Download Reversing Arithmetic mean – Geometric mean inequality

Reversing the Arithmetic mean – Geometric mean inequality
Tien Lam Nguyen
Abstract. In this paper, we discuss some inequalities which are obtained by adding a non-negative expression to one of
the sides of the AM-GM inequality. In this way, we can reverse the AM-GM inequality.
1. Introduction
The AM-GM inequality is one of the best inequalities and it is very useful for proving
inequalities. We will recall the AM-GM inequality as follows
Theorem 1. For all non-negative real numbers a1 , a2 ,..., an , we have
a1 + a2 + ... + an n
≥ a1a2 ...an , (1) where n is a natural number greater than 1 .
n
Equality holds if and only if a1 = a2 = ... = an .
Now, we will observe a question: “ How can we reverse AM-GM inequality ?”. The first
thing that springs to mind is adding a non-negative expression to the right-hand-side of the
AM-GM inequality. It is natural that the obtained inequality becomes equality if and only if
AM-GM inequality becomes equality, that is, if we add an expression M := f ( a1 , a2 ,..., an ) (
M is a function of n variables a1 , a2 ,..., an ) to the right-hand-side of AM-GM inequality,
then M must satisfy the two following conditions simultaneously
i) M := f ( a1 , a2 ,..., an ) ≥ 0
for non-negative numbers a1 , a2 ,..., an .
ii) M := f ( a1 , a2 ,..., an ) = 0
if and only if a1 = a2 = ... = an .
So, we can choose many expressions such as:
M 1 = k1 ⋅
∑
1≤i < j ≤ n
| ai − a j |
n
M 2 = k2 ⋅ ∑ | ai − ai +1 |
;
;
i =1
1
2
⎛
2⎞
M 4 = k4 ⋅ ⎜ ∑ ( ai − a j ) ⎟ , etc…
⎝ 1≤i < j ≤ n
⎠
(with the condition an +1 ≡ a1 ).
────────────────
The author thanks Thanh Ha Le for her help in the preparation of this paper.
1
M 3 = k3 ⋅ max {| ai − a j |}
1≤ i < j ≤ n
or
2. Two inequalities
Proposition 2. Given a natural number n ≥ 1 . Then, the smallest real number k such that for
all non-negative numbers a1 , a2 ,..., an we have the following inequality
a1 + a2 + ⋅⋅⋅ + an n
≤ a1a2 ⋅⋅⋅ an + k ⋅ ∑ | ai − a j | ,
(2)
n
1≤i < j ≤ n
1
is .
n
Proof. Suppose that (2) is satisfied for all a1 , a2 ,..., an ≥ 0 . Try a1 = a2 = ... = an −1 > 0 and
1
1
an = 0 , from (2) we deduce k ≥ . Now, we will prove that (2) is correct when k = , i.e.,
n
n
a1 + a2 + ⋅⋅⋅ + an n
1
≤ a1a2 ⋅⋅⋅ an + ⋅ ∑ | ai − a j | for all non-negative real numbers
n
n 1≤i < j ≤ n
a1 , a2 , ..., an .
Without losing generality, we assume that a1 ≥ a2 ≥ ⋅⋅⋅ ≥ an . Thus, the inequality we need
to prove is equivalent to
a1 + a2 + ⋅ ⋅ ⋅ + an n
1
≤ a1 a2 ⋅ ⋅ ⋅ an + ⋅ ( ( n − 1) a1 + ( n − 3) a2 + ( n − 5 ) a3 + ⋅ ⋅ ⋅ +
n
n
⎛
⎛
⎡n⎤ ⎞
⎡n⎤ ⎞
+ ⎜⎜ n − 2.⎢ ⎥ + 1⎟⎟a⎡ n ⎤ − ⎜⎜ n − 2.⎢ ⎥ − 1⎟⎟a⎡ n ⎤ − ... − (n − 1)an )
⎣ 2 ⎦ ⎠ ⎢⎣ 2 ⎥⎦ ⎝
⎣ 2 ⎦ ⎠ ⎢⎣ 2 ⎥⎦ +1
⎝
⎛
⎛
⎞
⎡n⎤⎞
⎡n⎤
or ⎜ n − 2 ⋅ ⎢ ⎥ ⎟ a⎡ n ⎤ + ⎜ n − 2 ⋅ ⎢ ⎥ − 2 ⎟ a⎡ n ⎤ + ⋅⋅⋅ + nan ≤
⎣ 2 ⎦ ⎠ ⎢⎣ 2 ⎥⎦ +1 ⎝
⎣2⎦
⎝
⎠ ⎢⎣ 2 ⎥⎦ + 2
⎛
⎡n⎤⎞
≤ ⎜ n − 2 ⋅ ⎢ ⎥ ⎟ a⎡ n ⎤ + ⋅⋅⋅ + ( n − 2 ) a1 + n ⋅ n a1a2 ⋅⋅⋅ an
⎣ 2 ⎦ ⎠ ⎢⎣ 2 ⎥⎦
⎝
where [ x ] is the largest integer which is smaller than x . It is easy to see that the last
inequality is correct since a1 ≥ a2 ≥ ⋅⋅⋅ ≥ an . The proof is completed.
Corollary 3. For all non-negative real numbers a1 , a2 ,..., an , we always have
⎛n⎞
⎜ ⎟
2
a1 + a2 + ⋅⋅⋅ + an n
≤ a1a2 ⋅⋅⋅ an + ⎝ ⎠ ⋅ max {| ai − a j |} , (3)
n
n 1≤i < j ≤ n
⎛n⎞
n!
where ⎜ ⎟ =
.
⎝ 2 ⎠ 2!( n − 2 ) !
⎛n⎞
Proof. For 1 ≤ i, j ≤ n , we have ⎜ ⎟ terms | ai − a j | and note that | ai − a j | is not greater
⎝2⎠
than max {| ai − a j |} , (3) follows immediately from (2).
1≤ i < j ≤ n
2
a1 + a2 + ⋅ ⋅ ⋅ + an n
⎛ n − 1 ⎞ n −1
≤ a1 a2 ⋅ ⋅ ⋅ an + ⎜
⎟ ⋅ ∑ | ai − ai +1 | ,
n
⎝ n ⎠ i =1
for all real numbers a1 , a2 ,..., an ≥ 0 .
Proof. Using the inequalities | ai − a j | ≤ | ai − ai +1 | + | ai +1 − ai + 2 | + ⋅ ⋅ ⋅ + | a j −1 − a j |
Corollary 4.
(4)
for all pairs ( i, j ) satisfying 1 ≤ i, j ≤ n , we can easily deduce that
n −1
∑
1≤ i < j ≤ n
| ai − a j | ≤ ( n − 1) ⋅ ∑ | ai − ai +1 | . Thus, from (2) we deduce (4).
i =1
Proposition 5. Given a natural number n greater than 1, the smallest real number k such
that for all non-negative numbers a1 , a2 , ..., an we have the inequality
a1 + a2 + ⋅⋅⋅ + an n
(5)
≤ a1a2 ⋅⋅⋅ an + k ⋅ max {| ai − a j |} ,
1≤i < j ≤ n
n
n −1
.
is
n
Proof. Since (5) is satisfied for all real numbers a1 , a2 ,..., an ≥ 0 trying a1 = a2 = ⋅⋅⋅ = an −1 > 0
n −1
.
and an = 0 , from (5) we deduce k ≥
n
n −1
We will prove that (5) is correct when k =
, that is,
n
a1 + a2 + ⋅⋅⋅ + an n
⎛ n −1 ⎞
≤ a1a2 ⋅⋅⋅ an + ⎜
⎟ ⋅ max {| ai − a j |} for all non-negative real numbers
n
⎝ n ⎠ 1≤i < j ≤ n
a1 , a2 ,..., an .
Without losing generality, we assume that a1 ≥ a2 ≥ ⋅ ⋅ ⋅ ≥ an ≥ 0 . Then,
max {| ai − a j |} = a1 − an . Hence, the inequality which needs be proved is equivalent to
1≤i < j ≤ n
a1 + a2 + ⋅⋅⋅ + an ≤ n ⋅ n a1a2 ⋅⋅⋅ an + ( n − 1)( a1 − an )
or
a2 + a3 + ⋅⋅⋅ + an −1 + nan ≤ ( n − 2 ) a1 + n ⋅ n a1a2 ⋅⋅⋅ an since a1 ≥ a2 ≥ ⋅⋅⋅ ≥ an ,
We have a1 + a2 + ⋅⋅⋅ + an −1 ≤ ( n − 2 ) a1 and an ≤ n a1a2 ⋅⋅⋅ an .
Therefore, a2 + a3 + ⋅ ⋅ ⋅ + an −1 + nan ≤ n ⋅ n a1 a2 ⋅ ⋅ ⋅ an + ( n − 2 ) a1 . This completes the proof.
Corollary 6. For all non-negative real numbers a1 , a2 ,..., an we have
a1 + a2 + ⋅ ⋅ ⋅ + an n
⎛ n −1⎞ n
≤ a1 a2 ⋅ ⋅ ⋅ an + ⎜
⎟ ⋅ ∑ | ai − ai +1 | , (6)
n
⎝ n ⎠ i =1
with the condition an +1 ≡ a1 .
Proof. Suppose that max {| ai − a j |} = a1 − an .
1≤ i , j ≤ n
Using the inequality | a1 − an | ≤ | a1 − a2 | + | a2 − a3 | + ⋅⋅⋅ + | an −1 − an | , from (5), we get
a1 + a2 + ⋅ ⋅ ⋅ + an n
⎛ n − 1 ⎞ n −1
≤ a1 a2 ⋅ ⋅ ⋅ an + ⎜
⎟ ⋅ ∑ | ai − ai +1 | .
n
⎝ n ⎠ i =1
a + a2 + ⋅ ⋅ ⋅ + an n
⎛ n −1⎞
On the other hand, we have 1
≤ a1 a2 ⋅ ⋅ ⋅ an + ⎜
⎟ ⋅ | a1 − an | .
n
⎝ n ⎠
3
Adding two above inequalities side by side, we get
⎛ a + a + ⋅⋅⋅ + an ⎞
⎛ n −1 ⎞ n
n a a ⋅⋅⋅ a +
2⋅⎜ 1 2
2
≤
⋅
1 2
n
⎜
⎟ ⋅ ∑ | ai − ai +1 |
⎟
n
⎝ n ⎠ i =1
⎝
⎠
a1 + a2 + ⋅ ⋅ ⋅ + an n
⎛ n −1⎞ n
or
≤ a1 a2 ⋅ ⋅ ⋅ an + ⎜
⎟ ⋅ ∑ | ai − ai +1 | .
n
⎝ 2n ⎠ i =1
Corollary 7. For all non-negative real numbers a1 , a2 ,..., an we have
a1 + a2 + ⋅ ⋅ ⋅ + an n
⎛ n −1⎞ ⎛ n
2 ⎞
≤ a1 a2 ⋅ ⋅ ⋅ an + ⎜
⎟ ⋅ ⎜ ∑ ( ai − ai +1 ) ⎟
n
⎠
⎝ 2 n ⎠ ⎝ i =1
with the condition an +1 ≡ a1 .
1
2
,
(7)
2
n
n
⎛ n
⎞
Proof. By using the Cauchy–Schwarz inequality ⎜ ∑ ai bi ⎟ ≤ ∑ ai2 ⋅ ∑ bi2 ,we have
i =1
i =1
⎝ i =1
⎠
1
2
⎛ n
2⎞
| ai − ai +1 | ≤ n ⋅ ⎜ ∑ ( ai − ai +1 ) ⎟ . So, from this and (6) we immediately have (7).
∑
i =1
⎝ i =1
⎠
n
3. Another proof of (2) in the case n = 3
In this section, we will show a new proof of (2) for the case n = 3 .
Lemma 8. For all non-negative real numbers a, b , we have
3
a 3 + b3
⎛ a+b⎞ 3 3 3
.
⎜
⎟ + ⋅| a −b | ≥
2
⎝ 2 ⎠ 8
Proof. Without losing generality, we assume that a ≥ b ≥ 0 . Then, (8) is equivalent to the
3
a 3 + b3
⎛a+b⎞ 3 3
3
following inequality ⎜
≥
+
⋅
(
a
−
b
)
⎟
2
⎝ 2 ⎠ 8
correct because a ≥ b ≥ 0 . The lemma is proved.
or 3ab ( a + b ) ≥ 6b3 . This is
Problem 9. For all non-negative real numbers a1 , a2 , a3 we have
a1 + a2 + a3 3
1
(9).
≤ a1a2 a3 + ⋅ (| a1 − a2 | + | a2 − a3 | + | a3 − a1 |) .
3
3
Proof. We introduce the following notations: b13 = a1 , b23 = a2 , b33 = a3 , then b1 , b2 , b3 ≥ 0
b13 + b23 + b33
1
≤ b1b2b3 + ⋅ (| b13 − b23 | + | b23 − b33 | + | b33 − b13 |) .
3
3
Using the lemma for three pairs of numbers ( b1 , b2 ) , ( b2 , b3 ) and ( b3 , b1 ) , we have
and (9) becomes
3
b13 + b23
⎛ b1 + b2 ⎞ 3 3 3
≥
,
+
⋅
|
b
−
b
|
1
2
⎜
⎟
2
⎝ 2 ⎠ 8
3
b23 + b33
⎛ b2 + b3 ⎞ 3 3
3
≥
,
+
⋅
|
b
−
b
|
3
⎜ 2 ⎟ 8 2
2
⎝
⎠
4
(8)
3
b33 + b13
⎛ b3 + b1 ⎞ 3 3
3
⎜ 2 ⎟ + 8 ⋅ | b3 − b1 | ≥ 2 ,
⎝
⎠
Adding above inequalities side by side, we get
3
3
3
⎛ b1 + b2 ⎞ ⎛ b2 + b3 ⎞ ⎛ b3 + b1 ⎞ 3
3
3
3
b1 + b2 + b3 ≤ ⎜
+⎜
+ ⋅ (| b13 − b23 | + | b23 − b33 | + | b33 − b13 |)
+⎜
⎟
⎟
⎟
⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 8
3
3
3
i =1
3
i =1
3
i =1
or 2 ⋅ ∑ bi3 ≤ ∑ bi bi +1 ⋅ ( bi + bi +1 ) + ∑ | bi − bi +1 | (with the condition bn +1 ≡ b1 when n = 3 )
Thus,
3
3
∑ bi3 + ∑ bi3 + 3b1b2b3 ≤ 3b1b2b3 + ∑ | bi − bi+1 | + ∑ bibi+1 ( bi + bi+1 ) (*)
i =1
i =1
i =1
i =1
3
According to Schur’s inequality, we have
3
∑ bi3 + 3b1b2b3 ≥ ∑ bibi+1 ( bi + bi+1 )
i =1
Hence, from (*) we deduce
3
∑b
i =1
3
or
3
i
i =1
3
≤ 3b1b2b3 + ∑ | bi3 − bi3+1 |
i =1
1
1 3
⋅ ∑ bi3 ≤ b1b2 b3 + ⋅ ∑ | bi3 − bi3+1 | .
3 i =1
3 i =1
4. An open problem.
The following inequality is a problem of USA-TST (2000)
a+b+c 3
− abc ≤ max
3
where a, b, c are non-negative real numbers.
Problem 10. Prove that
{(
) (
2
a− b ,
b− c
) ,(
2
c− a
)}
2
(10)
The inequality (10) is a beautiful inequality and there are some ways to prove it
(eg: see[2]). A natural question arises, can the power 2 of (10) be replaced by another power?
Now, we will observe the following problem.
Lemma 11. Let f ( x) = x − np n x + q , where p > 0, x ∈ [a, b], n ∈ N , n ≥ 2 then
f ( x) ≤ max{ f (a ), f (b)}.
Proof. The lemma is proved easily by using the property of convex function. It is left as an
exercise for the reader.
Problem 12. Prove that for all a1 , a 2 ,..., a n ≥ 0 we have the following inequality
2
a1 + a 2 + ... + a n n
⎛ n −1⎞
≤ a1 a 2 ...a n + ⎜
⎟ max ai − a j
n
⎝ n ⎠ 1≤i < j ≤ n
where n ≥ 2, n ∈ N .
Proof. Without losing generality, we assume that a1 ≥ a 2 ≥ ... ≥ a n . So
{(
max
1≤ i < j ≤ n
{( a − a ) }= ( a − a ) = a − 2
2
i
j
2
1
n
1
)}
(11)
a1 a n + a n .
Thus (11) is equivalent to the following inequality
a1 + a 2 + ... + a n ≤ n.n a1 a 2 ...a n + (n − 1) a1 − 2 a1 a n + a n
(
)
or a 2 + a3 + ... + a n −1 − n.n a1 a 2 ...a n −1 a n − (n − 2)a1 − (n − 2)a n + 2(n − 1) a1 a n ≤ 0. Put
F (a1 , a2 ,..., an )
:= a2 + a3 + ... + an −1 − n. n a1 a2 ...an −1 an − (n − 2)a1 − (n − 2)an + 2(n − 1) a1an .
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It is easy to see that the form of F (a1 , a 2 ,..., a n ) is the same as the form of f (x) of lemma
11. Note that ai ∈ [a n , a1 ], ∀i = 2, n − 1 , so using the lemma 11 for ai , i ∈ {2,3,..., n − 1} we
have F (a1 , a 2 ,..., a n ) ≤ max{F (a1 , a 2 ,..., ai −1 , a1 , ai +1 ,..., a n ), F (a1 , a 2 ,..., ai −1 , a n , ai +1 ,..., a n )} .
Using the above property for i = 2,3,..., n − 1, we get
F (a1 , a 2 ,..., a n ) ≤ max F (t1 , t 2 ,..., t n ) where t i ∈ {a1 , a n }, ∀i = 1,2...., n.
ti ∈{a1 , a n }
Suppose that there exist m numbers t i that equal a1 , (0 ≤ m ≤ n ) so
F (t1 , t 2 ,..., t n ) = (m − n + 2)a1 − ma n − n.n a1m +1 a nn − m −1 + 2(n − 1) a1 a n .
We will prove that F (t1 , t 2 ,..., t n ) ≤ 0. Indeed, if
n − m −1 n
> , we put
2
n
G (a1 ) = F (t1 , t 2 ,..., t n ) = (m − n + 2)a1 − ma n − n.n a1m +1 a nn − m −1 + 2(n − 1) a1 a n ,
G '(a1 ) = m − n + 2 − (m + 1). n
⎛ a
a n − m −1
= (m + 1) ⎜ n − n nn − m −1
⎜ a1
a1
⎝
ann − m −1
a
+ (n − 1) n
n − m −1
a1
a1
⎞
⎛ a
⎞
⎟ + (n − m − 2) ⎜ n − 1⎟
⎜ a
⎟
⎟
⎝ 1
⎠
⎠
an
a n − m −1
≤ n nn − m −1 and
a1
a1
n − m −1
that G (a1 ) ≤ G (a n ) = 0. If
≤
n
Since
an
≤1 ,
a1
an
≤ 1 . Hence G ' (a1 ) ≤ 0 .From this, we deduce
a1
1
, we put
2
H (a n ) = F (t1 , t 2 ,..., t n ) = (m − n + 2)a1 − ma n − n.n a1m +1 a nn − m −1 + 2(n − 1) a1 a n
Similarly, we also deduce that H (a n ) ≤ 0.
The proof is completed.
Problem 13. Find all real numbers k such that we have the inequality
a+b+c 3
− abc ≤ max | a − b |k ,| b − c |k ,| c − a |k
3
for all non-negative real numbers a, b, c .
{
}
(12)
Solution. The answer for this is k = 2 . Indeed, try b = c = 0 and a > 1 , from (11) we deduce
k
a
k ln 3
for every number a > 1 . Let a → ∞ , we get k ≥ 2 .
that ≤ a 2 . Hence, 1 − ≤
3
2 ln a
When k = 2 , (10) is correct (problem 10). We will prove that (11) is incorrect with k > 2 .
4
1
When k > 2 , try a = 2 , b = c = 2 (where n is positive real number), we get
n
n
3
2− 4
1⎫ 1
1
⎧1
≤ max ⎨ k , 0, k ⎬ = k . Hence, n k − 2 ≤
. This is not true as n tends to infinity.
2
n
n ⎭ n
2− 3 4
⎩n
Therefore, there is only real number k ( k = 2 ) satisfying (11).
It is clear that when k = 2 , the degree of left-hand-side of (11). Almost classical
inequalities are homogeneous inequalities of the first degree.
6
We conclude this paper with an open problem: for all non-negative real numbers
a1 , a2 ,..., an (where n is a natural greater than 1), suppose that there exists a function
f := f ( a1 , a2 ,..., an ) satisfying the following three conditions simultaneously
i) f ( a1 , a2 ,..., an ) ≥ 0 for all non-negative real numbers a1 , a2 ,..., an .
ii) f ( a1 , a2 ,..., an ) = 0 if and only if a1 = a2 = ... = an .
iii)
a1 + a2 + ⋅⋅⋅ + an n
≤ a1a2 ...an + f ( a1 , a2 ,..., an ) for all real numbers a1 , a2 ,..., an ≥ 0 .
n
Prove or disprove that f ( a1 , a2 ,..., an ) is a homogeneous function of the first degree.
References
[1] Kin Y Li, Schur’s Inequality, Mathematical Excalibur Vol.10, No.5 p2-4(2006), Hong Kong.
[2] N.V.Mau, Các bài toán nội suy và áp dụng, (in Vietnamese) Education Publishing House,Vietnam, 2007.
[3] P.V.Thuan, L.Vi, Bất đẳng thức: Suy luận và khám phá, (in Vietnamese) Vietnam national university
Publishing House, Hanoi, 2007.
Nguyễn Tiến Lâm, student of K50A1S
Department of Mathematices – Mechanics – Informatics
College of science, Vietnam national university, Hanoi
Email address : [email protected]
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