Download The stronger mixing variables method

The stronger mixing variables method
Pham Kim Hung
1. Stronger mixing variables - S.M.V theorem
The following result only uses elementary mathematics. Understanding the theorem’s
usage and its meaning is more important to you than remembering its detailed proof.
Lemma. (General mixing variables lemma). Suppose that (a1 , a2 , ..., an ) is an
arbitrary real sequence. Carry out the following transformation consecutively
1. Choosing i, j ∈ {1, 2, ..., n} to be two indices satisfying
ai = min(a1 , a2 , ..., an )
,
aj = max(a1 , a2 , ..., an ).
ai + aj
(but their orders don’t change).
2
After doing infinitely many of the above transformations, each number ai comes to
the same limit
a1 + a2 + ... + an
.
a=
n
Proof. Henceforward, the above transformation is called the ∆ transformation. Denote the first sequence as (a11 , a12 , ..., a1n ). After one transformation, we have a new sequence, denoted as (a21 , a22 , .., a2n ). Similarly, from the sequence (ak1 , ak2 , ..., akn ) we have
a new one denoted as (ak+1
, ak+1
, ..., ak+1
n ). Thus, for every integer i = 1, 2, ..., n, we
1
2
need to prove
a1 + a2 + ... + an
.
lim aki = a, a =
k→∞
n
2. Replacing ai and aj by
Let mk = min(ak1 , ak2 , ..., akn ) and Mk = max(ak1 , ak2 , ..., akn ).
Clearly, the transformations ∆ don’t make the value of Mk increase or the value of
mk decrease. Because both mk and Mk are bound, there exist
m = lim mk
k→∞
,
M = lim Mk .
k→∞
We must prove that m = M . By contradiction, suppose that M > m. Denote
dk = Mk − mk . We make a simple observation
Lemma. Suppose that after carrying out some transformations ∆, the sequence
M1 + m1
,
(a11 , a12 , ..., a1n ) turns into the new one (ak1 , ak2 , ..., akn ) satisfying that mk =
2
M1 + m1
then we will have m2 =
.
2
Indeed, without loss of generality may assume that M1 = a11 ≥ a12 ≥ ... ≥ a1n = m1 .
Mathematical Reflections 6 (2006)
1
a1 + an
and k is the smallest index
2
2
satisfying that equality then ai ≥ mk ∀i ∈ {1, 2, ..., n}. It follows from {mk } is a
a1 + an
non-decreasing sequences. Note that
is a term of the sequence (a21 , a22 , .., a2n ),
2
so we have done.
To be more brief, replace ai by a1i . If mk =
By the above property, we have a more important result. Denote
S = {k : ∃l > k|mk + Mk = 2ml } ⇒ S = {k|mk + Mk = 2mk+1 },
P = {k : ∃l > k|mk + Mk = 2Ml } ⇒ P = {k|mk + Mk = 2Mk+1 }.
If S or P has an infinite number of elements, assume |S| = ∞ then, for each k ∈ S
dk+1 = Mk+1 − mk+1 = Mk+1 −
mk + Mk
M k − mk
dk
≤
=
,
2
2
2
because (dr )+∞
lim dr = 0 and
r=1 is a decreasing sequence. Thus, if |S| = ∞ then r→∞
hence M = m, the conclusion follows.
Otherwise, we must have |S|, |P | < +∞. Hence, we can suppose that |S| = |P | = 0
a1 + an
without affecting the result of the problem. Then, for every k > 1 the number
2
can’t be the smallest or greatest number in the sequence (ak1 , ak2 , ..., akn ). So we can
consider the confined problem with n − 1 numbers when we rejected exactly one
number (a1 + an )/2. By a simple induction method, we have the desired result. q
From the above lemma, we have the direct result
Theorem 1 (Stronger mixing variables - S.M.V. theorem). If f : Rn → R is a
continuous, symmetric, under-limitary function satisfying
f (a1 , a2 , ..., an ) ≥ f (b1 , b2 , ..., bn ),
in which (b1 , b2 , ..., bn ) is a sequence obtained from the sequence (a1 , a2 , ..., an ) by the
transformation ∆, then we always have
f (a1 , a2 , ..., an ) ≥ f (a, a, ..., a)
a1 + a2 + ... + an
.
n
By this theorem, when using the mixing variables method, we only need to choose
the smallest and greatest numbers to perform. By using elementary knowledge, the
old mixing variables theorem is proved and improved to have a stronger result. So it
can be applied freely.
with
a=
Moreover,
the transformation ∆ can be different. For example, we can change it
r
√
a2 + b2
to ab,
or any arbitrary average form. Depending on the supposition of
2
problem, we can choose a suitable way of mixing variables.
2
2. S.M.V. theorem and some applications
If have never tried proving a difficult inequality which involves more than three variables, it’s not easy to understand the importance and significance of S.M.V. theorem.
The most useful application of S.M.V. theorem is for four-variable inequalities. Most
of the four-variable inequalities are solved more easily by this theorem.
For example, with a familiar problem in IMO shortlist, and the solution is very
brief
Problem 1. Suppose that a, b, c, d are non-negative real numbers whose sum is 1.
Prove the inequality
1
176
+
abcd.
27
27
(Nguyen Minh Duc, IMO Shortlist 1997)
abc + bcd + cda + dab ≤
Solution. Now, we’ll consider the most important content of this writing, this is
the stronger mixing variables method, or S.M.V. theorem. Without loss of generality,
assume that a ≤ b ≤ c ≤ d. Denote
176
abcd
f (a, b, c, d) = abc + bcd + cda + dab −
27 176
f (a, b, c, d) = ac(b + d) + bd a + c −
ac .
27
1
1
(a + b + c + d) = , hence
2
2
1 1
4
176
b+d
b+d
+ ≥
≥8≥
⇒ f (a, b, c, d) ≤ f a,
, c,
.
a c
a+c
27
2
2
From the supposition, we refer that a + c ≤
Considering the transformation ∆ for (b, c, d) and as the proved result, we obtain
f (a, b, c, d) ≤ f (a, t, t, t) ,
t=
b+c+d
.
3
Now, the problems becomes, if a + 3t = 1 then
176 3
1
+
at .
27
27
But it’s quite simple. Replacing a by 1 − 3t, we have an obviously true inequality
3at2 + t3 ≤
(1 − 3t)(4t − 1)2 (11t + 1) ≥ 0.
1
and the conclusion follows immediately. The equality occurs if a = b = c = d = or
4
1
a = b = c = , d = 0 up to permutation. q
3
Return to the introduced inequality T urkevici. To my knowledge, all the ways of
proving this inequality are complicated or too long. By using S.M.V. theorem in the
same manner as in example 3.1.14, it turns out to be very easy.
Mathematical Reflections 6 (2006)
3
Problem 2. Prove the below inequality for all positive real numbers a, b, c, d
a4 + b4 + c4 + d4 + 2abcd ≥ a2 b2 + b2 c2 + c2 d2 + d2 a2 + a2 c2 + b2 d2 .
(Turkevici’s Inequality)
Solution. Assume a ≥ b ≥ c ≥ d. Denote
f (a, b, c, d) = a4 + b4 + c4 + d4 + 2abcd − a2 b2 − b2 c2 − c2 d2 − d2 a2 − a2 c2 − b2 d2
= a4 + b4 + c4 + d4 + 2abcd − a2 c2 − b2 d2 − (a2 + c2 )(b2 + d2 ).
Hence
√
√
f (a, b, c, d) − f ( ac, b, ac, d) = (a2 − c2 )2 − (b2 + d2 )(a − c)2 ≥ 0.
By S.M.V. theorem with the transformation ∆ of (a, b, c), we only need to prove the
inequality when a = b = c = t ≥ d. In this case, the problem becomes
3t4 + d4 + 2t3 d ≥ 3t4 + 3t2 d2 ⇔ d4 + t3 d + t3 d ≥ 3t2 d2 .
By AM − GM Inequality, the problem is obviously true. The equality is taken if and
only if a = b = c = d or a = b = c, d = 0 or permutations. q
Problem 3. Let x, y, z, t be positive numbers satisfying the condition x+y +z +t = 4.
Prove that
(1 + 3x)(1 + 3y)(1 + 3z)(1 + 3t) ≤ 125 + 131xyzt.
(Pham Kim Hung)
Solution. It’s easy to check that the equality occurs if x = y = z = t = 1 or
x = y = z = 4/3, t = 0. So 131 is the greatest value of k for the following inequality
(1 + 3x)(1 + 3y)(1 + 3z)(1 + 3t) ≤ 256 + k(xyzt − 1).
Consider the expression
f (x, y, z, t) = (1 + 3x)(1 + 3y)(1 + 3z)(1 + 3t) − 131xyzt.
Without loss of generality, we may assume x ≥ y ≥ z ≥ t. Hence
x+z
x+z
, y,
,t
f (x, y, z, t) − f
2
2
(x + z)2
(x + z)2
= 9(1 + 3y)(1 + 3t) xz −
− 131yt xz −
.
4
4
Note that if y + t ≤ 2 then
9(1 + 3y)(1 + 3t) ≥ 131yt ⇔ 9 + 27(y + t) ≥ 50yt.
4
Because y + t ≤ 2, thus yt ≤ 1, Hence
√
9 + 27(y + t) ≥ 54 yt ≥ 54yt ≥ 50yt
x+z
x+z
which yields that f (x, y, z, t) ≤ f
, y,
, t . By S.M.V theorem, it’s enough
2
2
to prove the inequality in case x = y = z = a ≥ 1 ≥ t = 4 − 3z and in that case, we
obtain
(1 + 3a)3 (1 + 3(4 − 3a)) ≤ 125 + 131a3 (4 − 3a).
After expending and collecting terms, the above inequality becomes
150a4 − 416a3 + 270a2 + 108a − 112 ≤ 0
⇔ (a − 1)2 (3a − 4)(50a + 28) ≤ 0,
which is clearly true. We have equality if a = 1 or a = 4/3, which is equivalent to the
two cases of equality showed at the beginning of solution. q
The below problem is full of the color and character of this method.
Problem 4. Let a1 , a2 , ..., an be non-negative real numbers satisfying that a1 a2 ...an =
1. Prove that
1
1
1
3n
+
+ ... +
+
≥ n + 3,
a1
a2
an
a1 + a2 + ... + an
for all positive integer n ≥ 4.
(Pham Kim Hung)
Solution. Without loss of generality, we may assume that a1 ≥ a2 ≥ ... ≥ an .
Denote
1
1
1
3n
f (a1 , a2 , ..., an ) =
+
+ ... +
+
,
a1
a2
an
a1 + a2 + ... + an
We will prove that
√
√
f (a1 , a2 , ..., an ) ≥ f (a1 , a2 an , a2 an , a3 , a4 , ..., an−1 )
(∗)
Indeed, this one is equivalent to
√
√
f (a1 , a2 , ..., an ) − f (a1 , a2 an , a2 an , a3 , a4 , ..., an−1 )
2
√
√
3n( a2 − an )2
1
1
= √ −√
−
,
√
a2
an
(a1 + a2 + ... + an )(a1 + 2 a2 an + a3 + ... + an−1 )
hence, it suffices to prove that
√
(a1 + a2 + ... + an )(a1 + 2 a2 an + a3 + ... + an−1 ) ≥ 3na2 an .
Mathematical Reflections 6 (2006)
5
Because a1 ≥ a2 ≥ ... ≥ an , we deduce that
√
(a1 + a2 + ... + an )(a1 + 2 a2 an + a3 + ... + an−1 )
√
≥ (2a2 + (n − 2)an )(a2 + 2 a2 an + (n − 3)an )
p
√
≥ (2 + 2 n − 3)2 2(n − 2)a2 an ≥ 3na2 an ,
for all non-negative integer n ≥ 4. Otherwise, for n = 3, this one is still true (with
a2 ≥ a3 ) because
√
(a2 + 2 a2 a3 + a3 )(2a2 + a3 ) ≥ 9a2 a3 .
Thus (∗) is proved.
Furthermore, (∗) brings us a more important result. By S.M.V. theorem, we have
f (a1 , a2 , ..., an ) ≥ f (a1 , b, b, ..., b),
b=
√
n−1
a2 a3 ...an ,
and the rest is (before replacing n by n + 1 for aesthetics) proving that g(b) ≥ n + 4 .
n
3(n + 1)
n 3(n + 1)bn
n
+
=
b
+
+
b
nb + 1/bn
b
nbn+1 + 1
n−1
n+1
n
3(n + 1)(nb
(nb
+ 1) − (n + 1)nb2n )
g 0 (b) = nbn−1 − 2 +
b
(nbn+1 + 1)2
n
3n(n + 1)
g 0 (b) = nbn−1 − 2 +
(bn−1 − b2n ).
b
(nbn+1 + 1)2
g(b) = bn +
Hence
g 0 (b) = 0 ⇔ (bn+1 − 1)((nbn+1 + 1)2 − 3(n + 1)bn+1 ) = 0.
By AM − GM Inequality, we get (nbn+1 + 1)2 ≥ 4nbn+1 ≥ 3(n + 1)bn+1 .
Thus g 0 (b) ≤ 0 ∀b ≤ 1 and g 0 (1) = 0, imply g(b) ≥ g(1) = n + 4, which is exactly the
desired result. The equality holds for a1 = a2 = ... = an = 1. q
Notice that in the above problem, the best constant to replace 3n is 4(n − 1),
hence we need to add the condition n ≥ 4. But the solution for each one is similar.
Problem 5. Let a, b, c, d be positive real numbers adding up to 4 and k is a given
positive real number. Find the maximum value of the expression
(abc)k + (bcd)k + (cda)k + (dab)k .
(Pham Kim Hung)
Solution. Without loss of generality, we may assume a ≥ b ≥ c ≥ d. Firstly, suppose
a+c
a−c
that 1 ≤ k ≤ 3. Let t =
and u =
, we get a = t + u, c = t − u.
2
2
(abc)k + (bcd)k + (cda)k + (dab)k = (bk + dk )(ac)k + bk dk (ak + ck ).
6
Let s = b−k + d−k and consider the function
f (u) = s(ac)k + ak + ck = s(t2 − u2 )k + (t + u)k + (t − u)k .
We will prove that f (u) ≤ f (0). Indeed
f 0 (u) = −2kus(t2 − u2 )k−1 + k(t + u)k−1 − k(t − u)k−1
(t − u)−k+1 − (t + u)−k+1
= ku(t2 − u2 )k−1 −2s +
.
2u
Because a ≥ b ≥ c ≥ d, so d ≤ t − u. On the other hand, because k ≤ 3 and
δ(x) = x−k+1 is a decreasing function, so Lagrange Theorem implies that
(t + u)−k+1 − (t − u)−k+1
= δ 0 (β) ≥ (−k + 1)(t − u)−k
2u
(t − u)−k+1 − (t + u)−k+1
≤ (k − 1)(t − u)−k
⇒
2u
(t − u)−k+1 − (t + u)−k+1
−2
k−1
⇒ −2s +
≤ k +
≤ 0.
2u
d
(t − u)k
Thus f (u) ≤ f (0). By S.M.V. Theorem, we only need to prove the problem in case
a = b = c = t ≥ d. Consider the function
g(t) = t3k + 3t2k (4 − 3t)k ,
We will prove g(t) ≤ max g(1), g( 34 ) . Indeed,
g 0 (t) = 3kt3k−1 + 6kt2k−1 (4 − 3t)k − 9kt2k (4 − 3t)k−1
g 0 (t) = 0 ⇔ tk + 2(4 − 3t)k = 3t(4 − 3t)k
k
3t
t
+2=
.
⇔
4 − 3t
4 − 3t
Let r = r(t) =
t
⇒ r(t) is a monotonically increasing function, that yields
4 − 3t
g 0 (t) = 0 ⇔ rk + 2 = 3r.
Clearly, the above equation has no more than two positive real roots. Since g 0 (1) = 0,
we deduce that
3k !
4
4
g(t) ≤ max g(1), g
= max 4,
.
3
3
From the above result, we find out (by taking k = 1)
abc + bcd + cda + dab ≤ 4,
Mathematical Reflections 6 (2006)
7
Hence for all k ≤ 1 then
(abc)k + (bcd)k + (cda)k + (dab)k ≤ 4.
Consider the case k ≥ 3, we have
(abc)k + (bcd)k + (cda)k + (dab)k ≤ (ab)k (c + d)k
⇔ (ab)k (c + d)k − ck − dk ≥ (ak + bk )ck dk .
This last inequality is obviously true because (c + d)k − ck − dk ≥ kck−1 ≥ 2ck .
Moreover, applying the AM − GM Inequality
(ab)k (c + d)k ≤
a+b+c+d
3
3k
=
3k
4
.
3
Hence the inequality has been proved completely
(abc)k + (bcd)k + (cda)k + (dab)k ≤ max 4,
4 3k
3
. q
To conclude try applying the method to the following examples to improve your
skill.
Problem 6. Let a, b, c, d be non-negative real numbers such that a + b + c + d = 4.
Prove the following inequality
16 + 2abcd ≥ 3(ab + ac + ad + bc + bd + cd).
Problem 7. Let a, b, c, d, e ≥ 0 satisfy that a + b + c + d + e = 5. Prove that
4(a2 + b2 + c2 + d2 + e2 ) + 5abcde ≥ 25.
Problem 8. Let a, b, c, d be non-negative real numbers and a + b + c + d = 4. Prove
that
(1 + a2 )(1 + b2 )(1 + c2 )(1 + d2 ) ≥ (1 + a)(1 + b)(1 + c)(1 + d).
(Pham Kim Hung)
8