b − 1)3 (b + 1)
+
≥ 0,
b4
b2
(b − 1)2 (2b − 1)2 (b + 1)(5b + 1)
≥ 0.
b4
The proof is completed. The equality holds for a = b = c = d = 1, and also for a = 4,
b = 1, c = d = 1/2.
P 2.17. If a, b, c, d are positive real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
a bcd = 1,
then
p
a2 − a + 1 +
p
b2 − b + 1 +
p
c2 − c + 1 +
p
d 2 − d + 1 ≥ a + b + c + d.
(Vasile Cirtoaje, 2008)
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
d = ew,
we need to show that
f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s),
where
x ≥ y ≥ 0 ≥ z ≥ w,
f (u) =
p
s=
x + y +z+w
= 0,
4
e2u − eu + 1 − eu , u ∈ R.
We claim that f is convex for u ≥ 0. Since
e−u f 00 (u) =
4e3u − 6e2u + 9eu − 2
− 1,
4(e2u − eu + 1)3/2
HCF Method for Ordered Variables
129
we need to show that
(4t 3 − 6t 2 + 9t − 2)2 ≥ 16(t 2 − t + 1)3 ,
where t = eu ≥ 1. Indeed,
(4t 3 − 6t 2 + 9t − 2)2 − 16(t 2 − t + 1)3 = 12t 3 (t − 1) + 9t 2 + 12(t − 1) > 0.
By HCF-OV Theorem or RHCF-OV Corollary applied for n = 4 and m = 2, it suffices to
show that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y such that x + 2 y = 0; that is, to show
that
p
p
a2 − a + 1 + 2 b2 − b + 1 ≥ a + 2b
for all a, b > 0 such that a b2 = 1. This inequality is equivalent to
p
p
b4 − b2 + 1
1
+
2
b2 − b + 1 ≥ 2 + 2b,
2
b
b
p
b2 − 1
b4 − b2 + 1 + 1
Since
p
it suffices to show that
+p
b2 − 1
b4 − b2 + 1
2(1 − b)
b2 − b + 1 + b
≥
≥ 0.
b2 − 1
,
b2 + 1
b2 − 1
2(1 − b)
+p
≥ 0,
2
b +1
b2 − b + 1 + b
which is equivalent to
b+1
2
≥ 0,
−
p
b2 + 1
b2 − b + 1 + b
p
(b − 1) (b + 1) b2 − b + 1 − b2 + b − 2 ≥ 0,
(b − 1)
(b − 1)2 (3b2 − 2b + 3)
≥ 0.
p
(b + 1) b2 − b + 1 + b2 − b + 2
Clearly, the last inequality is true. The equality holds for a = b = c = d = 1.
P 2.18. If a, b, c, d are positive real numbers such that
a ≥ 1 ≥ b ≥ c ≥ d,
then
a3
a bcd = 1,
1
1
1
1
2
+ 3
+ 3
+ 3
≥ .
+ 3a + 2 b + 3b + 2 c + 3c + 2 d + 3d + 2 3
(Vasile Cirtoaje, 2007)
130
Vasile Cîrtoaje
Solution. Using the substitution
a = ex ,
b = ey,
c = ez ,
d = ew,
we need to show that
f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s),
where
x ≥ y ≥ 0 ≥ z ≥ w,
f (u) =
s=
x + y +z+w
= 0,
4
1
, u ∈ R.
e3u + 3eu + 2
We claim that f is convex for u ≥ 0. Indeed, denoting t = eu , t ≥ 1, we have
3t(3t 5 + 2t 3 − 6t 2 + 3t − 2)
(t 3 + 3t + 2)3
3t(t − 1)(3t 4 + 3t 3 + 5t 2 − t + 2)
=
≥ 0.
(t 3 + 3t + 2)3
f 00 (u) =
By HCF-OV Theorem or RHCF-OV Corollary applied for n = 4 and m = 1, it suffices to
show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0; that is, to show that
1
1
1
+
≥
a3 + 3a + 2 b3 + 3b + 2 3
for all a, b > 0 such that a b = 1. This inequality is equivalent to
(a − 1)4 (a2 + a + 1) ≥ 0,
which is clearly true. The equality holds for a = b = c = d = 1.
P 2.19. Let a1 , a2 , . . . , an be positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
If k ≥ 1, then
a1 a2 · · · an = 1.
1
1
1
n
+
+ ··· +
≥
.
1 + ka1 1 + ka2
1 + kan
1+k
(Vasile Cirtoaje, 2007)
HCF Method for Ordered Variables
131
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
x1 ≥ 0 ≥ x2 ≥ · · · ≥ x n,
s=
x1 + x2 + · · · + x n
= 0,
n
f (u) =
1
, u ∈ R.
1 + keu
f 00 (u) =
keu (keu − 1)
≥ 0,
(1 + keu )3
For u ≥ 0, we have
therefore f (u) is convex for u ≥ 0. By HCF-OV Theorem or RHCF-OV Corollary applied
for m = 1, it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that
x + y = 0; that is, to show that
1
1
2
+
≥
1 + ka 1 + k b
1+k
for all a, b > 0 such that a b = 1. This is true since
1
1
2
k(k − 1)(a − 1)2
+
−
=
≥ 0.
1 + ka 1 + k b 1 + k
(1 + ka)(a + k)
The equality holds for a1 = a2 = · · · = an = 1.
P 2.20. If a1 , a2 , . . . , a9 are positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ a9 ,
then
a1 a2 · · · a9 = 1,
1
1
1
+
+ ··· +
≥ 1.
(a1 + 2)2 (a2 + 2)2
(a9 + 2)2
(Vasile Cirtoaje, 2007)
Solution. Using the substitution
ai = e x i ,
i = 1, 2, . . . , 9,
we can write the inequality as
f (x 1 ) + f (x 2 ) + · · · + f (x 9 ) ≥ 9 f (s),
132
Vasile Cîrtoaje
where
x1 ≥ 0 ≥ x2 ≥ · · · ≥ x9,
f (u) =
(eu
For u ∈ [0, ∞), we have
f 00 (u) =
s=
x1 + x2 + · · · + x9
9
1
,
+ 2)2
= 0,
u ∈ R.
4eu (eu − 1)
≥ 0,
(eu + 2)4
hence f is convex on [0, ∞). According to HCF-OV Theorem or RHCF-OV Corollary
(case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that
x + y = 0; that is, to show that
1
1
2
+
≥
2
2
(a + 2)
(b + 2)
9
for all a, b > 0 such that a b = 1. Write this inequality as
b2
1
2
+
≥ ,
2
2
(2b + 1)
(b + 2)
9
which is equivalent to the obvious inequality
(b − 1)4 ≥ 0.
The equality holds for a1 = a2 = · · · = a9 = 1.
Remark. Similarly, we can prove the following generalization.
• Let a1 , a2 , · · · , an be positive real numbers such that
a1 a2 · · · an = 1.
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
If p, q ≥ 0 such that
p+q ≥1+
then
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
with equality for a1 = a2 = · · · = an = 1.
2pq
,
p + 4q
+ ··· +
1
n
≥
.
2
1 + pan + qan
1+p+q
HCF Method for Ordered Variables
133
P 2.21. If a1 , a2 , . . . , an (n ≥ 4) are positive real numbers such that
a1 a2 · · · an = 1,
a1 ≥ a2 ≥ a3 ≥ 1 ≥ a4 ≥ · · · ≥ an ,
then
1
1
1
n
+
+ ··· +
≥ .
2
2
2
(a1 + 1)
(a2 + 1)
(an + 1)
4
(Vasile Cirtoaje, 2007)
Solution. Using the substitution
ai = e x i ,
i = 1, 2, . . . , n,
we can write the inequality as
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
x1 ≥ x2 ≥ x3 ≥ 0 ≥ x4 ≥ · · · ≥ x n,
f (u) =
(eu
For u ∈ [0, ∞), we have
f 00 (u) =
1
,
+ 1)2
s=
x1 + x2 + · · · + x n
= 0,
n
u ∈ R.
2eu (2eu − 1)
> 0,
(eu + 1)4
hence f is convex on [0, ∞). According to HCF-OV Theorem or RHCF-OV Corollary
(case m = 3), it suffices to show that f (x) + 3 f ( y) ≥ 4 f (0) for all real x, y such that
x + 3 y = 0; that is, to show that
1
3
+
≥1
2
(a + 1)
(b + 1)2
for all a, b > 0 such that a b3 = 1. The inequality is equivalent to
3
b6
+
≥ 1.
3
2
(b + 1)
(b + 1)2
Using the Cauchy-Schwarz inequality, it suffices to show that
(b3 + 3)2
≥ 1,
(b3 + 1)2 + 3(b + 1)2
which is equivalent to the obvious inequality
(b − 1)2 (4b + 5) ≥ 0.
The equality holds for a1 = a2 = · · · = an = 1.
134
Vasile Cîrtoaje
P 2.22. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,
then
a1 a2 · · · an = 1,
1
1
n
1
+
+ ··· +
≤
.
2
2
2
(a1 + 3)
(a2 + 3)
(an + 3)
16
(Vasile Cirtoaje, 2007)
Solution. Using the substitution
ai = e x i ,
i = 1, 2, . . . , n,
we can write the inequality as
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
s=
x1 ≤ 0 ≤ x2 ≤ · · · ≤ x n,
f (u) =
−1
,
+ 3)2
(eu
x1 + x2 + · · · + x n
= 0,
n
u ∈ R.
For u ∈ (−∞, 0], we have
f 00 (u) =
2eu (3 − 2eu )
> 0,
(eu + 3)4
hence f is convex on (−∞, 0]. According to HCF-OV Theorem or LHCF-OV Corollary
(case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that
x + y = 0; that is, to show that
1
1
1
+
≤
2
2
(a + 3)
(b + 3)
8
for all a, b > 0 such that a b = 1. Write this inequality as
b2
1
1
+
≤ ,
2
2
(3b + 1)
(b + 3)
8
which is equivalent to the obvious inequality
(b2 − 1)2 + 12b(b − 1)2 ≥ 0.
The equality holds for a1 = a2 = · · · = an = 1.
Remark. Similarly, we can prove the following generalization:
HCF Method for Ordered Variables
135
• Let a1 , a2 , . . . , an be positive real numbers such that
a1 a2 · · · an = 1.
a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,
p
If k ≥ 1 + 2, then
1
1
1
n
+
+ ··· +
≤
,
(a1 + k)2 (a2 + k)2
(an + k)2
(1 + k)2
with equality for a1 = a2 = · · · = an = 1.
P 2.23. Let a1 , a2 , . . . , an be positive real numbers such that
a1 a2 · · · an = 1.
a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,
If p, q ≥ 0 such that p + q ≤ 1, then
1
1+
pa1 + qa12
+
1
1+
pa2 + qa22
+ ··· +
n
1
≤
.
1 + pan + qan2
1+p+q
(Vasile Cirtoaje, 2007)
Solution. Using the substitution
ai = e x i ,
i = 1, 2, . . . , n,
we can write the inequality as
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
x1 ≤ 0 ≤ x2 ≤ · · · ≤ x n,
f (u) =
s=
x1 + x2 + · · · + x n
= 0,
n
−1
,
1 + peu + qe2u
u ∈ R.
For u ≤ 0, we have
eu [−4q2 e3u − 3pqe2u + (4q − p2 )eu + p]
(1 + peu + qe2u )3
e2u [−4q2 − 3pq + (4q − p2 ) + p]
≥
(1 + peu + qe2u )3
e2u [(p + 4q)(1 − p − q) + 2pq]
=
≥ 0,
(1 + peu + qe2u )3
f 00 (u) =
136
Vasile Cîrtoaje
therefore f (u) is convex for u ≤ 0. According to HCF-OV Theorem or LHCF-OV Corollary
(case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that
x + y = 0; that is, to show that
1
1
2
+
≤
2
2
1 + pa + qa
1 + pb + qb
1+p+q
for all a, b > 0 such that a b = 1. Write this inequality as
(a − 1)2 [q(1 − p − q)a2 + (p + 2q − p2 − pq − 2q2 )a + q(1 − p − q)] ≥ 0,
which is true because
p + 2q − p2 − pq − 2q2 ≥ (p + 2q)(p + q) − p2 − pq − 2q2 = 2pq ≥ 0.
The equality holds for a1 = a2 = · · · = an = 1.
P 2.24. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≤ a2 ≤ 1 ≤ a3 ≤ · · · ≤ an ,
then
a1 a2 · · · an = 1,
1
1
1
n
+
+ ··· +
≤
.
2
2
2
(a1 + 5)
(a2 + 5)
(an + 5)
36
(Vasile Cirtoaje, 2007)
Solution. Using the substitution
ai = e x i ,
i = 1, 2, . . . , n,
we can write the inequality as
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
x1 ≤ x2 ≤ 0 ≤ x3 ≤ · · · ≤ x n,
f (u) =
s=
−1
,
(eu + 5)2
x1 + x2 + · · · + x n
= 0,
n
u ∈ R.
For u ∈ (−∞, 0], we have
f 00 (u) =
2eu (5 − 2eu )
> 0,
(eu + 5)4
HCF Method for Ordered Variables
137
hence f is convex on (−∞, 0]. According to HCF-OV Theorem or LHCF-OV Corollary
(case m = 2), it suffices to show that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y such that
x + 2 y = 0; that is, to show that
1
2
1
+
≤
2
2
(a + 5)
(b + 5)
12
for all a, b > 0 such that a b2 = 1. Since
1
b4
b4
b2
=
≤
=
,
(a + 5)2
(5b2 + 1)2
(4b2 + 2b)2
4(2b + 1)2
it suffices to show that
2
1
b2
+
≤
,
4(2b + 1)2 (b + 5)2
12
which is equivalent to the obvious inequality
(b − 1)2 (b2 + 16b + 1) ≥ 0.
The equality holds for a1 = a2 = · · · = an = 1.
Remark. Similarly, we can prove the following generalization:
• Let a1 , a2 , . . . , an be positive real numbers such that
a1 ≤ a2 ≤ 1 ≤ a3 ≤ · · · ≤ an ,
If k ≥ 2 +
a1 a2 · · · an = 1.
p
6, then
1
1
1
n
+
+ ··· +
≤
,
2
2
2
(a1 + k)
(a2 + k)
(an + k)
(1 + k)2
with equality for a1 = a2 = · · · = an = 1.
P 2.25. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
then
1
p
1 + 3a1
+p
1
1 + 3a2
a1 a2 · · · an = 1,
+ ··· + p
1
1 + 3an
≥
n
.
2
(Vasile Cirtoaje, 2007)
138
Vasile Cîrtoaje
Solution. Using the substitution
ai = e x i ,
i = 1, 2, . . . , n,
we can write the inequality as
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
x1 ≥ 0 ≥ x2 ≥ · · · ≥ x n,
s=
x1 + x2 + · · · + x n
= 0,
n
1
f (u) = p
,
1 + 3eu
u ∈ R.
For u ≥ 0, we have
eu (3eu − 2)
> 0,
2(1 + 3eu )5/2
hence f is convex on [0, ∞). According to HCF-OV Theorem or RHCF-OV Corollary
(case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that
x + y = 0; that is, to show that
f 00 (u) =
1
1
+p
≥1
p
1 + 3a
1 + 3b
for all a, b > 0 such that a b = 1. Write this inequality as
s
1
a
+
≥ 1.
p
a+3
1 + 3a
1
Substituting p
= t, 0 < t < 1, the inequality becomes
1 + 3a
v
t 1 − t2
≥ 1 − t.
8t 2 + 1
By squaring, we get
t(1 − t)(2t − 1)2 ≥ 0,
which is true. The equality holds for a1 = a2 = · · · = an = 1.
P 2.26. If a1 , a2 , . . . , an are positive real numbers such that
a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,
then
1
p
1 + 2a1
+p
1
1 + 2a2
a1 a2 · · · an = 1,
+ ··· + p
1
n
≤p .
3
1 + 2an
(Vasile Cirtoaje, 2007)
HCF Method for Ordered Variables
139
Solution. Using the substitution
ai = e x i ,
i = 1, 2, . . . , n,
we can write the inequality as
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
x1 ≤ 0 ≤ x2 ≤ · · · ≤ x n,
s=
x1 + x2 + · · · + x n
= 0,
n
−1
f (u) = p
,
1 + 2eu
For u ≤ 0, we have
f 00 (u) =
u ∈ R.
eu (1 − eu )
> 0,
(1 + 2eu )5/2
hence f is convex on (−∞, 0]. According to HCF-OV Theorem or LHCF-OV Corollary
(case m = 1), it suffices to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that
x + y = 0; that is, to show that
v
v
t 3
t 3
+
≤2
1 + 2a
1 + 2b
for all a, b > 0 such that a b = 1. By the Cauchy-Schwarz inequality, we get
v
v
v
t 3
t 3
t
3
3
+
≤
+1 1+
= 2.
1 + 2a
1 + 2b
1 + 2a
1 + 2b
The equality holds for a1 = a2 = · · · = an = 1.
140
Vasile Cîrtoaje
Chapter 3
Partially Convex Function Method
3.1
Theoretical Basis
Right Partially Convex Function Theorem (Vasile Cirtoaje, 2012). Let f be a function
defined on a real interval I and convex on [s, s0 ], where s, s0 ∈ I, s < s0 . In addition, f (u)
is decreasing on Iu≤s0 and
min f (u) = f (s0 ).
u∈I
The inequality
a + a + ··· + a 1
2
n
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
n
holds for all a1 , a2 , · · · , an ∈ I satisfying a1 + a2 + · · · + an = ns if and only if
f (x) + (n − 1) f ( y) ≥ n f (s)
for all x, y ∈ I such that x ≤ s ≤ y and x + (n − 1) y = ns.
Proof. The necessity is obvious. By Lemma below, to prove the sufficiency, it suffices to
consider that a1 , a2 , . . . , an ∈ J, where
J = Iu≤s0 .
Because f (u) is convex on Ju≥s , the desired inequality follows from HCF Theorem applied to the interval J.
Lemma. Let f be a function defined on a real interval I and convex on [s, s0 ], where
s, s0 ∈ I, s < s0 . In addition, f (u) is decreasing on Iu≤s0 and
min f (u) = f (s0 ).
u∈I
If the inequality
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s)
141
142
Vasile Cîrtoaje
holds for all a1 , a2 , · · · , an ∈ Iu≤s0 such that a1 + a2 + · · · + an = ns, then it holds for all
a1 , a2 , . . . , an ∈ I such that a1 + a2 + · · · + an = ns.
Proof. For i = 1, 2, . . . , n, define the numbers bi ∈ Iu≤s0 as follows
(
bi =
a i , a i ≤ s0
s0 ,
a i > s0 .
We have bi ≤ ai and f (bi ) ≤ f (ai ) for i = 1, 2, · · · , n. Therefore,
b1 + b2 + · · · + bn ≤ a1 + a2 + · · · + an = ns
and
f (b1 ) + f (b2 ) + · · · + f (bn ) ≤ f (a1 ) + f (a2 ) + · · · + f (an ).
Thus, it suffices to show that
f (b1 ) + f (b2 ) + · · · + f (bn ) ≥ n f (s)
for all b1 , b2 , . . . , bn ∈ Iu≤s0 such that b1 + b2 +· · ·+ bn ≤ ns. By hypothesis, this inequality
is true for b1 , b2 , . . . , bn ∈ Iu≤s0 and b1 + b2 + · · · + bn = ns. Since f (u) is decreasing on
Iu≤s0 , the more we have f (b1 ) + f (b2 ) + · · · + f (bn ) ≥ n f (s) for b1 , b2 , . . . , bn ∈ Iu≤s0
and b1 + b2 + · · · + bn ≤ ns.
Similarly, we can prove Left Partially Convex Function Theorem (Vasile Cirtoaje,
2012).
Left Partially Convex Function Theorem. Let f be a function defined on a real interval
I and convex on [s0 , s], where s0 , s ∈ I, s0 < s. In addition, f (u) is increasing on Iu≥s0 and
satisfies
min f (u) = f (s0 ).
u∈I
The inequality
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
a + a + ··· + a 1
2
n
n
holds for all a1 , a2 , · · · , an ∈ I satisfying a1 + a2 + · · · + an = ns if and only if
f (x) + (n − 1) f ( y) ≥ n f (s)
for all x, y ∈ I such that x ≥ s ≥ y and x + (n − 1) y = ns.
Remark 1. Let us denote
g(u) =
f (u) − f (s)
,
u−s
h(x, y) =
g(x) − g( y)
.
x−y
Partially Convex Function Method
143
As shown in Remark 1 from 1.1, we may replace the hypothesis condition in Right
Partially Convex Function Theorem (RPCF Theorem) and Left Partially Convex Function
Theorem (LPCF Theorem), namely
f (x) + (n − 1) f ( y) ≥ n f (s),
by the condition
h(x, y) ≥ 0 for all x, y ∈ I such that x + (n − 1) y = ns.
Remark 2. Assume that f is differentiable on I, and let
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
As shown in Remark 2 from 1.1, the inequalities in RPCF Theorem and LPCF Theorem
hold true by replacing the hypothesis
f (x) + (n − 1) f ( y) ≥ n f (s)
with the more restrictive condition
H(x, y) ≥ 0 for all x, y ∈ I such that x + (n − 1) y = ns.
Remark 3. The inequality in RPCF Theorem or LPCF Theorem becomes an equality for
a1 = a2 = · · · = an = s
and also for
a1 = x,
a2 = · · · = an = y
(or any cyclic permutation), where x, y ∈ I satisfy the equations
x + (n − 1) y = ns,
f (x) + (n − 1) f ( y) = n f (s).
For x 6= y, these equations are equivalent to
x + (n − 1) y = ns, h(x, y) = 0.
Remark 4. RPCF Theorem is also valid in the case when I = [a, b] \ {u0 } or I = (a, b) \
{u0 }, where a, b, u0 are real numbers such that
a < s < s0 < u0 < b.
Similarly, LPCF Theorem is also valid in the case when I = [a, b]\{u0 } or I = (a, b)\{u0 },
where a, b, u0 are real numbers such that
a < u0 < s0 < s < b.
144
Vasile Cîrtoaje
Remark 5. The desired inequality in RPCF Theorem holds true by replacing the condition
f is decreasing on Iu≤s0
with the sufficient condition
ns − (n − 1)s0 ≤ inf I.
To prove this claim, we define the function
(
f (u), u ≤ s0 , u ∈ I
f0 (u) =
f (s0 ), u ≥ s0 , u ∈ I,
which is convex for u ∈ I, u ≥ s. Taking into account that f0 (s) = f (s) and f0 (u) ≤ f (u)
for all u ∈ I, it suffices to prove that
f0 (x 1 ) + f0 (x 2 ) + · · · + f0 (x n ) ≥ n f0 (s)
for all x 1 , x 2 , · · · , x n ∈ I satisfying x 1 + x 2 + · · · + x n = ns. According to HCF Theorem
and Remark 5 from Section 1, we only need to show that
f0 (x) + (n − 1) f0 ( y) ≥ n f0 (s)
(*)
for all x, y ∈ I such that x ≤ s ≤ y and x +(n−1) y = ns. The case y > s0 is not possible
because
x = ns − (n − 1) y < ns − (n − 1)s0 ≤ inf I
involves x 6∈ I. For the possible case y ≤ s0 , (*) becomes f (x) + (n − 1) f ( y) ≥ n f (s),
which holds (by hypothesis) for all x, y ∈ I such that x + (n − 1) y = ns.
Similarly, we can show that the desired inequality in LPCF Theorem holds true by
replacing the condition
f is increasing on Iu≥s0
with the sufficient condition
ns − (n − 1)s0 ≥ sup I.
Partially Convex Function Method
3.2
145
Applications
3.1. If a, b, c are real numbers such that a + b + c = 3, then
16a − 5
16b − 5
16c − 5
+
+
≤ 1.
32a2 + 1 32b2 + 1 32c 2 + 1
3.2. If a, b, c, d are real numbers such that a + b + c + d = 4, then
18b − 5
18c − 5
18d − 5
18a − 5
+
+
+
≤ 4.
12a2 + 1 12b2 + 1 12c 2 + 1 12d 2 + 1
3.3. If a, b, c, d, e, f are real numbers such that a + b + c + d + e + f = 6, then
5f − 1
5a − 1
5b − 1
5c − 1
5d − 1
5e − 1
+
+
+
+
+
≤ 4.
5a2 + 1 5b2 + 1 5c 2 + 1 5d 2 + 1 5e2 + 1 5 f 2 + 1
3.4. If a1 , a2 , · · · , an (n ≥ 3) are real numbers such that a1 + a2 + · · · + an = n, then
n(n + 1) − 2a1
n2 + (n − 2)a12
+
n(n + 1) − 2a2
n2 + (n − 2)a22
+ ··· +
n(n + 1) − 2an
≤ n.
n2 + (n − 2)an2
3.5. If a, b, c, d are real numbers such that a + b + c + d = 4, then
a(a − 1) b(b − 1) c(c − 1) d(d − 1)
+
+ 2
+
≥ 0.
3a2 + 4
3b2 + 4
3c + 4
3d 2 + 4
3.6. Let a1 , a2 , · · · , an 6= −k be real numbers such that a1 + a2 + · · · + an = n, where
n
k≥ p
.
2 n−1
Then,
an (an − 1)
a1 (a1 − 1) a2 (a2 − 1)
+
+ ··· +
≥ 0.
2
2
(a1 + k)
(a2 + k)
(an + k)2
146
Vasile Cîrtoaje
3.7. Let a1 , a2 , . . . , an 6= −k be real numbers such that a1 + a2 + · · · + an = n. If
k ≥1+ p
then
a12 − 1
(a1 + k)2
+
a22 − 1
(a2 + k)2
n
n−1
+ ··· +
,
an2 − 1
(an + k)2
≥ 0.
3.8. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If
k > 1, then
a1
a1k
+ a2 + · · · + an
+
a2
a1 + a2k
+ · · · + an
+ ··· +
an
≤ 1.
a1 + a2 + · · · + ank
3.9. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If
k > 1, then
1
a1k
+ a2 + · · · + an
+
1
a1 + a2k
+ · · · + an
+ ··· +
1
≤ 1.
a1 + a2 + · · · + ank
3.10. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n,
then
an − 1
a1 − 1
a2 − 1
+
+ ··· +
≤ 0.
2
2
a1 + a2 + · · · + an2
a1 + a2 + · · · + an a1 + a2 + · · · + an
3.11. Let a1 , a2 , · · · , an be positive real numbers such that a1 + a2 + · · · + an = n. If
n
, then
0<k≤
n−1
k/a
k/a
a1 1 + a2 2 + · · · + ank/an ≤ n.
3.12. If a, b, c, d, e are nonzero real numbers such that a + b + c + d + e = 5, then
5 2
5 2
5 2
5 2
5 2
7−
+ 7−
+ 7−
+ 7−
+ 7−
≥ 20.
a
b
c
d
e
3.13. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
(1 − a + a4 )(1 − b + b4 )(1 − c + c 4 ) ≥ 1.
Partially Convex Function Method
147
3.14. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then
(1 − a + a3 )(1 − b + b3 )(1 − c + c 3 )(1 − d + d 3 ) ≥ 1.
3.15. If a1 , a2 , . . . , a10 are real numbers such that a1 + a2 + · · · + a10 = 10, then
2
(1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a10 + a10
) ≥ 1.
3.16. If a, b, c, d, e are nonzero real numbers such that a + b + c + d + e = 5, then
1
1
1
1
1
1 1 1 1 1
5 2 + 2 + 2 + 2 + 2 + 45 ≥ 14
+ + + +
.
a
b
c
d
e
a b c d e
3.17. If a, b, c are positive real numbers such that a bc = 1, then
7 − 6a 7 − 6b 7 − 6c
+
+
≥ 1.
2 + a2 2 + b2 2 + c 2
3.18. If a, b, c are positive real numbers such that a bc = 1, then
1
1
1
1
+
+
≤ .
a + 5bc b + 5ca c + 5a b
2
3.19. If a, b, c are positive real numbers such that a bc = 1, then
1
1
1
3
+
+
≤ .
2
2
2
4 − 3a + 4a
4 − 3b + 4b
4 − 3c + 4c
5
3.20. If a, b, c are positive real numbers such that a bc = 1, then
1
1
1
3
+
+
≤ .
2
2
2
(3a + 1)(3a − 5a + 3) (3b + 1)(3b − 5b + 3) (3c + 1)(3c − 5c + 3) 4
3.21. Let a1 , a2 , · · · , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If
p, q ≥ 0 such that p + 4q ≥ n − 1, then
1 − a1
1+
pa1 + qa12
+
1 − a2
1+
pa2 + qa22
+ ··· +
1 − an
≥ 0.
1 + pan + qan2
148
Vasile Cîrtoaje
3.22. If a, b, c are positive real numbers such that a bc = 1, then
1−a
1− b
1−c
+
+
≥ 0.
2
2
17 + 4a + 6a
17 + 4b + 6b
17 + 4c + 6c 2
3.23. If a1 , a2 , · · · , a8 are positive real numbers such that a1 a2 · · · a8 = 1, then
1 − a8
1 − a1
1 − a2
+
+ ··· +
≥ 0.
2
2
(1 + a1 )
(1 + a2 )
(1 + a8 )2
3.24. Let a, b, c be positive real numbers such that a bc = 1. If k ∈
a+k
b+k
c+k
3(1 + k)
+
+
≤
.
a2 + 1 b2 + 1 c 2 + 1
2
p
3.25. If a, b, c are positive real numbers and 0 < k ≤ 2 + 2 2, then
a3
b3
c3
a+b+c
+
+
≥
.
ka2 + bc k b2 + ca kc 2 + a b
k+1
3.26. If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that
a1 , a2 , . . . , an ≥
then
a12
a12 − a1 + 1
+
−1
,
n−2
a22
a22 − a2 + 1
a1 + a2 + · · · + an = n,
+ ··· +
an2
an2 − an + 1
≤ n.
3.27. If a1 , a2 , . . . , an (n ≥ 3) are nonzero real numbers such that
a1 , a2 , . . . , an ≥
then
1
a12
+
1
a22
−n
,
n−2
+ ··· +
a1 + a2 + · · · + an = n,
1
1
1
1
≥
+
+ ··· + .
2
an
a1 a2
an
−13 13
p , p , then
3 3 3 3
Partially Convex Function Method
149
3.28. If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that
a1 , a2 , . . . , an ≥
then
−(3n − 2)
,
n−2
a1 + a2 + · · · + an = n,
1 − an
1 − a1
1 − a2
+
+ ··· +
≥ 0.
2
2
(1 + a1 )
(1 + a2 )
(1 + an )2
3.29. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
2
n ≥ 3 and k ≥ 2 − , then
n
1 − an
1 − a1
1 − a2
+
+
·
·
·
+
≥ 0.
(1 − ka1 )2 (1 − ka2 )2
(1 − kan )2
150
Vasile Cîrtoaje
Partially Convex Function Method
3.3
151
Solutions
P 3.1. If a, b, c are real numbers such that a + b + c = 3, then
16b − 5
16c − 5
16a − 5
+
+
≤ 1.
32a2 + 1 32b2 + 1 32c 2 + 1
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a) + f (b) + f (c) ≥ 3 f (s),
where
f (u) =
5 − 16u
,
32u2 + 1
s=
a+b+c
= 1,
3
u ∈ R.
From
16(32u2 − 20u − 1)
,
(32u2 + 1)2
p
p 5 − 33
5 − 33
it follows that f is increasing on −∞,
∪[s0 , ∞) and decreasing on
, s0 ,
16
16
p
5 + 33
≈ 0.6715. Since
where s0 =
16
f 0 (u) =
lim f (u) = 0
u→−∞
and f (s0 ) < f (5/16) = 0, it follows that
min f (u) = f (s0 ).
u∈R
For u ∈ [5/16, 1], we have
1 00
−512u3 + 480u2 + 48u − 5
f (u) =
64
(32u2 + 1)3
2
512u (1 − u) + 32u(1 − u) + (16u − 5)
=
> 0.
(32u2 + 1)3
Therefore, f is convex on [1/2, 1], hence on [s0 , 1]. According to LPCF Theorem, we
only need to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3.
Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
152
Vasile Cîrtoaje
Indeed, we have
g(u) =
h(x, y) =
32(2u − 1)
,
3(32u2 + 1)
64(1 + 16x + 16 y − 32x y)
64(4x − 5)2
=
≥ 0.
3(32x 2 + 1)(32 y 2 + 1)
3(32x 2 + 1)(32 y 2 + 1)
Thus, the proof is completed. From x + 2 y = 3 and h(x, y) = 0, we get x = 5/4 and,
y = 7/8. Therefore, in accordance with Remark 3, the equality holds for a = b = c = 1,
and also for a = 5/4 and b = c = 7/8 (or any cyclic permutation).
P 3.2. If a, b, c, d are real numbers such that a + b + c + d = 4, then
18b − 5
18c − 5
18d − 5
18a − 5
+
+
+
≤ 4.
12a2 + 1 12b2 + 1 12c 2 + 1 12d 2 + 1
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
where
5 − 18u
,
12u2 + 1
f (u) =
s=
a+b+c+d
= 1,
4
u ∈ R.
From
6(36u2 − 20u − 3)
,
(12u2 + 1)2
p
p 5 − 52
5 − 52
∪[s0 , ∞) and decreasing on
, s0 ,
it follows that f is increasing on −∞,
18
18
p
5 + 52
≈ 0.678. Since
where s0 =
18
f 0 (u) =
lim f (u) = 0
u→−∞
and f (s0 ) < f (5/18) = 0, it follows that
min f (u) = f (s0 ).
u∈R
For u ∈ [5/18, 1], we have
1 00
−216u3 + 180u2 + 54u − 5
f (u) =
24
(12u2 + 1)3
216u2 (1 − u) + 36u(1 − u) + (18u − 5)
=
> 0.
(32u2 + 1)3
Partially Convex Function Method
153
Therefore, f is convex on [1/2, 1], hence on [s0 , 1]. According to LPCF Theorem and
Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ∈ R such that x + 3 y = 4. We
have
f (u) − f (1) 6(2u − 1)
g(u) =
=
,
u−1
12u2 + 1
h(x, y) =
g(x) − g( y) 12(1 + 6x + 6 y − 12x y)
12(2x − 3)2
=
=
≥ 0.
x−y
(12x 2 + 1)(12 y 2 + 1)
(12x 2 + 1)(12 y 2 + 1)
Thus, the proof is completed. From x + 3 y = 4 and h(x, y) = 0, we get x = 3/2 and,
y = 5/6. Therefore, in accordance with Remark 3, the equality holds for a = b = c =
d = 1, and also for a = 3/2 and b = c = d = 5/6 (or any cyclic permutation).
P 3.3. If a, b, c, d, e, f are real numbers such that a + b + c + d + e + f = 6, then
5f − 1
5b − 1
5c − 1
5d − 1
5e − 1
5a − 1
+ 2
+ 2
+ 2
+ 2
+
≤ 4.
2
5a + 1 5b + 1 5c + 1 5d + 1 5e + 1 5 f 2 + 1
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) + f (e) + f ( f ) ≥ 4 f (s),
where
f (u) =
1 − 5u
,
5u2 + 1
s=
a+b+c+d+e+ f
= 1,
6
u ∈ R.
From
5(5u2 − 2u − 1)
,
(5u2 + 1)2
p p
1− 6
1− 6
, s0 ,
it follows that f is increasing on −∞,
∪[s0 , ∞) and decreasing on
5
5
p
1+ 6
where s0 =
≈ 0.69. Since
5
f 0 (u) =
lim f (u) = 0
u→−∞
and f (s0 ) < f (1/5) = 0, it follows that
min f (u) = f (s0 ).
u∈R
154
Vasile Cîrtoaje
For u ∈ [1/5, 1], we have
−25u3 + 15u2 + 15u − 1
1 00
f (u) =
10
(5u2 + 1)3
2
25u (1 − u) + 10u(1 − u) + (5u − 1)
=
> 0.
(3u2 + 4)3
Therefore, f is convex on [1/5, 1], hence on [s0 , 1]. According to LPCF Theorem and
Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ∈ R such that x + 5 y = 6. We
have
f (u) − f (1)
5(2u − 1)
g(u) =
=
,
u−1
3(5u2 + 1)
h(x, y) =
g(x) − g( y) 5(2 + 5x + 5 y − 10x y)
10(x − 2)2
=
=
≥ 0.
x−y
3(5x 2 + 1)(5 y 2 + 1)
3(5x 2 + 1)(5 y 2 + 1)
Thus, the proof is completed. In accord with Remark 3, the equality holds for a = b =
c = d = e = f = 1, and also for a = 2 and b = c = d = e = f = 4/5 (or any cyclic
permutation).
P 3.4. If a1 , a2 , · · · , an (n ≥ 3) are real numbers such that a1 + a2 + · · · + an = n, then
n(n + 1) − 2a1
n2
+ (n − 2)a12
+
n(n + 1) − 2a2
n2
+ (n − 2)a22
+ ··· +
n(n + 1) − 2an
≤ n.
n2 + (n − 2)an2
(Vasile Cîrtoaje, 2008)
Solution. The desired inequality is true for a1 >
n(n + 1) − 2a1
n2 + (n − 2)a12
and
n(n + 1) − 2ai
n2
+ (n − 2)ai2
<
n(n + 1)
since
2
<0
n
, i = 2, 3, · · · , n.
n−1
The last inequality is equivalent to
n(n − 2)ai2 + 2(n − 1)ai + n > 0,
and is true because
n(n − 2)ai2 + 2(n − 1)ai + n ≥ (n − 1)ai2 + 2(n − 1)ai + n > (n − 1)(ai + 1)2 ≥ 0.
Partially Convex Function Method
155
Consider further that a1 , a2 , · · · , an ≤
n(n + 1)
and rewrite the desired inequality as
2
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
2u − n(n + 1)
,
f (u) =
(n − 2)u2 + n2
s=
a1 + a2 + · · · + an
= 1,
n
n(n + 1)
u ∈ I = −∞,
.
2
We have
f 0 (u)
n2 + n(n + 1)u − u2
=
2(n − 2)
[(n − 2)u2 + n2 ]2
and
f1 (u)
f 00 (u)
=
,
2(n − 2) [(n − 2)u2 + n2 ]3
where
f1 (u) = 2(n − 2)u3 − 3n(n + 1)(n − 2)u2 − 2n2 (2n − 3)u + n3 (n + 1).
From
of f 0 , it follows that f is decreasing on (−∞, s0 ] and increasing
the expression
n(n + 1)
, where
on s0 ,
2
s0 =
p
n
(n + 1 − n2 + 2n + 5 ) ∈ (−1, 0);
2
therefore,
min f (u) = f (s0 ).
u∈I
On the other hand, for −1 ≤ u ≤ 1, we have
f1 (u) > −2(n − 2) − 3n(n + 1)(n − 2) − 2n2 (2n − 3) + n3 (n + 1)
= n2 (n − 3)2 + 4(n + 1) > 0,
and hence f 00 (u) > 0. Since [s0 , 1] ⊂ [−1, 1], f is convex on [s0 , 1]. By LPCF Theorem
and Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ∈ R and x + (n − 1) y = n,
where
g(x) − g( y)
f (u) − f (1)
h(x, y) =
, g(u) =
.
x−y
u−1
Indeed, we have
g(u) =
(n − 2)u + n
(n − 2)u2 + n2
156
Vasile Cîrtoaje
and
n2 − n(x + y) − (n − 2)x y
h(x, y)
=
n−2
[(n − 2)x 2 + n2 ][(n − 2) y 2 + n2 ]
(n − 1)(n − 2) y 2
=
≥ 0.
[(n − 2)x 2 + n2 ][(n − 2) y 2 + n2 ]
The proof is completed. In accord with Remark 3, the equality holds for a1 = a2 = · · · =
an = 1, and also for a1 = n and a2 = · · · = an = 0 (or any cyclic permutation).
P 3.5. If a, b, c, d are real numbers such that a + b + c + d = 4, then
a(a − 1) b(b − 1) c(c − 1) d(d − 1)
+
+ 2
+
≥ 0.
3a2 + 4
3b2 + 4
3c + 4
3d 2 + 4
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
where
f (u) =
u2 − u
,
3u2 + 4
s=
a+b+c+d
= 1,
4
u ∈ R.
From
3u2 + 8u − 4
,
(3u2 + 4)2
p
p −4 − 2 7
−4 − 2 7
it follows that f is increasing on −∞,
∪[s0 , ∞) and decreasing on
, s0 ,
3
3
p
−4 + 2 7
≈ 0.43. Since
where s0 =
3
f 0 (u) =
lim f (u) = 1/3
u→−∞
and f (s0 ) < f (0) = 0, it follows that
min f (u) = f (s0 ).
u∈R
For u ∈ [0, 1], we have
1 00
−9u3 − 36u2 + 36u + 14
f (u) =
2
(3u2 + 4)3
9u2 (1 − u) + 45u(1 − u) + (16 − 9u)
=
> 0.
(3u2 + 4)3
Partially Convex Function Method
157
Therefore, f is convex on [0, 1], hence on [s0 , 1]. According to LPCF Theorem and
Remark 1, we only need to show that h(x, y) ≥ 0 for x, y ∈ R such that x + 3 y = 4. We
have
f (u) − f (1)
u)
g(u) =
= 2
,
u−1
3u + 4
h(x, y) =
g(x) − g( y)
4 − 3x y
(x − 2)2
=
=
≥ 0.
x−y
(3x 2 + 4)(3 y 2 + 4) (12x 2 + 1)(12 y 2 + 1)
The proof is completed. From x + 3 y = 4 and h(x, y) = 0, we get x = 2 and, y = 2/3.
In accord with Remark 3, the equality holds for a = b = c = d = 1, and also for a = 2
and b = c = d = 2/3 (or any cyclic permutation).
Remark. Similarly, we can prove the following generalization.
• If a1 , a2 , . . . , an are real numbers such that a1 + a2 + · · · + an = n, then
a1 (a1 − 1)
4(n − 1)a12
+ n2
+
a2 (a2 − 1)
4(n − 1)a22
+ n2
+ ··· +
an (an − 1)
≥ 0,
4(n − 1)an2 + n2
n
with equality for a1 = a2 = · · · = an = 1, and also for a1 = and a2 = a3 = · · · = an =
2
n
(or any cyclic permutation).
2(n − 1)
P 3.6. Let a1 , a2 , · · · , an 6= −k be real numbers such that a1 + a2 + · · · + an = n, where
n
k≥ p
.
2 n−1
Then,
an (an − 1)
a1 (a1 − 1) a2 (a2 − 1)
+
+ ··· +
≥ 0.
2
2
(a1 + k)
(a2 + k)
(an + k)2
(Vasile Cîrtoaje, 2008)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
s=
a1 + a2 + · · · + an
= 1,
n
u(u − 1)
, u ∈ I = R \ {−k}.
(u + k)2
From
f 0 (u) =
(2k + 1)u − k
,
(u + k)3
158
Vasile Cîrtoaje
it follows that f is increasing on (−∞, −k) ∪ [s0 , ∞) and decreasing on (−k, s0 ], where
s0 =
k
< 1.
2k + 1
Since
lim f (u) = 1
u→−∞
and f (s0 ) < f (1) = 0, we have
min f (u) = f (s0 ).
u∈I
From
1 00
k(k + 2) − (2k + 1)u
f (u) =
,
2
(u + k)4
k(k + 2)
, hence on [s0 , 1]. According to LPCF Theoit follows that f is convex on 0,
2k + 1
rem, Remark 4 and Remark 1, it suffices to show that h(x, y) ≥ 0 for all x, y ∈ I which
satisfy x + (n − 1) y = n, where
h(x, y) =
g(x) − g( y)
,
x−y
Indeed, we have
g(u) =
and
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
u
(u + k)2
k2 − x y
≥ 0,
(x + k)2 ( y + k)2
because
[2(n − 1) y − n]2
n2
−xy =
≥ 0.
4(n − 1)
4(n − 1)
n
The equality holds for a1 = a2 = · · · = an = 1. If k = p
, then the equality holds
2 n−1
n
n
(or any cyclic permutation).
also for a1 = and a2 = · · · = an =
2
2(n − 1)
k2 − x y ≥
P 3.7. Let a1 , a2 , . . . , an 6= −k be real numbers such that a1 + a2 + · · · + an = n. If
k ≥1+ p
then
a12 − 1
(a1 + k)2
+
a22 − 1
(a2 + k)2
n
n−1
+ ··· +
,
an2 − 1
(an + k)2
≥ 0.
(Vasile Cîrtoaje, 2008)
Partially Convex Function Method
159
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
s=
a1 + a2 + · · · + an
= 1,
n
u2 − 1
, u ∈ I = R \ {−k}.
(u + k)2
From
f 0 (u) =
2(ku + 1)
,
(u + k)3
it follows that f is increasing on (−∞, −k) ∪ [s0 , ∞) and decreasing on (−k, s0 ], where
s0 =
−1
∈ (−1, 0).
k
Since
lim f (u) = 1
u→−∞
and f (s0 ) < f (−1) = 0, we have
min f (u) = f (s0 ).
u∈I
From
f 00 (u) =
2(k2 − 3 − 2ku)
,
(u + k)4
it follows that f is convex on [−1, 1], hence on [s0 , 1], since
k2 − 3 − 2ku ≥ k2 − 3 − 2k = (k + 1)(k − 3) ≥ 0.
According to LPCF Theorem, Remark 4 and Remark 1, it suffices to show that h(x, y) ≥ 0
for x, y ∈ I which satisfy x + (n − 1) y = n. We have
g(u) =
h(x, y) =
f (u) − f (1)
u+1
=
,
u−1
(u + k)2
g(x) − g( y) (k − 1)2 − 1 − x − y − x y
=
≥ 0,
x−y
(x + k)2 ( y + k)2
since
[(n − 1) y − 1]2
n2
−1− x − y − xy =
≥ 0.
n−1
n−1
n
The equality holds for a1 = a2 = · · · = an = 1. If k = 1 + p
, then the equality
n−1
1
holds also for a1 = n − 1 and a2 = · · · = an =
(or any cyclic permutation).
n−1
(k − 1)2 − 1 − x − y − x y ≥
160
Vasile Cîrtoaje
P 3.8. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If
k > 1, then
a1
a1k
+ a2 + · · · + an
+
a2
a1 + a2k
+ · · · + an
+ ··· +
an
≤ 1.
a1 + a2 + · · · + ank
(Vasile Cîrtoaje, 2006)
Solution. According to the proof of P 2.106 from Volume 2, it suffices to consider the
case where a1 + a2 + · · · + an = n and k > n + 1. Write the desired inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
uk
−u
,
−u+n
We have
s=
a1 + a2 + · · · + an
= 1,
n
u ∈ I = [0, n].
f 0 (u) =
(k − 1)uk − n
(uk − u + n)2
f 00 (u) =
f1 (u)
,
(uk − u + n)3
and
where
f1 (u) = k(k − 1)uk−1 (uk − u + n) − 2(kuk−1 − 1)[(k − 1)uk − n].
From the expression of f 0 , it follows that f is decreasing on [0, s0 ] and increasing on
[s0 , n], where
n 1/k
s0 =
< 1.
k−1
On the other hand, for u ∈ [s0 , 1], we have
(k − 1)uk − n ≥ (k − 1)s0k − n = 0,
and hence
f1 (u) ≥ k(k − 1)uk−1 (uk − u + n) − 2kuk−1 [(k − 1)uk − n]
= kuk−1 [−(k − 1)(uk + u) + n(k + 1)]
≥ kuk−1 [−2(k − 1) + 2(k + 1)] = 4kuk−1 > 0.
Thus, f 00 (u) > 0, and hence f is convex on [s0 , 1]. By LPCF Theorem, we need to show
that f (x) + (n − 1) f ( y) ≥ n f (1) for all positive x, y which satisfy x ≥ 1 ≥ y ≥ 0 and
x + (n − 1) y = n. Consider the nontrivial case where x > 1 > y ≥ 0 and write the
inequality f (x) + (n − 1) f ( y) ≥ n f (1) as follows:
xk
(n − 1) y
x
+ k
≤ 1,
−x −n y − y+n
Partially Convex Function Method
161
xk − x + n ≥
xk − x ≥
x( y k − y + n)
,
yk − ny + n
(n − 1) y( y − y k )
.
yk − ny + n
Since y < 1, it suffices to show that
xk − x ≥
(n − 1)( y − y k )
,
yk − y + 1
which is equivalent to
h(x) ≥
y − yk
,
(1 − y)( y k − y + 1)
where
xk − x
.
x −1
By the weighted AM-GM inequality, we have
h(x) =
h0 (x) =
(k − 1)x k + 1 − k x k−1
> 0,
(x − 1)2
and hence h is strictly increasing. Since x − 1 = (n − 1)(1 − y) ≥ 1 − y, we get
h(x) ≥ h(2 − y) =
(2 − y)k − 2 + y
.
1− y
Thus, it suffices to show that
(2 − y)k − 2 + y ≥
y − yk
.
yk − y + 1
Putting 1 − y = t, 0 < t ≤ 1, we write this inequality as
(2 − y)k − 1 + y ≥
(1 + t)k − t ≥
1
,
yk − y + 1
1
,
(1 − t)k + t
(1 − t 2 )k + t(1 + t)k ≥ 1 + t 2 + t(1 − t)k .
By Bernoulli’s inequality,
(1 − t 2 )k + t(1 + t)k > 1 − kt 2 + t(1 + kt) = 1 + t.
So, we only need to show that t(1 − t) ≥ t(1 − t)k , which is clearly true. The proof is
completed. The equality holds for a1 = a2 = · · · = an = 1.
162
Vasile Cîrtoaje
P 3.9. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an ≥ n. If
k > 1, then
1
a1k
+ a2 + · · · + an
+
1
a1 + a2k
+ · · · + an
+ ··· +
1
≤ 1.
a1 + a2 + · · · + ank
(Vasile Cîrtoaje, 2006)
Solution. Clearly, it suffices to consider the case a1 + a2 +· · ·+ an = n, when the desired
inequality can be written as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
uk
−1
,
−u+n
From
f 0 (u) =
s=
a1 + a2 + · · · + an
= 1,
n
u ∈ I = [0, n].
kuk−1 − 1
,
(uk − u + n)2
it follows that f is decreasing on [0, s0 ] and increasing on [s0 , n], where
1
s0 = k 1−k < 1.
We will show that f is convex on [s0 , 1]. For u ∈ [s0 , 1], we have
f 00 (u) =
g(u)
−k(k + 1)u2k−2 + k(k + 3)uk−1 + nk(k − 1)uk−2 − 2
> k
,
k
3
(u − u + n)
(u − u + n)3
where
g(u) = −k(k + 1)u2k−2 + k(k + 3)uk−1 − 2.
Denoting
t = kuk−1 ,
we have 1 ≤ t ≤ k and
k g(u) = −(k + 1)t 2 + k(k + 3)t − 2k = (k + 1)(t − 1)(k − t) + (k − 1)(t + k) > 0.
By LPCF Theorem, it suffices to show that
xk
1
n−1
+ k
≤1
−x +n y − y+n
for x ≥ 1 ≥ y ≥ 0 and x + (n − 1) y = n. Since this inequality is trivial for x = y = 1,
assume next that x > 1 > y ≥ 0, and write the desired inequality as follows:
xk − x + n ≥
yk − y + n
,
yk − y + 1
Partially Convex Function Method
163
xk − x ≥
(n − 1)( y − y k )
,
yk − y + 1
y − yk
xk − x
≥
.
x −1
(1 − y)( y k − y + 1)
Let h(x) =
xk − x
, x > 1. By the weighted AM-GM inequality, we have
x −1
h0 (x) =
(k − 1)x k + 1 − k x k−1
> 0.
(x − 1)2
Therefore, h is increasing. Since x − 1 = (n − 1)(1 − y) ≥ 1 − y, hence x ≥ 2 − y > 1,
we get
(2 − y)k + y − 2
.
h(x) ≥ h(2 − y) =
1− y
Thus, it suffices to show that
(2 − y)k + y − 2 ≥
y − yk
,
yk − y + 1
which is equivalent to
(2 − y)k + y − 1 ≥
yk
1
.
− y +1
Substituting t = 1 − y, 0 < t ≤ 1, the inequality becomes
(1 + t)k − t ≥
1
,
(1 − t)k + t
(1 − t 2 )k + t(1 + t)k ≥ 1 + t 2 + t(1 − t)k .
By Bernoulli’s inequality,
(1 − t 2 )k + t(1 + t)k ≥ 1 − kt 2 + t(1 + kt) = 1 + t.
Thus, it suffices to show that t(1 − t) ≥ t(1 − t)k , which is clearly true. The proof is
completed. The equality holds for a1 = a2 = · · · = an = 1.
Remark. Using this result, we can get the following statement.
• Let x 1 , x 2 , . . . , x n be nonnegative real numbers such that x 1 + x 2 + · · · + x n ≥ n. If
k > 1, then
x 1k − x 1
x 1k + x 2 + · · · + x n
+
x 2k − x 2
x 1 + x 2k + · · · + x n
+ ··· +
x nk − x n
x 1 + x 2 + · · · + x nk
≥ 0.
164
Vasile Cîrtoaje
This inequality is equivalent to
1
x 1k
+ x2 + · · · + x n
+
1
x1 +
x 2k
+ · · · + xn
Using the substitutions
s=
+ ··· +
1
n
≤
.
k
x1 + x2 + · · · + x n
x1 + x2 + · · · + x n
x1 + x2 + · · · + x n
,
n
s ≥ 1,
and
xi
, i = 1, 2, . . . , n,
s
which yields a1 + a2 + · · · + an = n, the desired inequality becomes
ai =
1
X
s k−1 a1k
+ a2 + · · · + a n
≤ 1.
Since s k−1 ≥ 1, it suffices to show that
1
X
a1k
+ a2 + · · · + an
≤ 1,
which is just the inequality from P 3.9.
Since x 1 x 2 · · · x n ≥ 1 involves x 1 + x 2 + · · · + x n ≥ n, the inequality is also true
under the more restrictive condition x 1 x 2 · · · x n ≥ 1. For n = 3 and k = 5/2, we get the
inequality from IMO-2005:
• If x, y, z are nonnegative real numbers such that x yz ≥ 1, then
y5 − y2
x5 − x2
z5 − z2
+
+
≥ 0.
x 5 + y 2 + z2 x 2 + y 5 + z2 x 2 + y 2 + z5
P 3.10. If a1 , a2 , . . . , an are nonnegative real numbers such that a1 + a2 + · · · + an = n,
then
an − 1
a1 − 1
a2 − 1
+
+ ··· +
≤ 0.
2
2
a1 + a2 + · · · + an2
a1 + a2 + · · · + an a1 + a2 + · · · + an
(Vasile Cîrtoaje, 2006)
Solution. For n = 2, the inequality is equivalent to (a1 − a2 )2 ≥ 0. Consider further
that n ≥ 3 and write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
Partially Convex Function Method
where
f (u) =
u2
165
1−u
,
−u+n
u ∈ I = [0, n].
From
f 0 (u) =
u2 − 2u − n + 1
,
(u2 − u + n)2
it follows that f is decreasing on [0, s0 ] and increasing on [s0 , n], where
s0 = 1 +
p
n, s0 ∈ (1, n).
We will show that f is convex on [1, s0 ]. For u ∈ [1, s0 ], we have
f 00 (u) =
(u2
2g(u)
,
− u + n)3
g(u) = −u3 + 3u2 + 3(n − 1)u − 2n + 1.
Since
g 0 (u) = −3(u2 − 2u − n + 1) ≥ 0,
g is increasing on [1, s0 ], hence g(u) ≥ g(1) = n > 0 and f 00 (u) > 0. By RPCF Theorem,
it suffices to show that
(n − 1)(1 − y)
1− x
+
≥0
2
x −x +n
y2 − y + n
for 0 ≤ x ≤ 1 ≤ y and x + (n − 1) y = n. Since (n − 1)(1 − y) = x − 1, we have
(n − 1)(1 − y)
1− x
1
1
+
= (x − 1) − 2
+
x2 − x + n
y2 − y + n
x − x + n y2 − y + n
(x − 1)(x − y)(x + y − 1)
=
(x 2 − x + n)( y 2 − y + n)
n(x − 1)2 (x + y − 1)
=
≥ 0.
(n − 1)(x 2 − x + n)( y 2 − y + n)
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1.
P 3.11. Let a1 , a2 , · · · , an be positive real numbers such that a1 + a2 + · · · + an = n. If
n
0<k≤
, then
n−1
k/a
k/a
a1 1 + a2 2 + · · · + ank/an ≤ n.
(Vasile Cîrtoaje, 2006)
166
Vasile Cîrtoaje
Solution. According to the power mean inequality, we have
p/a1
a1
p/a2
+ a2
p/an
+ · · · + an
!1/p
q/a1
a1
≥
n
q/a2
+ a2
q/an
+ · · · + an
!1/q
n
for all p ≥ q > 0. Thus, it suffices to prove the desired inequality for
k=
n
,
n−1
1 < k ≤ 2.
Rewrite the desired inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = −uk/u ,
We have
u ∈ I = (0, n).
k
f 0 (u) = ku u −2 (ln u − 1),
k
f 00 (u) = ku u −4 [u + (1 − ln u)(2u − k + k ln u)].
For n = 2, when k = 2 and I = (0, 2), f is convex for u ∈ I, u ≥ 1, because
1 − ln u > 0,
2u − k + k ln u = 2u − 2 + 2 ln u ≥ 2u − 2 ≥ 0.
Therefore, we may apply HCF Theorem. Consider now that n ≥ 3. From the expression
of f 0 , it follows that f is decreasing on (0, s0 ] and increasing on [s0 , n), where s0 = e.
In addition, we claim that f is convex on [1, s0 ]. Indeed, since
1 − ln u ≥ 0,
2u − k + k ln u ≥ 2 − k > 0,
we have f 00 > 0 for u ∈ [1, s0 ]. Therefore, by HCF Theorem (for n = 2) and RPCF
Theorem (for n ≥ 3), we only need to show that
x k/x + (n − 1) y k/ y ≤ n
for 0 < x ≤ 1 ≤ y and x + (n − 1) y = n. We have k/x ≥ k > 1. Also, from
k
n
n
=
>
= 1,
y
(n − 1) y
x + (n − 1) y
we get
0<
k
− 1 ≤ 1.
y
k
2
≤ ≤ 2,
y
y
Partially Convex Function Method
167
Therefore, by Bernoulli’s inequality, we have
x k/x + (n − 1) y k/ y − n =
1
+ (n − 1) y · y k/ y−1 − n
1 k/x
x
≤
=
1
1+
k
x
1
x
k
− 1 ( y − 1) − n
+ (n − 1) y 1 +
y
−1
−(x − 1)2 [(k − 1)x + k(2 − k)]
x2
2
−
(k
−
1)x
−
(2
−
k)x
=
≤ 0.
x2 − kx + k
x2 − kx + k
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1
P 3.12. If a, b, c, d, e are nonzero real numbers such that a + b + c + d + e = 5, then
5 2
5 2
5 2
5 2
5 2
7−
+ 7−
+ 7−
+ 7−
+ 7−
≥ 20.
a
b
c
d
e
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s),
where
5 2
f (u) = 7 −
,
u
From
f 0 (u) =
s=
a+b+c+d+e
= 1,
5
u ∈ I = R \ {0}.
10(7u − 5)
,
u3
it follows that f is increasing on (−∞, 0) ∪ [s0 , ∞) and decreasing on (0, s0 ], where
s0 = 5/7. Since
lim f (u) = 49
u→−∞
and f (s0 ) = 0, we have
min f (u) = f (s0 ).
u∈I
Also, f is convex on [s0 , s] = [5/7, 1] because
f 00 (u) =
10(15 − 14u)
> 0.
u4
168
Vasile Cîrtoaje
According to LPCF Theorem and Remark 4, we only need to show that that f (x) +
4 f ( y) ≥ 5 f (1) for all nonzero real x, y such that x + 4 y = 5. Using Remark 1, it
suffices to prove that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
From
g(u) = 5
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
9
5
− 2 ,
u u
5(5x + 5 y − 9x y)
≥0
x2 y2
and
5x + 5 y − 9x y = (6 y − 5)2 ≥ 0,
it follows that h(x, y) ≥ 0. The proof is completed. In accordance with Remark 3, the
equality holds for a = b = c = d = e = 1, and also for a = 5/3 and b = c = d = e = 5/6
(or any cyclic permutation).
Remark. Similarly, we can prove the following generalization.
• Let a1 , a2 , . . . , an be nonzero real numbers such that a1 + a2 + · · · + an = n. If
n
, then
k=
p
n+ n−1
k 2
k 2
k 2
+ 1−
+ ··· + 1 −
≥ n(1 − k)2 ,
1−
a1
a2
an
with equality for a1 = a2 = · · · = an = 1, and also for a1 =
· · · = an =
n
(or any cyclic permutation).
p
n−1+ n−1
n
and a2 = a3 =
p
1+ n−1
P 3.13. If a, b, c are nonnegative real numbers such that a + b + c = 3, then
(1 − a + a4 )(1 − b + b4 )(1 − c + c 4 ) ≥ 1.
Solution. Write the inequality as
f (a) + f (b) + f (c) ≥ 3 f (s),
s=
a+b+c
= 1,
3
where
f (u) = ln(1 − u + u4 ),
u ∈ [0, 3].
Partially Convex Function Method
169
From
f 0 (u) =
4u3 − 1
,
1 − u + u4
p
3
it follows that f is decreasing on [0, s0 1] and increasing on [s0 , 4], where s0 = 1/ 4.
Also, f is convex for u ∈ [s0 , 1] because
f 00 (u) =
−4u6 − 4u3 + 12u2 − 1 −4u2 − 4u2 + 12u2 − 1
4u2 − 1
≥
=
> 0.
(1 − u + u4 )2
(1 − u + u4 )2
(1 − u + u4 )2
According to LPCF Theorem, we only need to show that f (x) + 2 f ( y) ≥ 3 f (1) for all
x, y ≥ 0 such that x + 2 y = 3. Using Remark 2, it suffices to prove that H(x, y) ≥ 0,
where
f 0 (x) − f 0 ( y)
.
H(x, y) =
x−y
We have
(x + y)(x − y)2 − 1 + 4(x 2 + y 2 + x y) − 2x y(x + y) − 4x 3 y 3
(1 − x + x 4 )(1 − y + y 4 )
−1 + 4(x 2 + y 2 + x y) − 2x y(x + y) − 4x 3 y 3
≥
(1 − x + x 4 )(1 − y + y 4 )
−1 + 4(x + y)2 − 2x y(x + y) − 4x y − 4x 3 y 3
=
.
(1 − x + x 4 )(1 − y + y 4 )
H(x, y) =
Thus, we only need to show that
−1 + 4(x + y)2 − 2x y(x + y) − 4x y − 4x 3 y 3 ≥ 0.
p
From 3 = x + 2 y ≥ 2 2x y and (1 − x)(1 − y) ≤ 0, we get
xy ≤
9
,
8
x + y ≥ 1 + x y.
Therefore,
− 1 + 4(x + y)2 − 2x y(x + y) − 4x y − 4x 3 y 3 ≥
≥ − 1 + 2(x + y)[2(x + y) − x y] − 4x y − 4x 3 y 3
≥ − 1 + 2(1 + x y)[2(1 + x y) − x y] − 4x y − 4x 3 y 3
=3 + 2x y + 2x 2 y 2 − 4x 3 y 3 ≥ 3 + 2x y + 2x 2 y 2 − 5x 2 y 2
=3 + 2x y − 3x 2 y 2 ≥ 3 + 2x y − 4x y = 3 − 2x y > 0.
The proof is completed. In accordance with Remark 3, the equality holds for a = b =
c = 1.
170
Vasile Cîrtoaje
P 3.14. If a, b, c, d are nonnegative real numbers such that a + b + c + d = 4, then
(1 − a + a3 )(1 − b + b3 )(1 − c + c 3 )(1 − d + d 3 ) ≥ 1.
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
s=
a+b+c+d
= 1,
4
where
f (u) = ln(1 − u + u3 ),
From
f 0 (u) =
u ∈ [0, 4].
3u2 − 1
,
1 − u + u3
p
it follows that f is decreasing on [0, s0 ] and increasing on [s0 , 4], where s0 = 1/ 34.
Also, f is convex for u ∈ [s0 , 1] because
f 00 (u) =
−3u4 + 6u − 1 −3u + 6u − 1
3u − 1
≥
=
> 0.
3
2
3
2
(1 − u + u )
(1 − u + u )
(1 − u + u3 )2
According to LPCF Theorem, we only need to show that f (x) + 3 f ( y) ≥ 4 f (1) for all
x, y ≥ 0 such that x + 3 y = 4. Using Remark 2, it suffices to prove that H(x, y) ≥ 0,
where
f 0 (x) − f 0 ( y)
.
H(x, y) =
x−y
We have
H(x, y) =
(x − y)2 + 3(x + y) − 1 − 3x 2 y 2
3(x + y) − 1 − 3x 2 y 2
≥
.
(1 − x + x 3 )(1 − y + y 3 )
(1 − x + x 3 )(1 − y + y 3 )
Thus, we only need to show that
3(x + y) − 1 − 3x 2 y 2 ≥ 0.
From 4 = x + 3 y ≥ 2
p
3x y and (1 − x)(1 − y) ≤ 0, we get
xy ≤
4
,
3
x + y ≥ 1 + x y.
Therefore,
3(x + y) − 1 − 3x 2 y 2 ≥ 3(1 + x y) − 1 − 3x 2 y 2 ≥ 3(1 + x y) − 1 − 4x y = 2 − x y > 0.
The proof is completed. The equality holds for a = b = c = d = 1.
Partially Convex Function Method
171
P 3.15. If a1 , a2 , . . . , a10 are real numbers such that a1 + a2 + · · · + a10 = 10, then
2
(1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a10 + a10
) ≥ 1.
(Vasile Cîrtoaje, 2006)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (a10 ) ≥ 10 f (s),
s=
a1 + a2 + · · · + a10
=1
10
where
f (u) = ln(1 − u + u2 ),
From
f 0 (u) =
u ∈ R.
2u − 1
,
1 − u + u2
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = 1/2.
Also, from
1 + 2u(1 − u)
f 00 (u) =
,
(1 − u + u2 )2
it follows that f 00 (u) > 0 for u ∈ [s0 , 1], hence f is convex on [s0 , 1]. According to LPCF
Theorem, we only need to show that f (x) + 9 f ( y) ≥ 10 f (1) for all real x, y such that
x + 9 y = 10. Using Remark 2, it suffices to prove that H(x, y) ≥ 0, where
H(x, y) =
Since
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
1 + x + y − 2x y
(1 − x + x 2 )(1 − y + y 2 )
and
1 + x + y − 2x y = 18 y 2 − 8 y + 1 = 2 y 2 + (4 y − 1)2 > 0,
the conclusion follows. The equality holds for a1 = a2 = · · · = a10 = 1.
Remark. By replacing a1 , a2 , . . . , a10 respectively with 1 − a1 , 1 − a2 , . . . , 1 − a10 , we get
the following statement.
• If a1 , a2 , . . . , a10 are real numbers such that a1 + a2 + · · · + a10 = 0, then
2
(1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a10 + a10
) ≥ 1,
with equality for a1 = a2 = · · · = an = 0.
172
Vasile Cîrtoaje
P 3.16. If a, b, c, d, e are nonzero real numbers such that a + b + c + d + e = 5, then
1
1
1
1
1 1 1 1 1
1
+ + + +
.
5 2 + 2 + 2 + 2 + 2 + 45 ≥ 14
a
b
c
d
e
a b c d e
(Vasile Cîrtoaje, 2013)
Solution. Write the desired inequality as
f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s), s =
where
f (u) =
a+b+c+d+e
= 1,
5
5
14
−
+ 9, u ∈ I = R \ {0}.
2
u
u
From
2(7u − 5)
,
u3
it follows that f is increasing on (−∞, 0) ∪ [s0 , ∞) and decreasing on (0, s0 ], where
f 0 (u) =
s0 =
5
.
7
Since
lim f (u) = 9
u→−∞
and f (s0 ) < f (1) = 0, we have
min f (u) = f (s0 ).
u∈I
From
2(15 − 14u)
,
u4
it follows that f is convex on [s0 , 1]. By LPCF Theorem, Remark 4 and Remark 1, it
suffices to show that h(x, y) ≥ 0 for all x, y ∈ I which satisfy x + 4 y = 5, where
f 00 (u) =
h(x, y) =
g(x) − g( y)
,
x−y
Indeed, we have
g(u) =
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
9
5
− 2,
u u
(6 y − 5)2
5x + 5 y − 9x y
=
≥ 0.
x2 y2
x2 y2
In accordance with Remark 3, the equality holds for a = b = c = d = e = 1, and also
5
5
for a = and b = c = d = e = (or any cyclic permutation).
3
6
Partially Convex Function Method
173
P 3.17. If a, b, c are positive real numbers such that a bc = 1, then
7 − 6a 7 − 6b 7 − 6c
+
+
≥ 1.
2 + a2 2 + b2 2 + c 2
(Vasile Cîrtoaje, 2008)
Solution. Using the substitution
a = e x , b = e y , c = ez ,
we need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
where
f (u) =
From
f 0 (u) =
s=
x + y +z
= 0,
3
7 − 6eu
, u ∈ R.
2 + e2u
2(3eu + 2)(eu − 3)
,
(2 + e2u )2
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln 3.
Also, we have
2t · h(t)
f 00 (u) =
, t = eu ,
(2 + t 2 )3
where
h(t) = −3t 4 + 14t 3 + 36t 2 − 28t − 12.
We will show that h(t) > 0 for t ∈ [1, 3], hence f is convex on [0, s0 ]. We have
h(t) = 3(t 2 − 1)(9 − t 2 ) + 14t 3 + 6t 2 − 28t + 15
= 3(t 2 − 1)(9 − t 2 ) + 14t 2 (t − 1) + 14(t − 1)2 + 6t 2 + 1 > 0.
By RPCF Theorem, we only need to prove that f (x) + 2 f ( y) ≥ 3 f (0) for all real x, y
such that x + 2 y = 0. That is, to show that the original inequality holds for b = c := t
and a = 1/t 2 , where t > 0. Write this inequality as
t 2 (7t 2 − 6) 2(7 − 6t)
+
≥ 1,
2t 4 + 1
2 + t2
(t − 1)2 (t − 2)2 (5t 2 + 6t + 3) ≥ 0.
Since the last inequality is true, the proof is completed. In accordance with Remark 3,
the equality holds for a = b = c = 1, and also for a = 1/4 and b = c = 2 (or any cyclic
permutation).
174
Vasile Cîrtoaje
P 3.18. If a, b, c are positive real numbers such that a bc = 1, then
1
1
1
1
+
+
≤ .
a + 5bc b + 5ca c + 5a b
2
(Vasile Cîrtoaje, 2008)
Solution. Write the inequality as
a2
a
b
c
1
+ 2
+ 2
≤ .
+5 b +5 c +5 2
Using the substitution
a = e x , b = e y , c = ez ,
we need to show that
f (x) + g( y) + g(z) ≥ 3 f (s),
where
f (u) =
s=
x + y +z
= 0,
3
−eu
, u ∈ R.
e2u + 5
From
f 0 (u) =
eu (e2u − 5)
,
(e2u + 5)2
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 =
Also, from
f 00 (u) =
eu (−e4u + 30e2u − 25)
,
(e2u + 5)3
it follows that f is convex on [0, s0 ], because 0 ≤ u ≤ s0 involves 1 ≤ eu ≤
1
ln 5.
2
p
5, hence
−e4u + 30e2u − 25 ≥ e2u (5 − e2u ) + 25(e2u − 1) > 0.
By RPCF Theorem, we only need to prove the original inequality for b = c := t and
a = 1/t 2 , where t > 0. Write this inequality as
t2
2t
1
+
≤ ,
5t 4 + 1 t 2 + 5 2
(t − 1)2 (5t 4 − 10t 3 − 2t 2 + 6t + 5) ≥ 0,
(t − 1)2 [5(t − 1)4 + 2t(5t 2 − 16t + 13)] ≥ 0.
Since the last inequality is true, the proof is completed. The equality holds for a = b =
c = 1.
Partially Convex Function Method
175
P 3.19. If a, b, c are positive real numbers such that a bc = 1, then
1
1
3
1
+
+
≤ .
4 − 3a + 4a2 4 − 3b + 4b2 4 − 3c + 4c 2
5
(Vasile Cirtoaje, 2008)
Solution. Let
a = ex ,
b = ey,
c = ez .
We need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
where
f (u) =
s=
x + y +z
= 0,
3
−1
, u ∈ R.
4 − 3eu + 4e2u
From
f 0 (u) =
eu (8eu − 3)
,
(4 − 3eu + 4e2u )2
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where
s0 = ln
3
< 0.
8
We claim that f is convex on [s0 , 0]. Since
f 00 (u) =
eu (−64e3u + 36e2u + 55eu − 12)
,
(4 − 3eu + 4e2u )3
we need to show that
−64x 3 + 36x 2 + 55x − 12 ≥ 0,
where x = eu ∈ [3/8, 1]. Indeed,
−64x 3 + 36x 2 + 55x − 12 > −72x 3 + 36x 2 + 48x − 12 = 12(1 − x)(6x 2 + 3x − 1) ≥ 0.
By LPCF Theorem, we only need to prove the original inequality for b = c := t and
a = 1/t 2 , where t > 0. Write this inequality as follows:
t4
2
3
+
≤ ,
4
2
2
4t − 3t + 4 4 − 3t + 4t
5
28t 6 − 21t 5 − 48t 4 + 27t 3 + 42t 2 − 36t + 8 ≥ 0,
(t − 1)2 (28t 4 + 35t 3 − 6t 2 − 20t + 8) ≥ 0.
It suffices to show that
7(4t 4 + 5t 3 − t 2 − 3t + 1) ≥ 0.
176
Vasile Cîrtoaje
Indeed,
4t 4 + 5t 3 − t 2 − 3t + 1 = t 2 (2t − 1)2 + 9t 3 − 2t 2 − 3t + 1
and
9t 3 − 2t 2 − 3t + 1 = t(3t − 1)2 + (2t − 1)2 > 0.
The equality holds for a = b = c = 1.
Remark. Since
1
1
1
≥
=
,
4 − 3a + 4a2
4 − 3a + 4a2 + (1 − a)2
5(1 − a + a2 )
we get the following known inequality
1
1
1
+
+
≤ 3.
2
2
1−a+a
1− b+ b
1 − c + c2
P 3.20. If a, b, c are positive real numbers such that a bc = 1, then
1
1
1
3
+
+
≤ .
(3a + 1)(3a2 − 5a + 3) (3b + 1)(3b2 − 5b + 3) (3c + 1)(3c 2 − 5c + 3) 4
Solution. Let
a = ex ,
b = ey,
c = ez .
We need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
where
f (u) =
(3eu
From
f 0 (u) =
s=
x + y +z
= 0,
3
−1
, u ∈ R.
+ 1)(3e2u − 5eu + 3)
(3eu − 2)(9eu − 2)
,
(3eu + 1)2 (3e2u − 5eu + 3)2
it follows that f is increasing on (−∞, s1 ] ∪ [s0 , ∞) and decreasing on [s1 , s0 ], where
s1 = ln 2 − ln 9, s0 = ln 2 − ln 3,
s1 < s0 < 0.
Since
lim f (u) = f (s0 ) = −1/3,
u→−∞
Partially Convex Function Method
177
we get
min f (u) = f (s0 ).
u∈R
We claim that f is convex on [s0 , 0]. We have
f 00 (u) =
(3t
t · h(t)
,
− 5t + 3)3
+ 1)3 (3t 2
where
t = eu , h(t) = −729t 5 + 1188t 4 − 648t 3 + 387t 2 − 160t + 12.
Since the polynomial h(t) has the real roots
t 1 ≈ 0.0933, t 2 ≈ 0.5072, t 3 ≈ 1.11008,
it follows that h(t) > 0 for t ∈ [2/3, 1], hence f is convex for u ∈ [s0 , 0]. By LPCF
Theorem, we only need to prove the original inequality for b = c := t and a = 1/t 2 ,
where t ∈ (0, 1]. Write this inequality as follows:
2
3
t6
+
≤ .
2
4
2
2
(t + 3)(3t − 5t + 3) (3t + 1)(3t − 5t + 3) 4
Since
t 2 + 3 ≥ 2(t + 1)
and
3t 4 − 5t 2 + 3 ≥ t(3t 2 − 5t + 3),
it suffices to prove that
t5
2
3
+
≤ .
2(t + 1)(3t 2 − 5t + 3) (3t + 1)(3t 2 − 5t + 3) 4
This is equivalent to the obvious inequality
(1 − t)2 (1 + 15t + 5t 2 − 14t 3 − 6t 4 ) ≥ 0.
The equality holds for a = b = c = 1.
P 3.21. Let a1 , a2 , · · · , an (n ≥ 3) be positive real numbers such that a1 a2 · · · an = 1. If
p, q ≥ 0 such that p + 4q ≥ n − 1, then
1 − a1
1+
pa1 + qa12
+
1 − a2
1+
pa2 + qa22
+ ··· +
1 − an
≥ 0.
1 + pan + qan2
(Vasile Cîrtoaje, 2008)
178
Vasile Cîrtoaje
Solution. For q = 0, we get a known inequality (see Remark 2 from the proof of P 1.51).
Consider further that q > 0. Using the substitutions ai = e x i for i = 1, 2, . . . , n, we need
to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
s=
x1 + x2 + · · · + x n
= 0,
n
f (u) =
1 − eu
, u ∈ R.
1 + peu + qe2u
f 0 (t) =
eu (qe2u − 2qeu − p − 1)
,
(1 + peu + qe2u )2
From
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where
v
t
p+1
.
s0 = ln r0 > 0, r0 = 1 + 1 +
q
Also, we have
f 00 (u) =
t · h(t)
,
(1 + pt + qt 2 )3
where
h(t) = −q2 t 4 + q(p + 4q)t 3 + 3q(p + 2)t 2 + (p − 4q + p2 )t − p − 1,
t = eu .
We will show that h(t) ≥ 0 for t ∈ [1, r0 ], hence f is convex on [0, s0 ]. We have
h0 (t) = −4q2 t 3 + 3q(p + 4q)t 2 + 6q(p + 2)t + p − 4q + p2 ,
h00 (t) = 6q[−2qt 2 + (p + 4q)t + p + 2].
Since
h00 (t) = 6q[2(−qt 2 + 2qt + p + 1) + p(t − 1)] ≥ 12q(−qt 2 + 2qt + p + 1) ≥ 0,
h0 (t) is increasing,
h0 (t) ≥ h0 (1) = p2 + 9pq + 8q2 + p + 8q > 0,
h is increasing, hence
h(t) ≥ h(1) = p2 + 4pq + 3q2 + 2q − 1 = (p + 2q)2 − (q − 1)2
= (p + q + 1)(p + 3q − 1).
Therefore, f is convex on [0, s0 ] if p + 3q − 1 ≥ 0. Indeed,
p + 3q − 1 ≥ p + 3q −
p + 2q
p + 4q
=
> 0.
n−1
2
Partially Convex Function Method
179
By RPCF Theorem, we only need to prove the original inequality for a2 = · · · = an := t
and a1 = 1/t n−1 , where t ≥ 1. Write this inequality as
(n − 1)(1 − t)
t n−1 (t n−1 − 1)
+
≥ 0,
t 2n−2 + pt n−1 + q
1 + pt + qt 2
or
pA + qB ≥ C,
where
A = t n−1 (t n − nt + n − 1),
B = t 2n − t n+1 − (n − 1)(t − 1),
C = t n−1 [(n − 1)t n − nt n−1 + 1].
Since p+4q ≥ n−1 and C ≥ 0 (by the AM-GM inequality applied to n positive numbers),
it suffices to show that
(p + 4q)C
pA + qB ≥
,
n−1
which is equivalent to
p[(n − 1)A − C] + q[(n − 1)B − 4C] ≥ 0.
Clearly, this is true if (n − 1)A − C ≥ 0 and (n − 1)B − 4C ≥ 0 for t ≥ 1. By the AM-GM
inequality, we have
(n − 1)A − C = nt n−1 [t n−1 − (n − 1)t + n − 2] ≥ 0.
For n = 3 and n = 4.... Consider further that n ≥ 5. Since
t − 1 ≤ t n−1 (t − 1),
we have
B ≥ t 2n − t n+1 − (n − 1)t n−1 (t − 1)
= t n−1 [t n+1 − t 2 − (n − 1)t + n − 1]
t2 + 1
n−1
n+1
2
≥t
t
− t − (n − 1)
+n−1
2
1
= t n−1 [2t n+1 − (n + 1)t 2 + n − 1]
2
hence
2(n − 1)B − 8C ≥ (n − 1)t n−1 [2t n+1 − (n + 1)t 2 + n − 1] − 8C
= (n − 1)t n−1 h(t),
180
Vasile Cîrtoaje
where
h(t) = 2t n+1 − 8t n +
8nt n−1 − 8
− (n + 1)t 2 + n − 1.
n−1
We have
h0 (t) = 2th1 (t),
h1 (t) = (n + 1)t n−1 − 4nt n−2 + 4nt n−3 − n − 1,
h01 (t) = t n−4 g(t),
g(t) = (n2 − 1)t 2 + 4n(n − 3) − 4n(n − 2)t,
where
Æ
(n2 − 1)t 2 · 4n(n − 3) − 4n(n − 2)t
p
p Æ
=4 n
(n2 − 1)(n − 3) − (n − 2) n t.
g(t) ≥ 2
Since
(n2 − 1)(n − 3) − n(n − 2)2 = n(n − 5) + 3 > 0,
we have g(t) > 0, h01 (t) > 0, h1 (t) is increasing for t ≥ 1, h1 (t) ≥ h1 (1) = 0, h0 (t) ≥ 0,
h(t) is increasing for t ≥ 1, h(t) ≥ h(1) = 0, hence (n − 1)B − 4C ≥ 0. Thus, the proof is
completed. In accordance with Remark 3, the equality holds for a1 = a2 = · · · = an = 1.
Remark 1. For p = 0 and q = 1, we get the following inequality (Vasile Cîrtoaje, 2006):
1−a
1− b
1−c
1−d
1−e
+
+
+
+
≥ 0,
2
2
2
2
1+a
1+ b
1+c
1+d
1 + e2
where a, b, c, d, e are positive real numbers such that a bcd e = 1. Replacing a, b, c, d, e
by 1/a, 1/b, 1/c, 1/d, 1/e, we get the inequality
1+a
1+ b
1+c
1+d
1+e
+
+
+
+
≤ 5,
1 + a2 1 + b2 1 + c 2 1 + d 2 1 + e2
where a, b, c, d, e are positive real numbers such that a bcd e = 1.
Notice that the inequality
1 − a1
1 + a12
+
1 − a2
1 + a22
+
1 − a3
1 + a32
+
1 − a4
1 + a42
+
1 − a5
1 + a52
+
1 − a6
1 + a62
≥0
is not true for all positive numbers a1 , a2 , a3 , a4 , a5 , a6 satisfying a1 a2 a3 a4 a5 a6 = 1.
Indeed, for a2 = a3 = a4 = a5 = a6 = 2, the inequality becomes
1 − a1
1 + a12
which is false for a1 > 0.
− 1 ≥ 0,
Partially Convex Function Method
181
P 3.22. If a, b, c are positive real numbers such that a bc = 1, then
1−a
1− b
1−c
+
+
≥ 0.
2
2
17 + 4a + 6a
17 + 4b + 6b
17 + 4c + 6c 2
(Vasile Cîrtoaje, 2008)
Solution. Using the substitution
a = e x , b = e y , c = ez ,
we need to show that
f (x) + g( y) + g(z) ≥ 3 f (s),
where
f (u) =
s=
x + y +z
= 0,
3
1 − eu
, u ∈ R,
1 + peu + qe2u
with
p=
4
,
17
q=
6
.
17
As we have shown in the proof of the preceding P 3.21, f is decreasing on (−∞, s0 ]
and increasing on [s0 , ∞), where
s0 = ln r0 > 0,
r0 = 1 +
v
t
v
t9
p+1
1+
=1+
.
q
2
In addition, f is convex on [0, s0 ] if p + 3q − 1 ≥ 0. Indeed,
p + 3q − 1 =
5
> 0.
17
By RPCF Theorem, we only need to prove the original inequality for b = c := t and
a = 1/t 2 , where t > 0. Write this inequality as follows:
t 2 (t 2 − 1)
2(1 − t)
+
≥ 0,
4
2
t + pt + q 1 + pt + qt 2
pA + qB ≥ C,
where
A = t 2 (t − 1)2 (t + 2),
B = (t − 1)2 (t 4 + 2t 3 + 2t 2 + 2t + 2),
C = t 2 (t − 1)2 (2t + 1).
182
Vasile Cîrtoaje
Indeed, we have
pA + qB − C =
3(t − 1)2 (t − 2)2 (2t 2 + 2t + 1)
≥ 0.
17
In accordance with Remark 3, the equality holds for a = b = c = 1, and also for a = 1/4
and b = c = 2 (or any cyclic permutation).
P 3.23. If a1 , a2 , · · · , a8 are positive real numbers such that a1 a2 · · · a8 = 1, then
1 − a8
1 − a1
1 − a2
+
+
·
·
·
+
≥ 0.
(1 + a1 )2 (1 + a2 )2
(1 + a8 )2
(Vasile Cîrtoaje, 2006)
Solution. Using the substitutions ai = e x i for i = 1, 2, . . . , 8, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x 8 ) ≥ 8 f (s),
where
f (u) =
s=
x1 + x2 + · · · + x8
= 0,
8
1 − eu
, u ∈ R.
(1 + eu )2
From
f 0 (t) =
eu (eu − 3)
,
(1 + eu )3
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where
s0 = ln 3 > 1.
Also, we have
f 00 (u) =
eu (8eu − e2u − 3)
.
(1 + eu )4
We will show that 8eu − e2u − 3 > 0 for u ∈ [0, s0 ], hence f is convex on [0, s0 ]. Indeed,
8eu − e2u − 3 ≥ 8eu − 3eu − 3 = 5eu − 3 > 5 − 3 > 0.
By RPCF Theorem, we only need to prove the original inequality for a2 = · · · = a8 := t
and a = 1/t 7 , where t ≥ 1. For the nontrivial case t > 1, write this inequality as follows:
t 7 (t 7 − 1) 7(t − 1)
≥
.
(t 7 + 1)2
(t + 1)2
Partially Convex Function Method
183
t 7 (t 7 − 1)(t + 1)2
≥ 7,
(t − 1)(t 7 + 1)2
t 7 (t 6 + t 5 + t 4 + t 3 + t 2 + t + 1)
≥ 7.
(t 6 − t 5 + t 4 − t 3 + t 2 − t + 1)2
Since
t 6 − t 5 + t 4 − t 3 + t 2 − t + 1 = t 4 (t 2 − t + 1) − (t − 1)(t 2 + 1) > t 4 (t 2 − t + 1),
it suffices to show that
t6 + t5 + t4 + t3 + t2 + t + 1
≥ 7,
t(t 2 − t + 1)2
which is equivalent to the obvious inequality (t − 1)6 ≥ 0. Thus, the proof is completed.
The equality holds for a1 = a2 = · · · = a8 = 1.
Remark. The inequality
1 − a9
1 − a1
1 − a2
+
+
·
·
·
+
≥0
(1 + a1 )2 (1 + a2 )2
(1 + a9 )2
is not true for all positive numbers a1 , a2 , · · · , a9 satisfying a1 a2 · · · a9 = 1. Indeed, for
a2 = a3 = · · · = a9 = 3, the inequality becomes
1 − a1
− 1 ≥ 0,
(1 + a1 )2
which is false for a1 > 0.
−13 13
P 3.24. Let a, b, c be positive real numbers such that a bc = 1. If k ∈ p , p , then
3 3 3 3
a+k
b+k
c+k
3(1 + k)
+ 2
+ 2
≤
.
2
a +1 b +1 c +1
2
(Vasile Cîrtoaje, 2012)
Solution. The desired inequality is equivalent to
X (a − 1)2
a2 + 1
≥k
X
2
−
3
.
a2 + 1
184
Vasile Cîrtoaje
13
Thus, it suffices to prove this inequality for only |k| = p . On the other hand, replacing
3 3
a, b, c by 1/a, 1/b, 1/c, the inequality becomes
X (a − 1)2
a2 + 1
X
≥ k 3−
2
.
a2 + 1
Thus, we only need to prove the desired inequality for
13
k= p .
3 3
Using the substitution
a = e x , b = e y , c = ez ,
we need to show that
f (x) + g( y) + g(z) ≥ 3 f (s),
where
f (u) =
s=
x + y +z
= 0,
3
−eu − k
, u ∈ R.
e2u + 1
From
f 0 (t) =
e2u + 2keu − 1
,
(e2u + 1)2
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where
s0 = ln r0 < 0,
1
r0 = p .
3 3
Also, we have
f 00 (u) =
t · h(t)
,
(1 + t 2 )3
where
h(t) = −t 4 − 4kt 3 + 6t 2 + 4kt − 1,
t = eu .
We will show that h(t) > 0 for t ∈ [r0 , 1], hence f is convex on [s0 , 0]. Indeed, since
52t
52
4kt = p ≥
> 1,
3 3 27
we have
h(t) = −t 4 + 6t 2 − 1 + 4kt(1 − t 2 ) ≥ −t 4 + 6t 2 − 1 + (1 − t 2 ) = t 2 (5 − t 2 ) > 0.
Partially Convex Function Method
185
By LPCF Theorem, we only need to prove the original inequality for b = c := t and
a = 1/t 2 , where t > 0. Write this inequality as
t 2 (kt 2 + 1) 2(t + k) 3(1 + k)
+ 2
≤
,
t4 + 1
t +1
2
3t 6 − 4t 5 + t 4 + t 2 − 4t + 3 − k(1 − t 2 )3 ≥ 0,
(t − 1)2 [(3 + k)t 4 + 2(1 + k)t 3 + 2t 2 + 2(1 − k)t + 3 − k] ≥ 0
p
p
p
p
(t − 1)2 (t − 2 + 3)2 [(27 + 13 3)t 2 + 24(2 + 3)t + 33 + 17 3] ≥ 0.
The last inequality is clearly true. Thus, the proof is completed. The equality holds
p
13
for a = b = c = 1. If k = p , then the equality holds also for a = 7 + 4 3 and
3 3
p
−13
b = c = 2 − 3 (or any cyclic permutation). If k = p , then the equality holds also
3 3
p
p
for a = 7 − 4 3 and b = c = 2 + 3 (or any cyclic permutation).
p
P 3.25. If a, b, c are positive real numbers and 0 < k ≤ 2 + 2 2, then
b3
c3
a+b+c
a3
+
+
≥
.
ka2 + bc k b2 + ca kc 2 + a b
k+1
(Vasile Cîrtoaje, 2011)
p
p
Solution. For k < 2 + 2 2, the proof is similar to the one of the main case k = 2 + 2 2.
For this reason, we consider further only the case where
p
k = 2 + 2 2.
Due to homogeneity, we may assume that a bc = 1. On this hypothesis,
X
X a4
a3
1 X
a
1 X a4 − a
−
a
=
−
=
.
ka2 + bc k + 1
ka3 + 1 k + 1
k+1
ka3 + 1
Thus, we can write the inequality as
a4 − a
b4 − b
c4 − c
+
+
≥ 0.
ka3 + 1 k b3 + 1 kc 3 + 1
Using the substitution
a = e x , b = e y , c = ez ,
186
Vasile Cîrtoaje
we need to show that
f (x) + g( y) + g(z) ≥ 3 f (s),
where
f (u) =
From
f 0 (t) =
s=
x + y +z
= 0,
3
e4u − eu
, u ∈ R.
ke3u + 1
ke6u + 2(k + 2)e3u − 1
,
(ke3u + 1)2
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where
s0 = ln r0 < 0,
r0 =
v
p
u
t
3 −k − 2 +
(k + 1)(k + 4)
k
≈ 0.4149.
Also, we have
f 00 (u) =
t · h(t)
,
(kt 3 + 1)3
where
h(t) = k2 t 9 − k(4k + 1)t 6 + (13k + 16)t 3 − 1,
t = eu .
We will show that h(t) > 0 for t ∈ [r0 , 1], hence f is convex on [s0 , 0]. Indeed, we have
h(t) ≥ 0 for t ∈ [t 1 , t 2 ], where t 1 ≈ 0.2345 and t 2 ≈ 1.02. Since [r0 , 1] ⊂ [t 1 , t 2 ], the
conclusion follows. By LPCF Theorem, we only need to prove the original inequality for
b = c. Due to homogeneity, we may consider that b = c = 1. Thus, we need to show
that
a3
2
a+2
+
≥
,
2
ka + 1 a + k
k+1
which is equivalent to the obvious inequality
p
(a − 1)2 (a − 2)2 ≥ 0.
Thus, the proof is completed. In accordance with Remark 3, the equality holds for
p
a
a = b = c. If k = 2 + 2 2, then the equality holds also for p = b = c (or any cyclic
2
permutation).
P 3.26. If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that
a1 , a2 , . . . , an ≥
−1
,
n−2
a1 + a2 + · · · + an = n,
Partially Convex Function Method
then
a12
a12 − a1 + 1
+
187
a22
a22 − a2 + 1
+ ··· +
an2
an2 − an + 1
≤ n.
(Vasile Cirtoaje, 2012)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
1−u
−1 n2 − n − 1
f (u) = 2
, u∈I=
,
.
u −u+1
n−2
n−2
From
f 0 (u) =
u(u − 2)
,
(u2 − u + 1)2
f 00 (u) =
2(3u2 − u3 − 1) 2u2 (2 − u) + 2(u2 − 1)
=
,
(u2 − u + 1)3
(u2 − u + 1)3
it follow that f is convex on [1, s0 ], decreasing for 1 ≤ u ≤ s0 and increasing for u ≥ s0 ,
where s0 = 2 ∈ I. Since f is not decreasing on Iu≤s0 , we can not apply RPCF Theorem in
its original form. However, according to Remark 5, we may replace this condition with
the condition ns − (n − 1)s0 ≤ inf I. Indeed, we have
ns − (n − 1)s0 = n − 2(n − 1) = −n + 2 ≤
−1
= inf I.
n−2
So, it suffices to show that f (x)+(n−1) f ( y) ≥ n f (1) for all x, y ∈ I such that x ≤ 1 ≤ y
and x + (n − 1) y = n. According to Remark 1, we only need to show that h(x, y) ≥ 0,
where
f (u) − f (1)
g(x) − g( y)
g(u) =
, h(x, y) =
.
u−1
x−y
We have
g(u) =
h(x, y) =
(x 2
u2
−1
,
−u+1
x + y −1
(n − 2)x + 1
=
≥ 0.
2
2
− x + 1)( y − y + 1) (n − 1)(x − x + 1)( y 2 − y + 1)
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1, and also for
−1
n−1
a1 =
and a2 = a3 = · · · = an =
(or any cyclic permutation).
n−2
n−2
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Vasile Cîrtoaje
P 3.27. If a1 , a2 , . . . , an (n ≥ 3) are nonzero real numbers such that
a1 , a2 , . . . , an ≥
then
1
a12
+
1
a22
−n
,
n−2
+ ··· +
a1 + a2 + · · · + an = n,
1
1
1
1
≥
+
+ ··· + .
an2
a1 a2
an
(Vasile Cirtoaje, 2012)
Solution. According to P 2.25-(a) in Volume 1, the inequality is true for n = 3. Assume
further that n ≥ 4 and write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
s=
a1 + a2 + · · · + an
= 1,
n
1
1
−n n(2n − 3)
f (u) = 2 − , u ∈ I =
,
\ {0}.
u
u
n−2
n−2
From
f 0 (u) =
u−2
,
u3
f 00 (u) =
2(3 − u)
,
u4
it follow that f is convex on [1, 3], decreasing for 1 ≤ u ≤ s0 and increasing for u ≥ s0 ,
where s0 = 2 ∈ I. We see that f is not decreasing on Iu≤s0 . Therefore, according to
Remark 4 and Remark 5, we may replace this condition in RPCF Theorem with the
condition ns − (n − 1)s0 ≤ inf I. Indeed, for n ≥ 4, we have
ns − (n − 1)s0 = n − 2(n − 1) = −n + 2 ≤
−n
= inf I.
n−2
So, it suffices to show that f (x)+(n−1) f ( y) ≥ n f (1) for all x, y ∈ I such that x ≤ 1 ≤ y
and x + (n − 1) y = n. According to Remark 1, we only need to show that h(x, y) ≥ 0,
where
f (u) − f (1)
g(x) − g( y)
g(u) =
, h(x, y) =
.
u−1
x−y
We have
g(u) =
x+y
−1
(n − 2)x + n
, h(x, y) = 2 2 =
≥ 0.
2
u
x y
(n − 1)x 2 y 2
The proof is completed. In accordance with Remark 3, the equality holds for a1 = a2 =
−n
n
· · · = an = 1, and also for a1 =
and a2 = a3 = · · · = an =
(or any cyclic
n−2
n−2
permutation).
Remark. Similarly, we can prove the following generalization.
Partially Convex Function Method
• Let a1 , a2 , · · · , an ≥
189
−n
such that a1 + a2 + · · · + an = n. If n ≥ 3 and k ≥ 0, then
n−2
1 − a1
k + a12
+
1 − a2
k + a22
+ ··· +
1 − an
≥ 0,
k + an2
−n
with equality for a1 = a2 = · · · = an = 1, and also for a1 =
and a2 = a3 = · · · =
n−2
n
(or any cyclic permutation).
an =
n−2
P 3.28. If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that
a1 , a2 , . . . , an ≥
then
−(3n − 2)
,
n−2
a1 + a2 + · · · + an = n,
1 − an
1 − a1
1 − a2
+
+ ··· +
≥ 0.
2
2
(1 + a1 )
(1 + a2 )
(1 + an )2
(Vasile Cîrtoaje, 2014)
Solution. According to P 2.25-(b) in Volume 1, the inequality is true for n = 3. Assume
further that n ≥ 4 and write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
s=
a1 + a2 + · · · + an
= 1,
n
−(3n − 2) 4n2 − 7n + 2
1−u
f (u) =
, u∈I=
,
\ {−1}.
(1 + u)2
n−2
n−2
From
f 0 (u) =
u−3
,
(u + 1)3
f 00 (u) =
2(5 − u)
,
(u + 1)4
it follow that f is convex on [1, 5], decreasing for 1 ≤ u ≤ s0 and increasing for u ≥ s0 ,
where s0 = 3 ∈ I. We see that f is not decreasing on Iu≤s0 . Therefore, according to
Remark 4 and Remark 5, we may apply RPCF Theorem by replacing this condition with
the sufficient condition ns − (n − 1)s0 ≤ inf I. Indeed, for n ≥ 4, we have
ns − (n − 1)s0 = n − 3(n − 1) = −2n + 3 ≤
−(3n − 2)
= inf I.
n−2
Thus, it suffices to show that f (x) + (n − 1) f ( y) ≥ n f (1) for all x, y ∈ I such that
x ≤ 1 ≤ y and x + (n − 1) y = n. According to Remark 1, we only need to show that
h(x, y) ≥ 0, where
g(u) =
f (u) − f (1)
,
u−1
h(x, y) =
g(x) − g( y)
.
x−y
190
Vasile Cîrtoaje
We have
g(u) =
x + y +2
(n − 2)x + 3n − 2
−1
, h(x, y) =
=
≥ 0.
2
2
2
(u + 1)
(1 + x) (1 + y)
(n − 1)(1 + x)2 (1 + y)2
The proof is completed. In accordance with Remark 3, the equality holds for a1 = a2 =
−(3n − 2)
n+2
· · · = an = 1, and also for a1 =
and a2 = a3 = · · · = an =
(or any cyclic
n−2
n−2
permutation).
P 3.29. Let a1 , a2 , . . . , an be nonnegative real numbers such that a1 + a2 + · · · + an = n. If
2
n ≥ 3 and k ≥ 2 − , then
n
1 − an
1 − a1
1 − a2
+
+ ··· +
≥ 0.
(1 − ka1 )2 (1 − ka2 )2
(1 − kan )2
(Vasile Cîrtoaje, 2012)
Solution. According to P 3.96 in Volume 1, the inequality is true for n = 3. Assume
further that n ≥ 4 and write the inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
where
f (u) =
From
f 0 (u) =
s=
a1 + a2 + · · · + an
= 1,
n
1−u
, u ∈ I = [0, 3] \ {1/k}.
(1 − ku)2
−ku + 2k − 1
,
(1 − ku)3
f 00 (u) =
2k(−ku + 3k − 2)
,
(1 − ku)4
it follow that f is convex on [1, s0 ], increasing on [0, 1/k) ∪ [s0 , 3] and decreasing on
(1/k, s0 ], where
3n − 4
s0 = 2 − 1/k ≥
> 1.
2(n − 1)
Since f is not decreasing on Iu≤s0 , we can not apply RPCF Theorem in its original form.
However, according to Remark 5, we may replace this condition with the condition
s1 ≤ inf I, where s1 = ns − (n − 1)s0 . Indeed, we have
s1 ≤ n − (n − 1) ·
3n − 4
4−n
=
≤ 0 = inf I.
2(n − 1)
2
Partially Convex Function Method
191
So, it suffices to show that f (x)+(n−1) f ( y) ≥ n f (1) for all x, y ∈ I such that x ≤ 1 ≤ y
and x + (n − 1) y = n. According to Remark 1, we only need to show that h(x, y) ≥ 0,
where
f (u) − f (1)
g(x) − g( y)
g(u) =
, h(x, y) =
.
u−1
x−y
Since
g(u) =
h(x, y) =
−1
,
(1 − ku)2
k[k(x + y) − 2]
,
(1 − k x)2 (1 − k y)2
we need to show that k(x + y) − 2 ≥ 0. Indeed,
k(x + y) =
kn
k[(n − 2)x + n]
≥
≥ 2.
n−1
n−1
2
The proof is completed. The equality holds for a1 = a2 = · · · = an = 1. If k = 2 − ,
n
n
then the equality also holds for a1 = 0 and a2 = a3 = · · · = an =
(or any cyclic
n−1
permutation).
192
Vasile Cîrtoaje
Chapter 4
Partially Convex Function Method
for Ordered Variables
4.1
Theoretical Basis
Right Partially Convex Function Theorem for Ordered Variables (Vasile Cirtoaje,
2012). Let f be a function defined on a real interval I and convex on [s, s0 ], where s, s0 ∈ I,
s < s0 . In addition, f (u) is decreasing on Iu≤s0 and
min f (u) = f (s0 ).
u∈I
The inequality
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
a + a + ··· + a 1
2
n
n
holds for all a1 , a2 , . . . , an ∈ I such that a1 + a2 + · · · + an = ns and at least n − m of
a1 , a2 , . . . , an are smaller than or equal to s if and only if
f (x) + m f ( y) ≥ (1 + m) f (s)
for all x, y ∈ I such that x ≤ s ≤ y and x + m y = (1 + m)s.
Proof. The necessity is obvious. By Lemma from section 3, to prove the sufficiency, it
suffices to consider that a1 , a2 , . . . , an ∈ J, where
J = Iu≤s0 .
Because f (u) is convex on Ju≥s , the desired inequality follows from HCF-OV Theorem
applied to the interval J.
Similarly, we can prove Left Partially Convex Function Theorem for Ordered Variables (Vasile Cirtoaje, 2012).
193
194
Vasile Cîrtoaje
Left Partially Convex Function Theorem for Ordered Variables. Let f be a function
defined on a real interval I and convex on [s0 , s], where s0 , s ∈ I, s0 < s. In addition, f (u)
is increasing on Iu≥s0 and satisfies
min = f (s0 ).
u∈I
The inequality
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
a + a + ··· + a 1
2
n
n
holds for all a1 , a2 , · · · , an ∈ I such that a1 + a2 + · · · + an = ns and at least n − m of
a1 , a2 , . . . , an are greater than or equal to s if and only if
f (x) + m f ( y) ≥ (1 + m) f (s)
for all x, y ∈ I such that x ≥ s ≥ y and x + m y = (1 + m)s.
Remark 1. Let us denote
g(u) =
f (u) − f (s)
,
u−s
h(x, y) =
g(x) − g( y)
.
x−y
We may replace the hypothesis condition in RPCF-OV Theorem and LPCF-OV Theorem,
namely
f (x) + m f ( y) ≥ (1 + m) f (s),
by the condition
h(x, y) ≥ 0 for all x, y ∈ I such that x + m y = (1 + m)s.
Remark 2. Assume that f is differentiable on I, and let
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
Then, the desired inequality of Jensen’s type in RPCF-OV Theorem and LPCF-OV Theorem holds true by replacing the hypothesis
f (x) + m f ( y) ≥ (1 + m) f (s)
with the more restrictive condition
H(x, y) ≥ 0 for all x, y ∈ I such that x + m y = (1 + m)s.
Remark 3. RPCF-OV Theorem is a generalization of LPCF Theorem, because the last
theorem can be obtained from the first theorem for m = n − 1. Similarly, LPCF-OV
Theorem is a generalization of LPCF Theorem.
PCF Method for Ordered Variables
195
Remark 4. If
a1 ≥ · · · ≥ am ≥ s ≥ am+1 ≥ · · · ≥ an ,
then at least n − m of a1 , a2 , . . . , an are smaller than or equal to s. Also, if
a1 ≥ · · · ≥ an−m ≥ s ≥ an−m+1 ≥ · · · ≥ an ,
then at least n − m of a1 , a2 , . . . , an are greater than or equal to s. Thus, from RPCF-OV
Theorem and LPCF-OV Theorem, we get the following corollaries.
RPCF-OV Corollary. Let f be a function defined on a real interval I and convex on [s, s0 ],
where s, s0 ∈ I, s < s0 . In addition, f (u) is decreasing on Iu≤s0 and
min f (u) = f (s0 ).
u∈I
The inequality
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
a + a + ··· + a 1
2
n
n
holds for all a1 , a2 , . . . , an ∈ I satisfying
a1 ≥ · · · ≥ am ≥ s ≥ am+1 ≥ · · · ≥ an ,
a1 + a2 + · · · + an = ns,
if and only if
f (x) + m f ( y) ≥ (1 + m) f (s)
for all x, y ∈ I such that x ≤ s ≤ y and x + m y = (1 + m)s.
LPCF-OV Corollary. Let f be a function defined on a real interval I and convex on [s0 , s],
where s0 , s ∈ I, s0 < s. In addition, f (u) is increasing on Iu≥s0 and satisfies
min f (u) = f (s0 ).
u∈I
The inequality
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f
a + a + ··· + a 1
2
n
n
holds for all a1 , a2 , . . . , an ∈ I satisfying
a1 ≤ · · · ≤ am ≤ s ≤ am+1 ≤ · · · ≤ an ,
a1 + a2 + · · · + an = ns,
if and only if
f (x) + m f ( y) ≥ (1 + m) f (s)
for all x, y ∈ I such that x ≥ s ≥ y and x + m y = (1 + m)s.
196
Vasile Cîrtoaje
Remark 5. The inequality in RPCF-OV Corollary becomes an equality for
a1 = a2 = · · · = an = s
and also for
a1 = · · · = am = y,
am+1 = · · · = an−1 = s,
an = x (x < y),
where x, y ∈ I satisfy the equations
x + m y = (1 + m)s,
f (x) + m f ( y) = (1 + m) f (s).
For x 6= y, these equations are equivalent to
x + m y = (1 + m)s, h(x, y) = 0.
Remark 6. The inequality in LPCF-OV Corollary becomes an equality for
a1 = a2 = · · · = an = s
and also for
a1 = x,
a2 = · · · = an−m = s,
an−m+1 = · · · = an = y (x > y),
where x, y ∈ I satisfy the equations
x + m y = (1 + m)s,
f (x) + m f ( y) = (1 + m) f (s).
For x 6= y, these equations are equivalent to
x + m y = (1 + m)s, h(x, y) = 0.
Remark 7. RPCF-OV Theorem and RPCF-OV Corollar are also valid in the case when
I = [a, b] \ {u0 } or I = (a, b) \ {u0 }, where a, b, u0 are real numbers such that
a < s < s0 < u0 < b.
Similarly, LPCF Theorem is also valid in the case when I = [a, b]\{u0 } or I = (a, b)\{u0 },
where a, b, u0 are real numbers such that
a < u0 < s0 < s < b.
PCF Method for Ordered Variables
4.2
197
Applications
4.1. If a, b, c, d are real numbers such that
a ≤ 1 ≤ b ≤ c ≤ d,
then
a + b + c + d = 4,
b
c
d
a
+ 2
+ 2
+ 2
≤ 1.
+ 1 3b + 1 3c + 1 3d + 1
3a2
4.2. If a, b, c, d are real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
then
a + b + c + d = 4,
16b − 5
16c − 5
16d − 5
4
16a − 5
+
+
+
≤ .
32a2 + 1 32b2 + 1 32c 2 + 1 32d 2 + 1 3
4.3. If a, b, c, d, e are real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d ≥ e,
then
a + b + c + d + e = 5,
18a − 5
18b − 5
18c − 5
18d − 5
18e − 5
+
+
+
+
≤ 5.
2
2
2
2
12a + 1 12b + 1 12c + 1 12d + 1 12e2 + 1
4.4. If a, b, c, d, e are real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d ≥ e,
then
a + b + c + d + e = 5,
a(a − 1) b(b − 1) c(c − 1) d(d − 1) e(e − 1)
+
+ 2
+
+ 2
≥ 0.
3a2 + 4
3b2 + 4
3c + 4
3d 2 + 4
3e + 4
4.5. Let a1 , a2 , . . . , a2n 6= −k be real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
If k ≥
a1 + a2 + · · · + a2n = 2n.
n+1
p , then
2 n
a2n (a2n − 1)
a1 (a1 − 1) a2 (a2 − 1)
+
+ ··· +
≥ 0.
2
2
(a1 + k)
(a2 + k)
(a2n + k)2
198
Vasile Cîrtoaje
4.6. Let a1 , a2 , . . . , a2n 6= −k be real numbers such that
a1 + a2 + · · · + a2n = 2n.
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
n+1
If k ≥ 1 + p , then
n
a12 − 1
(a1 + k)2
+
a22 − 1
(a2 + k)2
+ ··· +
2
a2n
−1
(a2n + k)2
≥ 0.
4.7. If a1 , a2 , · · · , an are positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
then
2/a1
a1
2/a2
+ a2
a1 + a2 + · · · + an = n,
+ · · · + an2/an ≤ n.
4.8. If a1 , a2 , · · · , a11 are real numbers such that
a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ a11 ,
a1 + a2 + · · · + a11 = 11,
then
2
) ≥ 1.
(1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a11 + a11
4.9. If a1 , a2 , · · · , a8 are nonzero real numbers such that
a1 ≥ a2 ≥ a3 ≥ a4 ≥ 1 ≥ a5 ≥ a6 ≥ a7 ≥ a8 ,
then
5
1
a12
+
1
a22
+ ··· +
1
a82
a1 + a2 + · · · + a8 = 8,
1
1
1
+ 72 ≥ 14
+
+ ··· +
.
a1 a2
a8
4.10. If a, b, c, d are positive real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
then
a bcd = 1,
7 − 6a 7 − 6b 7 − 6c 7 − 6d
4
+
+
+
≥ .
2
2
2
2
2+a
2+ b
2+c
2+d
3
PCF Method for Ordered Variables
199
4.11. If a, b, c are positive real numbers such that
a ≥ 1 ≥ b ≥ c,
then
a bc = 1,
7 − 4a 7 − 4b 7 − 4c
+
+
≥ 3.
2 + a2 2 + b2 2 + c 2
4.12. Let a1 , a2 , · · · , an be positive real numbers such that
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
a1 a2 · · · an = 1.
If p, q ≥ 0 such that p + 3q ≥ 1, then
1 − a1
1+
pa1 + qa12
+
1 − a2
1+
pa2 + qa22
+ ··· +
1 − an
≥ 0.
1 + pan + qan2
200
Vasile Cîrtoaje
PCF Method for Ordered Variables
4.3
201
Solutions
P 4.1. If a, b, c, d are real numbers such that
a ≤ 1 ≤ b ≤ c ≤ d,
then
a + b + c + d = 4,
b
c
d
a
+ 2
+ 2
+ 2
≤ 1.
+ 1 3b + 1 3c + 1 3d + 1
3a2
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
where
f (u) =
−u
,
3u2 + 1
From
s=
a+b+c+d
= 1,
4
u ∈ R.
3u2 − 1
,
(3u2 + 1)2
f 0 (u) =
it follows
p that f is increasing on (−∞, −s0 ]∪[s0 , ∞) and decreasing on [−s0 , s0 ], where
s0 = 1/ 3. Since
lim f (u) = 0
u→−∞
and f (s0 ) < 0, it follows that
min f (u) = f (s0 ).
u∈R
From
f 00 (u) =
18u(1 − u2 )
,
(3u2 + 1)3
it follows that f is convex on [0, 1], hence on [s0 , 1]. Therefore, we apply LPCF-OV
Theorem or LPCF-OV Corollary for n = 4 and m = 1. We only need to show that
f (x) + f ( y) ≥ 2 f (1) for all real x, y such that x + y = 2. Using Remark 1, it suffices to
prove that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
Indeed, we have
g(u) =
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
3u − 1
,
4(3u2 + 1)
3(1 + x + y − 3x y)
9(1 − x y)
=
≥ 0,
2
2
2
4(3x + 1)(3 y + 1) 4(3x + 1)(3 y 2 + 1)
202
Vasile Cîrtoaje
since
4(1 − x y) = (x + y)2 − 4x y = (x − y)2 ≥ 0.
Thus, the proof is completed. The equality holds for a = b = c = d = 1.
Remark. Similarly, we can prove the following generalization.
• If a1 , a2 , . . . , an are real numbers such that
a1 + a2 + · · · + an = n,
a1 ≤ 1 ≤ a2 ≤ · · · ≤ an ,
then
a1
3a12
+1
+
a2
3a22
+1
+ ··· +
an
n
≤ ,
3an2 + 1 4
with equality for a1 = a2 = · · · = an = 1.
P 4.2. If a, b, c, d are real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
then
a + b + c + d = 4,
16a − 5
16b − 5
16c − 5
16d − 5
4
+
+
+
≤ .
2
2
2
2
32a + 1 32b + 1 32c + 1 32d + 1 3
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) ≥ 4 f (s),
s=
a+b+c+d
= 1,
4
where
5 − 16u
, u ∈ R.
32u2 + 1
As shown in the proof of P 3.1, f is convex on [s0 , 1], increasing for u ≥ s0 and
f (u) =
min f (u) = f (s0 ),
u∈R
where
p
5 + 33
s0 =
≈ 0.6715.
16
Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for n = 4 and m = 2.
We only need to show that f (x) + 2 f ( y) ≥ 3 f (1) for all real x, y such that x + 2 y = 3.
Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
PCF Method for Ordered Variables
203
Indeed, we have
g(u) =
h(x, y) =
32(2u − 1)
,
3(32u2 + 1)
64(1 + 16x + 16 y − 32x y)
64(4x − 5)2
=
≥ 0.
3(32x 2 + 1)(32 y 2 + 1)
3(32x 2 + 1)(32 y 2 + 1)
Thus, the proof is completed. From x + 2 y = 3 and h(x, y) = 0, we get x = 5/4 and,
y = 7/8. Therefore, in accordance with Remark 6, the equality holds for a = b = c =
d = 1, and also for a = 5/4, b = 1 and c = d = 7/8.
Remark. Similarly, we can prove the following generalization.
• If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that
a1 ≥ · · · ≥ an−2 ≥ 1 ≥ an−1 ≥ an ,
then
16a1 − 5
32a12
+1
+
16a2 − 5
32a22
+1
+ ··· +
a1 + a2 + · · · + an = n,
16an − 5
n
≤ ,
2
32an + 1 3
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 5/4, a2 = · · · = an−2 = 1 and
an−1 = an = 7/8.
P 4.3. If a, b, c, d, e are real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d ≥ e,
then
a + b + c + d + e = 5,
18a − 5
18b − 5
18c − 5
18d − 5
18e − 5
+
+
+
+
≤ 5.
2
2
2
2
12a + 1 12b + 1 12c + 1 12d + 1 12e2 + 1
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s),
where
f (u) =
5 − 18u
,
12u2 + 1
s=
a+b+c+d+e
= 1,
5
u ∈ R.
As shown in the proof of P 3.2, f is convex on [s0 , 1], increasing for u ≥ s0 and
min f (u) = f (s0 ),
u∈R
204
Vasile Cîrtoaje
where
p
5 + 52
s0 =
≈ 0.678.
18
Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for n = 5 and m = 3.
We only need to show that f (x) + 3 f ( y) ≥ 4 f (1) for all real x, y such that x + 3 y = 4.
Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
Indeed, we have
g(u) =
h(x, y) =
6(2u − 1)
,
12u2 + 1
12(1 + 6x + 6 y − 12x y)
12(2x − 3)2
=
≥ 0.
(12x 2 + 1)(12 y 2 + 1)
(12x 2 + 1)(12 y 2 + 1)
Thus, the proof is completed. From x + 3 y = 4 and h(x, y) = 0, we get x = 3/2 and,
y = 5/6. Therefore, in accordance with Remark 6, the equality holds for a = b = c =
d = e = 1, and also for a = 3/2, b = 1 and c = d = e = 5/6.
Remark. Similarly, we can prove the following generalization.
• If a1 , a2 , . . . , an (n ≥ 4) are real numbers such that
a1 ≥ · · · ≥ an−3 ≥ 1 ≥ an−2 ≥ an−1 ≥ an ,
then
18a1 − 5
12a12
+1
+
18a2 − 5
12a22
+1
+ ··· +
a1 + a2 + · · · + an = n,
18an − 5
≤ n,
12an2 + 1
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 3/2, a2 = · · · = an−3 = 1 and
an−2 = an−1 = an = 5/6.
P 4.4. If a, b, c, d, e are real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d ≥ e,
a + b + c + d + e = 5,
then
a(a − 1) b(b − 1) c(c − 1) d(d − 1) e(e − 1)
+
+ 2
+
+ 2
≥ 0.
3a2 + 4
3b2 + 4
3c + 4
3d 2 + 4
3e + 4
(Vasile Cîrtoaje, 2012)
PCF Method for Ordered Variables
205
Solution. Write the inequality as
f (a) + f (b) + f (c) + f (d) + f (e) ≥ 5 f (s),
s=
a+b+c+d+e
= 1,
5
where
u2 − u
, u ∈ R.
3u2 + 4
As shown in the proof of P 3.5, f is convex on [s0 , 1], increasing for u ≥ s0 and
f (u) =
min f (u) = f (s0 ),
u∈R
where
p
−4 + 2 7
s0 =
≈ 0.43.
3
Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for n = 5 and m = 3.
We only need to show that f (x) + 3 f ( y) ≥ 4 f (1) for all real x, y such that x + 3 y = 4.
Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
f (u) − f (1)
.
u−1
Indeed, we have
g(u) =
h(x, y) =
u)
,
+4
3u2
4 − 3x y
(x − 2)2
=
≥ 0.
(3x 2 + 4)(3 y 2 + 4) (12x 2 + 1)(12 y 2 + 1)
Thus, the proof is completed. From x + 3 y = 4 and h(x, y) = 0, we get x = 2 and,
y = 2/3. Therefore, in accordance with Remark 6, the equality holds for
a = b = c = d = e = 1,
and also for
a = 2, b = 1, c = d = e = 2/3.
Remark. Similarly, we can prove the following generalizations.
• If a1 , a2 , . . . , an (n ≥ 4) are real numbers such that
a1 ≥ · · · ≥ an−3 ≥ 1 ≥ an−2 ≥ an−1 ≥ an ,
then
a1 (a1 − 1)
3a12
+4
+
a2 (a2 − 1)
3a22
+4
+ ··· +
a1 + a2 + · · · + an = n,
an (an − 1)
≥ 0,
3an2 + 4
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Vasile Cîrtoaje
with equality for a1 = a2 = · · · = an = 1, and also for a1 = 2, a2 = · · · = an−3 = 1 and
an−2 = an−1 = an = 2/3.
• If a1 , a2 , . . . , an (n ≥ 3) are real numbers such that
a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ an ,
a1 + a2 + · · · + an = n,
then
a1 (a1 − 1)
4(n − 2)a12
+ (n − 1)2
+
a2 (a2 − 1)
4(n − 2)a22
+ (n − 1)2
+ ··· +
an (an − 1)
≥ 0,
4(n − 2)an2 + (n − 1)2
with equality for a1 = a2 = · · · = an = 1, and also for a1 =
an =
n−1
.
2(n − 2)
n−1
, a2 = 1 and a3 = · · · =
2
P 4.5. Let a1 , a2 , . . . , a2n 6= −k be real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
If k ≥
a1 + a2 + · · · + a2n = 2n.
n+1
p , then
2 n
a2n (a2n − 1)
a1 (a1 − 1) a2 (a2 − 1)
+
+ ··· +
≥ 0.
2
2
(a1 + k)
(a2 + k)
(a2n + k)2
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (a2n ) ≥ 2n f (s),
where
f (u) =
s=
a1 + a2 + · · · + a2n
= 1,
2n
u(u − 1)
, u ∈ I = R \ {−k}.
(u + k)2
As shown in the proof of P 3.6, f is convex on [s0 , 1], increasing for u ≥ s0 and
min f (u) = f (s0 ),
u∈I
where
s0 =
k
< 1.
2k + 1
PCF Method for Ordered Variables
207
Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for 2n real numbers
and m = n. We only need to show that f (x) + n f ( y) ≥ (n + 1) f (1) for all real x, y such
that x + n y = n + 1. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where
h(x, y) =
g(x) − g( y)
,
x−y
Indeed, we have
g(u) =
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
u
,
(u + k)2
k2 − x y
≥ 0,
(x + k)2 ( y + k)2
because
[2n y − n − 1]2
(n + 1)2
−xy =
≥ 0.
4n
4n
n+1
The equality holds for a1 = a2 = · · · = an = 1. If k = p , then the equality holds also
2 n
n+1
n+1
for a1 =
, a2 = · · · = an = 1 and an+1 = · · · = a2n =
.
2
2n
k2 − x y ≥
P 4.6. Let a1 , a2 , . . . , a2n 6= −k be real numbers such that
a1 ≥ · · · ≥ an ≥ 1 ≥ an+1 ≥ · · · ≥ a2n ,
a1 + a2 + · · · + a2n = 2n.
n+1
If k ≥ 1 + p , then
n
a12 − 1
(a1 + k)2
a22 − 1
+
(a2 + k)2
+ ··· +
2
a2n
−1
(a2n + k)2
≥ 0.
(Vasile Cîrtoaje, 2012)
Solution. Write the inequality as
f (a1 ) + f (a2 ) + · · · + f (a2n ) ≥ 2n f (s),
where
f (u) =
s=
a1 + a2 + · · · + a2n
= 1,
2n
u2 − 1
, u ∈ I = R \ {−k}.
(u + k)2
As shown in the proof of P 3.7, f is convex on [s0 , 1], increasing for u ≥ s0 and
min f (u) = f (s0 ),
u∈I
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Vasile Cîrtoaje
where
−1
∈ (−1, 0).
k
Therefore, we may apply LPCF-OV Theorem or LPCF-OV Corollary for 2n real numbers
and m = n. We only need to show that f (x) + n f ( y) ≥ (n + 1) f (1) for all real x, y such
that x + n y = n + 1. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where
s0 =
h(x, y) =
g(x) − g( y)
,
x−y
Indeed, we have
g(u) =
h(x, y) =
g(u) =
f (u) − f (1)
.
u−1
u+1
,
(u + k)2
(k − 1)2 − 1 − x − y − x y
≥ 0,
(x + k)2 ( y + k)2
because
(k − 1)2 − 1 − x − y − x y ≥
(n y − 1)2
(n + 1)2
−1− x − y − xy =
≥ 0.
n
n
n+1
The equality holds for a1 = a2 = · · · = an = 1. If k = 1 + p , then the equality holds
n
1
also for a1 = n, a2 = · · · = an = 1 and an+1 = · · · = a2n = .
n
P 4.7. If a1 , a2 , · · · , an are positive real numbers such that
a1 + a2 + · · · + an = n,
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
then
2/a1
a1
2/a2
+ a2
+ · · · + an2/an ≤ n.
(Vasile Cîrtoaje, 2012)
Solution. Rewrite the desired inequality as
f (a1 ) + f (a2 ) + · · · + f (an ) ≥ n f (s),
s=
a1 + a2 + · · · + an
= 1,
n
where
f (u) = −u2/u ,
We have
u ∈ I = (0, n).
2
f 0 (u) = 2u u −2 (ln u − 1),
PCF Method for Ordered Variables
209
2
f 00 (u) = 2u u −4 [u + 2(1 − ln u)(u − 1 + ln u)].
From the expression of f 0 , it follows that f is decreasing on (0, s0 ] and increasing on
[s0 , n), where s0 = e. In addition, we claim that f is convex on [1, s0 ]. Indeed, since
1 − ln u ≥ 0 and
u − 1 + ln u ≥ u − 1 ≥ 0,
we have f 00 > 0 for u ∈ [1, s0 ]. Therefore, we may apply RPCF-OV Theorem or RPCF-OV
Corollary for m = 1. We only need to show that f (x) + f ( y) ≥ 2 f (1) for all x, y > 0
such that x + y = 2. The inequality f (x) + f ( y) ≥ 2 f (1) is equivalent to
x 2/x + y 2/ y ≤ 2,
which is just the inequality in P 3.27 from Volume 2. The equality holds for a1 = a2 =
· · · = an = 1.
P 4.8. If a1 , a2 , · · · , a11 are real numbers such that
a1 ≥ a2 ≥ 1 ≥ a3 ≥ · · · ≥ a11 ,
a1 + a2 + · · · + a11 = 11,
then
2
) ≥ 1.
(1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a11 + a11
(Vasile Cîrtoaje, 2012)
Solution. Rewrite the desired inequality as
f (a1 ) + f (a2 ) + · · · + f (a11 ) ≥ 11 f (s),
s=
a1 + a2 + · · · + a11
= 1,
11
where
f (u) = ln(1 − u + u2 ),
From
f 0 (u) =
u ∈ R.
2u − 1
,
1 − u + u2
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = 1/2.
Also, from
1 + 2u(1 − u)
f 00 (u) =
,
(1 − u + u2 )2
it follows that f 00 (u) > 0 for u ∈ [s0 , 1], hence f is convex on [s0 , 1]. Therefore, we may
apply LPCF-OV Theorem or LPCF-OV Corollary for n = 11 and m = 9. We only need to
210
Vasile Cîrtoaje
show that f (x) + 9 f ( y) ≥ 9 f (1) for all real x, y such that x + 9 y = 10. Using Remark
2, it suffices to prove that H(x, y) ≥ 0, where
H(x, y) =
Since
H(x, y) =
f 0 (x) − f 0 ( y)
.
x−y
1 + x + y − 2x y
(1 − x + x 2 )(1 − y + y 2 )
and
1 + x + y − 2x y = 18 y 2 − 8 y + 1 = 2 y 2 + (4 y − 1)2 > 0,
the conclusion follows. The equality holds for a1 = a2 = · · · = a11 = 1.
Remark. By replacing a1 , a2 , . . . , a11 respectively with 1 − a1 , 1 − a2 , . . . , 1 − a11 , we get
the following statement.
• If a1 , a2 , . . . , a11 are real numbers such that
a1 ≤ a2 ≤ 0 ≤ a3 ≤ · · · ≤ a11 ,
a1 + a2 + · · · + a11 = 0,
then
2
) ≥ 1,
(1 − a1 + a12 )(1 − a2 + a22 ) · · · (1 − a11 + a11
with equality for a1 = a2 = · · · = an = 0.
P 4.9. If a1 , a2 , · · · , a8 are nonzero real numbers such that
a1 ≥ a2 ≥ a3 ≥ a4 ≥ 1 ≥ a5 ≥ a6 ≥ a7 ≥ a8 ,
then
5
1
a12
+
1
a22
+ ··· +
1
a82
+ 72 ≥ 14
a1 + a2 + · · · + a8 = 8,
1
1
1
+
+ ··· +
.
a1 a2
a8
(Vasile Cîrtoaje, 2012)
Solution. Write the desired inequality as
f (a1 ) + f (a2 ) + · · · + f (a8 ) ≥ 8 f (s), s =
where
a1 + a2 + · · · + a8
= 1,
8
5
14
−
+ 9, u ∈ I = R \ {0}.
2
u
u
As shown in the proof of P 3.16, f is convex on [s0 , 1], increasing for u ≥ s0 and
f (u) =
min f (u) = f (s0 ),
u∈I
PCF Method for Ordered Variables
211
where
5
.
7
Taking into account Remark 7, we may apply LPCF-OV Theorem or LPCF-OV Corollary
for n = 8 and m = 4. We only need to show that f (x) + 4 f ( y) ≥ 5 f (1) for all x, y ∈ I
such that x + 4 y = 5. Using Remark 1, it suffices to prove that h(x, y) ≥ 0, where
s0 =
h(x, y) =
g(x) − g( y)
,
x−y
g(u) =
Indeed, we have
g(u) = g(u) =
h(x, y) =
f (u) − f (1)
.
u−1
5
9
− 2,
u u
5x + 5 y − 9x y
(6 y − 5)2
=
≥ 0.
x2 y2
x2 y2
In accordance with Remark 6, the equality holds for a1 = a2 = · · · = a8 = 1, and also
5
5
for a1 = , a2 = a3 = a4 = 1 and a5 = a6 = a7 = a8 = (or any cyclic permutation).
3
6
P 4.10. If a, b, c, d are positive real numbers such that
a ≥ b ≥ 1 ≥ c ≥ d,
then
a bcd = 1,
7 − 6a 7 − 6b 7 − 6c 7 − 6d
4
+
+
+
≥
.
2 + a2 2 + b2 2 + c 2 2 + d 2
3
(Vasile Cîrtoaje, 2012)
Solution. Using the substitution
a = e x , b = e y , c = ez , d = e w
we need to show that
f (x) + f ( y) + f (z) + f (w) ≥ 4 f (s),
where
x ≥ y ≥ 0 ≥ z ≥ w,
s=
x + y +z+w
= 0,
4
7 − 6eu
, u ∈ R.
2 + e2u
As shown in the proof of P 3.17, f is convex on [0, s0 ], is decreasing on (−∞, s0 ] and
increasing on [s0 , ∞), where s0 = ln 3. Therefore, we may apply RPCF-OV Theorem or
f (u) =
212
Vasile Cîrtoaje
RPCF-OV Corollary for n = 4 and m = 2. We only need to show that f (x) + 2 f ( y) ≥
3 f (0) for all real x, y such that x + 2 y = 0. That is, to prove that
7 − 6a 2(7 − 6b)
+
≥1
2 + a2
2 + b2
for all a, b > 0 such that a b2 = 1. This is equivalent to
(b − 1)2 (b − 2)2 (5b2 + 6bt + 3) ≥ 0,
which is clearly true. In accordance with Remark 5, the equality holds for a = b = c =
d = 1, and also for a = b = 2, c = 1 and d = 1/4.
P 4.11. If a, b, c are positive real numbers such that
a ≥ 1 ≥ b ≥ c,
then
a bc = 1,
7 − 4a 7 − 4b 7 − 4c
+
+
≥ 3.
2 + a2 2 + b2 2 + c 2
(Vasile Cîrtoaje, 2012)
Solution. Using the substitution
a = e x , b = e y , c = ez ,
we need to show that
f (x) + f ( y) + f (z) ≥ 3 f (s),
where
x ≥ 1 ≥ y ≥ z,
f (u) =
From
f 0 (u) =
s=
x + y +z
= 0,
3
7 − 4eu
, u ∈ R.
2 + e2u
2eu (2eu + 1)(eu − 4)
,
(2 + e2u )2
it follows that f is decreasing on (−∞, s0 ] and increasing on [s0 , ∞), where s0 = ln 4.
Also, we have
4t · h(t)
f 00 (u) =
, t = eu ,
(2 + t 2 )3
where
h(t) = −t 4 + 7t 3 + 12t 2 − 14t − 4.
PCF Method for Ordered Variables
213
We will show that h(t) ≥ 0 for t ∈ [1, 4], hence f is convex on [0, s0 ]. Indeed,
h(t) = (t − 1)[t 2 (−t + 6) + 18t + 4] ≥ 0.
Therefore, we may apply RPCF-OV Theorem or RPCF-OV Corollary for n = 3 and m = 1.
We only need to show that f (x) + f ( y) ≥ 2 f (0) for all real x, y such that x + y = 0.
That is, to prove that
7 − 4a 7 − 4b
+
≥2
2 + a2 2 + b2
for all a, b > 0 such that a b = 1. This is equivalent to
(a − 1)4 ≥ 0.
The equality holds for a = b = c = 1.
P 4.12. Let a1 , a2 , · · · , an be positive real numbers such that
a1 a2 · · · an = 1.
a1 ≥ 1 ≥ a2 ≥ · · · ≥ an ,
If p, q ≥ 0 such that p + 3q ≥ 1, then
1 − a1
1+
pa1 + qa12
+
1 − a2
1+
pa2 + qa22
+ ··· +
1 − an
≥ 0.
1 + pan + qan2
(Vasile Cîrtoaje, 2012)
Solution. For q = 0, we need to show that p ≥ 1 involves
1 − an
1 − a1
1 − a2
+
+ ··· +
≥ 0.
1 + pa1 1 + pa2
1 + pan
This is just the inequality from P 2.19. Consider next that q > 0. Using the substitutions
ai = e x i for i = 1, 2, . . . , n, we need to show that
f (x 1 ) + f (x 2 ) + · · · + f (x n ) ≥ n f (s),
where
x1 ≥ 1 ≥ x2 ≥ · · · ≥ x n,
f (u) =
s=
x1 + x2 + · · · + x n
= 0,
n
1 − eu
, u ∈ R.
1 + peu + qe2u
As shown in the proof of P 3.21, f is convex on [0, s0 ] if p + 3q − 1 ≥ 0, where
v
t
p+1
s0 = ln r0 > 0, r0 = 1 + 1 +
.
q
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Vasile Cîrtoaje
Clearly, the condition p+3q−1 ≥ 0 is satisfied by hypothesis. In addition, f is decreasing
on (−∞, s0 ] and increasing on [s0 , ∞). Therefore, we may apply RPCF-OV Theorem
or RPCF-OV Corollary for m = 1. We only need to show that f (x) + f ( y) ≥ 2 f (0) for
all real x, y such that x + y = 0. That is, to prove that
1−a
1− b
+
≥0
2
1 + pa + qa
1 + p b + q b2
for all a, b > 0 such that a b = 1. This is equivalent to
(a − 1)2 [(p − 1)a + q(a2 + a + 1)] ≥ 0,
which is true because
(p − 1)a + q(a2 + a + 1) ≥ (p − 1)a + q(3a) = (p + 3q − 1)a ≥ 0.
The equality holds for a1 = a2 = · · · = an = 1.
Chapter 5
Bibliography
215
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