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Losonczi We generalize and sharpen Acz´el’s inequality and Popoviciu’s inequality by means of two classical inequalities, a unified improvement of Acz´el’s inequality and Popoviciu’s ine

Volume 2007, Article ID 72173, 9 pages

doi:10.1155/2007/72173

Research Article

Improvement of Aczél’s Inequality and Popoviciu’s Inequality

Shanhe Wu

Received 30 December 2006; Accepted 24 April 2007

Recommended by Laszlo I Losonczi

We generalize and sharpen Acz´el’s inequality and Popoviciu’s inequality by means of two classical inequalities, a unified improvement of Acz´el’s inequality and Popoviciu’s inequality is given As application, an integral inequality of Acz´el-Popoviciu type is es-tablished

Copyright © 2007 Shanhe Wu This is an open access article distributed under the Cre-ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

In 1956, Acz´el [1] proved the following result:



a2−

n



i =2

a2

i



b2−

n



i =2

b2

i



≤



a1b1−

n



i =2

a i b i

 2

wherea i,b i(i =1, 2, , n) are positive numbers such that a2−n

i =2a2i > 0 or b2−n

i =2b2i >

0 This inequality is called Acz´el’s inequality

It is well known that Acz´el’s inequality has important applications in the theory of functional equations in non-Euclidean geometry In recent years, this inequality has at-tracted the interest of many mathematicians and has motivated a large number of re-search papers involving different proofs, various generalizations, improvements, and ap-plications (see [2–11] and references therein) We state here a brief history on improve-ment of Acz´el’s inequality

Popoviciu [12] first presented an exponential extension of Acz´el’s inequality, as fol-lows

Theorem 1.1 Let p > 0, q > 0, 1/ p + 1/q = 1, and let a i , b i (i =1, 2, , n) be positive numbers such that a1p −n

i =2a i p > 0 and b q1−n

i =2b i q > 0 Then



a1p −

n



i =2

a i p

 1/ p

b1q −

n



i =2

b q i

 1/q

≤ a1b1−

n



i =2

Wu and Debnath [13] generalized inequality (1.2) in the following form

Theorem 1.2 Let p > 0, q > 0, and let a i , b i(i =1, 2, , n) be positive numbers such that

a p1−n

i =2a p i > 0 and b1q −n

i =2b q i > 0 Then



a1p −

n



i =2

a i p

 1/ p

b q1−

n



i =2

b i q

 1/q

≤ n1−min{ p −1 +q −1 ,1} a1b1−

n



i =2

a i b i (1.3)

In a recent paper [14], Wu established a sharp and generalized version of Popoviciu’s inequality as follows

Theorem 1.3 Let p > 0, q > 0, 1/ p + 1/q ≥ 1, and let a i , b i (i =1, 2, , n) be positive numbers such that a1p −n

i =2a i p > 0 and b q1−n

i =2b i q > 0 Then



a p1−

n



i =2

a i p

 1/ p

b1q −

n



i =2

b q i

 1/q

≤ a1b1−

 n



i =2

a i b i



− a1b1 max{ p, q, 1 }

 n



i =2



a i p

a1p

− b

q i

b1q

 2

.

(1.4)

In this paper, we show a new sharp and generalized version of Popoviciu’s inequal-ity, which is a unified improvement of Acz´el’s inequality and Popoviciu’s inequality In

Section 4, the obtained result will be used to establish an integral inequality of Acz´el-Popoviciu type

2 Lemmas

In order to prove the theorem inSection 3, we first introduce the following lemmas Lemma 2.1 (generalized H¨older inequality [15, page 20]) Let a i j > 0, λ j ≥0 (i =1, 2, ,

n, j =1, 2, , m), and let λ1+λ2+···+λ m = 1 Then

m



j =1

n

i =1

a i j

λ j

≥

n



i =1

m



j =1

a λ j

with equality holding if and only if a11/a1j = a21/a2j = ··· = a n1 /a n j (j =2, 3, , m) for

λ1λ2··· λ n = 0.

Lemma 2.2 (mean value inequality [16, page 17]) Let x i > 0, λ i > 0 (i =1, 2, , n) and let

λ1+λ2+···+λ n = 1 Then

n



i =1

λ i x i ≥

n



i =1

x λ i

with equality holding if and only if x1= x2= ··· = x n

Lemma 2.3 Let p1≥ p2≥ ··· ≥ p m > 0, 1/ p1+ 1/ p2+···+ 1/ p m = 1, 0 < x j < 1 ( j =

1, 2, , m), and let x m+1 = x1, p m+1 = p1 Then

m



j =1

x j+

m



j =1



1− x p j j

 1/ p j

≤1− 1

2 1

m



j =1



x p j

j − x p j+1 j+1

 2

(2.3)

with equality holding if and only if x1p1 = x2p2 = ··· = x p m

m Proof From hypotheses inLemma 2.3, it is easy to verify that

1

p m −1 ≥ ··· ≥ 1

p1

> 0,

1

2 2− 1

2 1 ≥0, 1

2 3− 1

2 2 ≥0, , 1

2 m −1 ≥0, 1

2 1 ≥0, 1

2 1+ 1

2 1+

1

2 2+ 1

2 2+

1

2 3− 1

2 m −2+ 1

2 m −2 +

2 m −1− 1

2 m −2 + 1

2 m −1+ 1

2 m −1+

1

2 m −1 + 1

2 1+ 1

2 1+

1

2 1

= p1

1+ 1

p2+···+ 1

p m =1.

(2.4) Hence, by usingLemma 2.1we obtain

x1p1+

1− x2p2 1/2p1

x2p2+

1− x1p1 1/2p1

x2p2+

1− x2p2 1/2p2 −1/2p1

×x2p2+

1− x3p3

 1/2p2

x3p3+

1− x2p2

 1/2p2

x3p3+

1− x3p3

 1/2p3 −1/2p2

×x p m −2

m −2+

1− x p m −1

m −1

 1/2p m −2

×x p m −1

m −1+

1− x p m −2

m −2

 1/2p m −2

x p m −1

m −1+

1− x p m −1

m −1

 1/2p m −1−1/2p m −2

×x p m −1

m −1+

1− x p m

m  1/2p m −1

x p m

m +

1− x p m −1

m −1

 1/2p m −1

x p m

m +

1− x p m

m  1/2p m −1/2p m −1

×x p m

m +

1− x1p1 1/2p1

x1p1+

1− x p m

m  1/2p1

x p m

m +

1− x p m

m  1/2p m −1/2p1

≥ x1p1/2p1 x2p2/2p1 x2p2/2p2 − p2/2p1 x2p2/2p2 ··· x p m −1/2pm −2

m −1 x p m −1/2pm −1− p m −1/2pm −2

m −1 x p m −1/2pm −1

m −1

× x p m /2p m −1

m x p m /2p m − p m /2p m −1

m x p m /2p1

m x p m /2p m − p m /2p1

m x1p1/2p1

+

1− x1p1

 1/2p1

1− x2p2

 1/2p1

1− x2p2

 1/2p2 −1/2p1

1− x2p2

 1/2p2

···1− x p m −1

m −1

 1/2p m −2 

1− x p m −1

m −1

 1/2p m −1−1/2p m −2 

1− x p m −1

m −1

 1/2p m −1

×1− x p m

m  1/2p m −1 

1− x p m

m  1/2p m −1/2p m −1 

1− x p m

m  1/2p1

1− x p m

m  1/2p m −1/2p1

×1− x1p1

 1/2p1

,

(2.5)

which is equivalent to

1−x1p1 − x2p2

 2 1/2p1

1−x2p2 − x3p3

 2 1/2p2

···1−(x p m −1

m −1− x p m

m  2 1/2p m −1

1−x p m

m − x1p1 2 1/2p1

≥ x1x2··· x m+

1− x1p1 1/ p1

1− x2p2 1/ p2

···1− x p m

m  1/ p m

.

(2.6)

On the other hand, it follows fromLemma 2.2that

1

2 1

1−x1p1 − x2p2 2

2 2

1−x2p2 − x3p3 2

2 m −1

1−x p m −1

m −1− x p m

m  2

2 1

1−x p m

m − x1p1 2

+

1

2 2+ 1

2 3+···+ 1

2 m −1+ 1

≥1−x1p1 − x2p2 2 1/2p1

1−x2p2 − x3p3 2 1/2p2

···1−x p m −1

m −1− x p m

m  2 1/2p m −1

1−x p m

m − x1p1 2 1/2p1

,

(2.7)

this yields

1−x1p1 − x2p2 2 1/2p1

1−x2p2 − x3p3 2 1/2p2

···1−x p m −1

m −1− x p m

m  2 1/2p m −1

1−x p m

m − x1p1 2 1/2p1

≤

1

p1+ 1

p2+···+ 1

2 1



x1p1 − x2p2 2

2 2



x2p2 − x3p3 2

2 m −1



x p m −1

m −1− x p m

m  2

2 1



x p m

m − x1p1 2

≤1− 1

2 1



x1p1 − x2p2 2

+

x2p2 − x3p3 2

+···+

x p m −1

m −1+x p m

m  2 +

x p m

m − x1p1 2

.

(2.8) Combining inequalities (2.6) and (2.8) leads to inequality (2.3) In addition, from Lemmas2.1and2.2, we can easily deduce that the equality holds in both (2.6) and (2.8)

if and only ifx1p1 = x2p2 = ··· = x p m

m , and thus we obtain the condition of equality in (2.3)

3 Improvement of Acz´el’s inequality and Popoviciu’s inequality

Theorem 3.1 Let p1≥ p2≥ ··· ≥ p m > 0, 1/ p1+ 1/ p2+···+ 1/ p m = 1, a i j > 0, a p j

1j −

n

i =2a p j

i j > 0 (i =1, 2, , n, j =1, 2, , m), and let p m+1 = p1, a im+1 = a i1(i =1, 2, , n) Then one has the following inequality:

m



j =1



a p j

1j −

n



i =2

a p j

i j

 1/ p j

≤

m



j =1

a1j −

n



i =2

m



j =1

a i j − a11a12··· a1m

2 1

m



j =1

n

i =2



a p j

i j

a1p j j − a

p j+1

i j+1

a1p j+1 j+1

 2

.

(3.1)

Equality holds in ( 3.1 ) if and only if a11p1 /a p j

1j = a p121/a p j

2j = ··· = a n1 p1 /a p j

n j(j =2, 3, , m).

Proof Since by hypotheses inTheorem 3.1we have

0<



a p j

1j −n

i =2a p j

i j

 1/ p j



a p j

1j

it follows fromLemma 2.3, with a substitution x j =(a p j

1j −n

i =2a p j

i j)1/ p j /(a p j

1j)1/ p j (j =

1, 2, , m) in (2.3), that

m



j =1



a p1j j −n

i =2a p i j j

a p j

1j

 1/ p j

+

m



j =1

 n

i =2a i j p j

a p j

1j

 1/ p j

≤1− 1

2 1

m



j =1



a p j

1j −n

i =2a p j

i j

a1p j j − a

p j+1

1j+1 −n

i =2a p j+1

i j+1

a1p j+1 j+1

 2 ,

(3.3)

which is equivalent to

m



j =1



a p j

1j −

n



i =2

a p j

i j

 1/ p j

≤

m



j =1

a1j −

m



j =1

n

i =2

a p j

i j

 1/ p j

− a11a12··· a1m

2 1

m



j =1

n

i =2



a p j

i j

a p j

1j

− a

p j+1

i j+1

a p j+1

1j+1

 2 , (3.4) where equality holds if and only if (n

i =2a p j

i j)/a p j

1j =(n

i =2a p j+1

i j+1)/a p j+1

1j+1(j =1, 2, , m), that

is, if and only ifa p111/a p j

1j =(n

i =2a i1 p1)/(n

i =2a p j

i j) (j =2, 3, , m).

On the other hand, usingLemma 2.1gives

m



j =1

n

i =2

a i j p j

 1/ p j

≥

n



i =2

m



j =1

where equality holds if and only ifa p121/a p j

2j = a31p1 /a p j

3j = ··· = a n1 p1 /a p j

n j(j =2, 3, , m).

Combining inequalities (3.4) and (3.5) leads to the desired inequality (3.1) By means

of the conditions of equality in (3.4) and (3.5), it is easy to conclude that there is equality

in (3.1) if and only ifa11p1 /a1p j j = a21p1 /a2p j j = ··· = a p1 n1 /a n j p j (j =2, 3, , m) This completes

As a consequence ofTheorem 3.1, puttingm =2,p1= p, p2= q, a i1 = a i,a i2 = b i(i =

1, 2, , n) in (3.1), we get the following

Corollary 3.2 Let p ≥ q > 0, 1/ p + 1/q = 1, and let a i , b i(i =1, 2, , n) be positive num-bers such that a1p −n

i =2a i p > 0 and b q1−n

i =2b i q > 0 Then



a1p −

n



i =2

a i p

 1/ p

b q1−

n



i =2

b i q

 1/q

≤ a1b1−

n

i =2

a i b i



− a1b1

p

n

i =2



a i p

a1p

− b

q i

b q1

 2 (3.6)

with equality holding if and only if a p /b q = a p /b q = ··· = a n p /b q n

A simple application ofCorollary 3.2yields the following sharp version of Popoviciu’s inequality

Corollary 3.3 Let p > 0, q > 0, 1/ p + 1/q = 1, and let a i , b i(i =1, 2, , n) be positive numbers such that a1p −n

i =2a i p > 0 and b q1−n

i =2b i q > 0 Then



a1p −

n



i =2

a i p

 1/ p

b q1−

n



i =2

b i q

 1/q

≤ a1b1−

n

i =2

a i b i



− a1b1 max{ p, q }

n

i =2



a i p

a1p − b

q i

b q1

 2 , (3.7)

with equality holding if and only if a1p /b q1= a p2/b q2= ··· = a n p /b q n

Obviously, inequalities (3.1), (3.6), and (3.7) are the improvement of Acz´el’s inequality and Popoviciu’s inequality

4 Integral version of Acz´el-Popoviciu-type inequality

As application of Theorem 3.1, we establish here an interesting integral inequality of Acz´el-Popoviciu type

Theorem 4.1 Let p1≥ p2≥ ··· ≥ p m > 0, 1/ p1+ 1/ p2+···+ 1/ p m = 1, B j > 0 ( j =

1, 2, , m), let f j be positive Riemann integrable functions on [a, b] such that B p j

j −

b

a f j p j(x)dx > 0 for all j =1, 2, , m, and let B m+1 = B1, p m+1 = p1, f m+1 = f1 Then one has the following inequality:

m



j =1



B p j

j −

b

a f p j

j (x)dx

 1/ p j

≤

m



j =1

B j −

b

a

m

j =1

f j(x)



dx − B1B2··· B m

2 1

m



j =1

b

a



f p j

j (x)

B p j j − f

p j+1 j+1 (x)

B p j+1 j+1



dx

 2

.

(4.1)

Proof For any positive integer n, we choose an equidistant partition of [a, b] as

a < a + b − a

n < ··· < a + b − a

n i < ··· < a + b − a

n (n −1)< b,

Δx i = b − a

n , i =1, 2, , n.

(4.2)

Since the hypothesisB p j j − a b f j p j(x)dx > 0 ( j =1, 2, , m) implies that

B p j

j −lim

n →∞

n



i =1

f p j j

a + i(b − a) n

b − a

there exists a positive integerN such that

B p j

j −

n



i =1

f p j j

a + i(b − a) n

n > 0 ∀ n > N, j =1, 2, , m. (4.4)

ApplyingTheorem 3.1, one obtains for anyn > N the following inequality:

m



j =1



B p j

j −

n



i =1

f p j j

a + i(b − a) n

b − a n

 1/ p j

≤

m



j =1

B j −

n



i =1

m

j =1

f j

a + i(b − a) n

b − a n

1/ p1+1/ p2+···+1/ p m

− B1B2··· B m

2 1

m



j =1

n

i =1



1

B p j j

f p j j

a + i(b − a) n

n

B p j+1 j+1 f

p j+1 j+1

a + i(b − a) n

b − a n

 2

.

(4.5)

Note that 1/ p1+ 1/ p2+···+ 1/ p m =1, the above inequality can be transformed to

m



j =1



B p j

j −

n



i =1

f p j j

a + i(b − a) n

b − a n

 1/ p j

≤

m



j =1

B j −

n



i =1

m

j =1

f j

a + i(b − a) n

b − a n

− B1B2··· B m

2 1

m



j =1

n

i =1



1

B p j j f

p j j

a + i(b − a) n

B p j+1 j+1

f j+1 p j+1

a + i(b − a) n

b − a n

 2 , (4.6)

where equality holds if and only iff p j

j (a + i(b − a)/n)/B p j

j = f p j+1 j+1 (a + i(b − a)/n)/B p j+1

j+1 for alli =1, 2, , n ( j =1, 2, , m).

In view of the hypotheses that f jare positive Riemann integrable functions on [a, b]

and p j > 0 ( j =1, 2, , m), we conclude thatm

j =1f j and f p j

j (j =1, 2, , m) are also

integrable on [a, b] Passing the limit as n → ∞in both sides of inequality (4.6), we obtain

Conjecture 4.3 Suppose that p1≥ p2≥ ··· ≥ p m > 0, 1/ p1+ 1/ p2+···+ 1/ p m =1,B j >

0 (j =1, 2, , m), suppose also that f j ∈ L p j[a, b], B p j

j − a b | f j(x) | p j dx > 0 for all j =

1, 2, , m, let B m+1 = B1,p m+1 = p1, m+1 = f1 Then the following inequality holds true:

m



j =1



B p j j −

b

a

f j(x)p j

dx

 1/ p j

≤

m



j =1

B j −

b

a

 m



j =1

f j(x)dx − B1B2··· B m

2 1

m



j =1

b

a

 f j(x)p j

B p j j

−f j+1(x)p j+1

B p j+1 j+1



dx

 2 (4.7)

with equality holding if and only if| f j(x) | p j /B p j

j = | f j+1(x) | p j+1 /B p j+1

j+1 (j =1, 2, , m)

al-most everywhere on [a, b].

As a consequence ofTheorem 4.1, puttingm =2,p1= p, p2= q, B1= A, B2= B, f1=

f , f2= g in (4.1), we obtain the following

Corollary 4.4 Let p ≥ q > 0, 1/ p + 1/q = 1, A > 0, B > 0, and let f , g be positive Riemann integrable functions on [a, b] such that A p − b

a f p(x)dx > 0 and B q − b

a g q(x)dx > 0 Then



A p −

b

a f p(x)dx

 1/ p

B q −

b

a g q(x)dx

 1/q

≤ AB −

b

p

b

a

f p(x)

A p − g q(x)

 2

.

(4.8)

Further, fromCorollary 4.4we have the following

Corollary 4.5 Let p > 0, q > 0, 1/ p + 1/q = 1, A > 0, B > 0, and let f , g be positive Riemann integrable functions on [a, b] such that A p − a b f p(x)dx > 0 and B q − a b g q(x)dx >

0 Then



A p −

b

a f p(x)dx

 1/ p

B q −

b

a g q(x)dx

 1/q

≤ AB −

b

max{ p, q }

b

a

f p(x)

A p − g q(x)

 2

.

(4.9)

Acknowledgment

The author would like to express hearty thanks to the anonymous referees for valuable comments on this paper

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Inequalities and Application, vol 10, no 3, 2007.

[15] E F Beckenbach and R Bellman, Inequalities, Springer, New York, NY, USA, 1983.

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Shanhe Wu: Department of Mathematics, Longyan College, Longyan, Fujian 364012, China

Email address:wushanhe@yahoo.com.cn

A simple application ofCorollary 3.2yields the following sharp version of Popoviciu’s inequality

Corollary... (3.1), (3.6), and (3.7) are the improvement of Acz´el’s inequality and Popoviciu’s inequality

4 Integral version of Acz´el-Popoviciu-type inequality< /b>

As application of Theorem... m

m , and thus we obtain the condition of equality in (2.3)

3 Improvement of Acz´el’s inequality and Popoviciu’s inequality< /b>

Theorem 3.1