Skip to main content

New refinements of generalized Aczél inequality

Abstract

In this article, we present several new refinements of the generalized Aczél inequality. As an application, an integral type of the generalized Aczél-Vasić-Pečarić inequality is refined.

MSC:26D15, 26D10.

1 Introduction

In 1956, Aczél [1] established the following inequality, which is called the Aczél inequality.

Theorem A Let a i >0, b i >0 (i=1,2,,n), a 1 2 i = 2 n a i 2 >0, b 1 2 i = 2 n b i 2 >0. Then

( a 1 2 i = 2 n a i 2 ) ( b 1 2 i = 2 n b i 2 ) ( a 1 b 1 i = 2 n a i b i ) 2 .
(1)

As is well known, the Aczél inequality plays an important role in the theory of functional equations in non-Euclidean geometry, and many authors (see [26] and references therein) have given considerable attention to this inequality and its refinements.

In 1959, Popoviciu [3] generalized the Aczél inequality (1) in the form asserted by Theorem B below.

Theorem B Let p>1, q>1, 1 p + 1 q =1, let a i >0, b i >0 (i=1,2,,n), a 1 p i = 2 n a i p >0, b 1 q i = 2 n b i q >0. Then

( a 1 p i = 2 n a i p ) 1 p ( b 1 q i = 2 n b i q ) 1 q a 1 b 1 i = 2 n a i b i .
(2)

Later, in 1982, Vasić and Pečarić [7] presented the reversed version of inequality (2), which is stated in the following theorem. The inequality is called the Aczél-Vasić-Pečarić inequality.

Theorem C Let p<1 (p0), 1 p + 1 q =1, and let a i >0, b i >0 (i=1,2,,n), a 1 p i = 2 n a i p >0, b 1 q i = 2 n b i q >0. Then

( a 1 p i = 2 n a i p ) 1 p ( b 1 q i = 2 n b i q ) 1 q a 1 b 1 i = 2 n a i b i .
(3)

In another paper, Vasić and Pečarić [8] presented an interesting generalization of inequality (2). The inequality is called the generalized Aczél-Vasić-Pečarić inequality.

Theorem D Let a r j >0, λ j >0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m, and let j = 1 m 1 λ j 1. Then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j j = 1 m a 1 j r = 2 n j = 1 m a r j .
(4)

In 2012, Tian [5] gave the reversed version of inequality (4) in the following form.

Theorem E Let λ 1 0, λ j <0 (j=2,3,,m), j = 1 m 1 λ j 1, and let a r j >0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m. Then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j j = 1 m a 1 j r = 2 n j = 1 m a r j .
(5)

Moreover, in [5] Tian established an integral type of generalized Aczél-Vasić-Pečarić inequality.

Theorem F Let λ 1 >0, λ j <0 (j=2,3,,m), j = 1 m λ j =1, let A j >0 (j=1,2,,m), and let f j (x) (j=1,2,,m) be positive Riemann integrable functions on [a,b] such that A j λ j a b f j λ j (x)dx>0. Then

j = 1 m ( A j λ j a b f j λ j ( x ) d x ) 1 λ j j = 1 m A j a b j = 1 m f j (x)dx.
(6)

The main object of this paper is to give several new refinements of inequality (4) and (5). As an application, a new refinement of inequality (6) is given.

2 New refinements of generalized Aczél inequality

In order to prove the main results in this section, we need the following lemmas.

Lemma 2.1 [5]

Let a r j >0 (r=1,2,,n, j=1,2,,m), let λ 1 be a real number, λ j 0 (j=2,3,,m), and let β=max{ j = 1 m λ j ,1}. Then

r = 1 n j = 1 m a r j λ j n 1 β j = 1 m ( r = 1 n a r j ) λ j .
(7)

Lemma 2.2 [9]

Let a r j >0 (r=1,2,,n, j=1,2,,m), let λ j 0 (j=1,2,,m), and let γ=min{ j = 1 m λ j ,1}. Then

r = 1 n j = 1 m a r j λ j n 1 γ j = 1 m ( r = 1 n a r j ) λ j .
(8)

Lemma 2.3 [10]

If x>1, α>1 or α<0, then

( 1 + x ) α 1+αx.
(9)

The inequality is reversed for 0<α<1.

Lemma 2.4 [10]

Let A 1 , A 2 ,, A m be real numbers, let m be a natural number, and let m2. Then

1 i < j m ( A i A j ) 2 =m ( i = 1 m A i 2 ) ( i = 1 m A i ) 2 .
(10)

Lemma 2.5 Let λ 1 λ 2 λ m <0, let X j >1 (j=1,2,,m), and let m2. Then

j = 1 m ( 1 X j λ j ) 1 λ j + j = 1 m X j { 1 2 m ( m 1 ) [ m ( j = 1 m X j 2 λ j ) ( j = 1 m X j λ j ) 2 ] } m 2 λ 1 .
(11)

Proof From the assumptions in Lemma 2.5, we find

1 ( m 1 ) λ i <0, 1 ( m 1 ) λ j 1 ( m 1 ) λ i 0(1i<jm),

and

1 i < j m [ 1 ( m 1 ) λ i + 1 ( m 1 ) λ i + 1 ( m 1 ) λ j 1 ( m 1 ) λ i ] = 1 i < j m [ 1 ( m 1 ) λ i + 1 ( m 1 ) λ j ] = 1 λ 1 + 1 λ 2 + + 1 λ m .
(12)

Thus, by using inequality (7) we have

1 i < j m [ 1 ( X i λ i X j λ j ) 2 ] 1 ( m 1 ) λ i = 1 i < j m { [ X i λ i + ( 1 X j λ j ) ] 1 ( m 1 ) λ i [ X j λ j + ( 1 X i λ i ) ] 1 ( m 1 ) λ i × [ X j λ j + ( 1 X j λ j ) ] 1 ( m 1 ) λ j 1 ( m 1 ) λ i } 1 i < j m [ ( X i λ i ) 1 ( m 1 ) λ i ( X j λ j ) 1 ( m 1 ) λ i ( X j λ j ) 1 ( m 1 ) λ j 1 ( m 1 ) λ i ] + 1 i < j m [ ( 1 X j λ j ) 1 ( m 1 ) λ i ( 1 X i λ i ) 1 ( m 1 ) λ i ( 1 X j λ j ) 1 ( m 1 ) λ j 1 ( m 1 ) λ i ] = 1 i < j m X i 1 m 1 X j 1 m 1 + 1 i < j m [ ( 1 X i λ i ) 1 ( m 1 ) λ i ( 1 X j λ j ) 1 ( m 1 ) λ j ] = j = 1 m X j + j = 1 m ( 1 X j λ j ) 1 λ j .
(13)

Noting the fact that there are m ( m 1 ) 2 product terms in the expression 1 i < j m [1 ( X i λ i X j λ j ) 2 ], and using the arithmetic-geometric mean’s inequality, we obtain

1 i < j m [ 1 ( X i λ i X j λ j ) 2 ] { 2 m ( m 1 ) 1 i < j m [ 1 ( X i λ i X j λ j ) 2 ] } m ( m 1 ) 2 = [ 1 2 m ( m 1 ) 1 i < j m ( X i λ i X j λ j ) 2 ] m ( m 1 ) 2 .
(14)

Therefore, we have

1 i < j m [ 1 ( X i λ i X j λ j ) 2 ] 1 ( m 1 ) λ i { 1 i < j m [ 1 ( X i λ i X j λ j ) 2 ] } 1 ( m 1 ) λ 1 [ 1 2 m ( m 1 ) 1 i < j m ( X i λ i X j λ j ) 2 ] m 2 λ 1 .
(15)

On the other hand, from Lemma 2.4 we have

[ 1 2 m ( m 1 ) 1 i < j m ( X i λ i X j λ j ) 2 ] m 2 λ 1 = { 1 2 m ( m 1 ) [ m ( j = 1 m X j 2 λ j ) ( j = 1 m X j λ j ) 2 ] } m 2 λ 1 .
(16)

Consequently, from (13), (15), and (16), we obtain the desired inequality (11). □

Lemma 2.6 Let λ m >0, λ 1 λ 2 λ m 1 <0, let 0< X m <1, X j >1 (j=1,2,,m1), and let α=max{ j = 1 m 1 λ j ,1}. If m>2, then

j = 1 m ( 1 X j λ j ) 1 λ j + j = 1 m X j n 1 α { 1 2 ( m 1 ) ( m 2 ) [ ( m 1 ) ( j = 1 m 1 X j 2 λ j ) ( j = 1 m 1 X j λ j ) 2 ] } m 1 2 λ 1 .
(17)

If m=2, then

j = 1 2 ( 1 X j λ j ) 1 λ j + j = 1 2 X j n 1 α { 1 [ 2 ( j = 1 2 X j 2 λ j ) ( j = 1 2 X j λ j ) 2 ] } 1 λ 1 .
(18)

Proof Case I. When m>2. Let us consider the following product:

1 i < j m 1 { [ X i λ i + ( 1 X j λ j ) ] 1 ( m 2 ) λ i [ X j λ j + ( 1 X i λ i ) ] 1 ( m 2 ) λ i × [ X j λ j + ( 1 X j λ j ) ] 1 ( m 2 ) λ j 1 ( m 2 ) λ i } .
(19)

From the hypotheses of Lemma 2.6, it is easy to see that

1 ( m 2 ) λ i <0, 1 ( m 2 ) λ j 1 ( m 2 ) λ i 0(1i<jm1),

and

1 i < j m 1 [ 1 ( m 2 ) λ i + 1 ( m 2 ) λ i + 1 ( m 2 ) λ j 1 ( m 2 ) λ i ] = 1 i < j m 1 [ 1 ( m 2 ) λ i + 1 ( m 2 ) λ j ] = 1 λ 1 + 1 λ 2 + + 1 λ m 1 .
(20)

Then, applying inequality (7), we have

1 i < j m 1 [ 1 ( X i λ i X j λ j ) 2 ] 1 ( m 2 ) λ i = [ X m λ m + ( 1 X m λ m ) ] 1 λ m 1 i < j m 1 { [ X i λ i + ( 1 X j λ j ) ] 1 ( m 2 ) λ i × [ X j λ j + ( 1 X i λ i ) ] 1 ( m 2 ) λ i [ X j λ j + ( 1 X j λ j ) ] 1 ( m 2 ) λ j 1 ( m 2 ) λ i } n α 1 { X m λ m λ m 1 i < j m 1 [ ( X i λ i ) 1 ( m 2 ) λ i ( X j λ j ) 1 ( m 2 ) λ i ( X j λ j ) 1 ( m 2 ) λ j 1 ( m 2 ) λ i ] + ( 1 X m λ m ) 1 λ m 1 i < j m 1 [ ( 1 X j λ j ) 1 ( m 2 ) λ i ( 1 X i λ i ) 1 ( m 2 ) λ i × ( 1 X j λ j ) 1 ( m 2 ) λ j 1 ( m 2 ) λ i ] } = n α 1 { X m 1 i < j m 1 X i 1 m 2 X j 1 m 2 + ( 1 X m λ m ) 1 λ m 1 i < j m 1 [ ( 1 X i λ i ) 1 ( m 2 ) λ i ( 1 X j λ j ) 1 ( m 2 ) λ j ] } = n α 1 [ j = 1 m X j + j = 1 m ( 1 X j λ j ) 1 λ j ] .
(21)

There are ( m 1 ) ( m 2 ) 2 product terms in the expression 1 i < j m 1 [1 ( X i λ i X j λ j ) 2 ], and then we derive from the arithmetic-geometric mean’s inequality that

1 i < j m 1 [ 1 ( X i λ i X j λ j ) 2 ] { 2 ( m 1 ) ( m 2 ) 1 i < j m 1 [ 1 ( X i λ i X j λ j ) 2 ] } ( m 1 ) ( m 2 ) 2 = [ 1 2 ( m 1 ) ( m 2 ) 1 i < j m 1 ( X i λ i X j λ j ) 2 ] ( m 1 ) ( m 2 ) 2 .
(22)

Therefore, we have

1 i < j m 1 [ 1 ( X i λ i X j λ j ) 2 ] 1 ( m 2 ) λ i { 1 i < j m 1 [ 1 ( X i λ i X j λ j ) 2 ] } 1 ( m 2 ) λ 1 [ 1 2 ( m 1 ) ( m 2 ) 1 i < j m 1 ( X i λ i X j λ j ) 2 ] m 1 2 λ 1 .
(23)

On the other hand, from Lemma 2.4 we find

[ 1 2 ( m 1 ) ( m 2 ) 1 i < j m 1 ( X i λ i X j λ j ) 2 ] m 1 2 λ 1 = { 1 2 ( m 1 ) ( m 2 ) [ ( m 1 ) ( j = 1 m 1 X j 2 λ j ) ( j = 1 m 1 X j λ j ) 2 ] } m 1 2 λ 1 .
(24)

Combining inequalities (21), (23), and (24) yields the desired inequality (17).

Case II. When m=2. By the same method as in Lemma 2.5, it is easy to obtain the desired inequality (18). So we omit the proof. The proof of Lemma 2.6 is completed. □

Lemma 2.7 Let λ 1 λ 2 λ m >0, let 0< X j <1 (j=1,2,,m), and let m2, ρ=min{ j = 1 m 1 λ j ,1}. Then

j = 1 m ( 1 X j λ j ) 1 λ j + j = 1 m X j n 1 ρ { 1 2 m ( m 1 ) [ m ( j = 1 m X j 2 λ j ) ( j = 1 m X j λ j ) 2 ] } m 2 λ 1 .
(25)

Proof By the same method as in Lemma 2.5, applying Lemma 2.2, it is easy to obtain the desired inequality (25). So we omit the proof. □

Lemma 2.8 Let λ 1 , λ 2 ,, λ m <0, let X j >1 (j=1,2,,m), and let m2. Then

j = 1 m ( 1 X j λ j ) 1 λ j + j = 1 m X j [ 1 2 m ( m 1 ) 1 i < j m ( X i λ i X j λ j ) 2 ] m 2 min { λ 1 , λ 2 , , λ m } .
(26)

Proof After simply rearranging, we write by λ j 1 λ j 2 λ j m the component of λ 1 , λ 2 ,, λ m in increasing order, where j 1 , j 2 ,, j m is a permutation of 1,2,,m.

Then from Lemma 2.5 and Lemma 2.4 we get

j = 1 m ( 1 X j λ j ) 1 λ j + j = 1 m X j = ( 1 X j 1 λ j 1 ) 1 λ j 1 ( 1 X j 2 λ j 2 ) 1 λ j 2 ( 1 X j m λ j m ) 1 λ j m + X j 1 X j 2 X j m { 1 2 m ( m 1 ) [ m ( k = 1 m X j k 2 λ j k ) ( k = 1 m X j k λ j k ) 2 ] } m 2 λ j 1 = { 1 2 m ( m 1 ) [ m ( k = 1 m X j k 2 λ j k ) ( k = 1 m X j k λ j k ) 2 ] } m 2 min { λ 1 , λ 2 , , λ m } = [ 1 2 m ( m 1 ) 1 i < j m ( X i λ i X j λ j ) 2 ] m 2 min { λ 1 , λ 2 , , λ m } .
(27)

The proof of Lemma 2.8 is completed. □

By the same method as in Lemma 2.8, we obtain the following two lemmas.

Lemma 2.9 Let λ m >0, λ 1 , λ 2 ,, λ m 1 <0, let 0< X m <1, X j >1 (j=1,2,,m1), and let α=max{ j = 1 m 1 λ j ,1}. If m>2, then

j = 1 m ( 1 X j λ j ) 1 λ j + j = 1 m X j n 1 α [ 1 2 ( m 1 ) ( m 2 ) 1 i < j m 1 ( X i λ i X j λ j ) 2 ] m 1 2 min { λ 1 , λ 2 , , λ m } .
(28)

If m=2, then

j = 1 2 ( 1 X j λ j ) 1 λ j + j = 1 2 X j n 1 α [ 1 1 i < j 2 ( X i λ i X j λ j ) 2 ] 1 λ 1 .
(29)

Lemma 2.10 Let λ 1 , λ 2 ,, λ m >0, let 0< X j <1 (j=1,2,,m), and let m2, ρ=min{ j = 1 m 1 λ j ,1}. Then

j = 1 m ( 1 X j λ j ) 1 λ j + j = 1 m X j n 1 ρ [ 1 2 m ( m 1 ) 1 i < j m ( X i λ i X j λ j ) 2 ] m 2 max { λ 1 , λ 2 , , λ m } .
(30)

Now, we give the refinement and generalization of inequality (5).

Theorem 2.11 Let a r j >0, λ j <0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m, and let m2. Then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j { 1 2 m ( m 1 ) 1 i < j m [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 } m 2 min { λ 1 , λ 2 , , λ m } j = 1 m a 1 j r = 2 n j = 1 m a r j j = 1 m a 1 j r = 2 n j = 1 m a r j .
(31)

Proof From the assumptions in Theorem 2.11, it is easy to verify that

( a 1 j λ j r = 2 n a r j λ j ) 1 λ j ( a 1 j λ j ) 1 λ j >1(j=1,2,,m).
(32)

It thus follows from Lemma 2.8 with the substitution X j = ( a 1 j λ j r = 2 n a r j λ j a 1 j λ j ) 1 λ j in (26) that

j = 1 m ( r = 2 n a r j λ j a 1 j λ j ) 1 λ j + j = 1 m ( a 1 j λ j r = 2 n a r j λ j a 1 j λ j ) 1 λ j { 1 2 m ( m 1 ) 1 i < j m [ ( 1 r = 2 n a r i λ i a 1 i λ i ) ( 1 r = 2 n a r j λ j a 1 j λ j ) ] 2 } m 2 min { λ 1 , λ 2 , , λ m } = { 1 2 m ( m 1 ) 1 i < j m [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 } m 2 min { λ 1 , λ 2 , , λ m } ,
(33)

which implies

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j { 1 2 m ( m 1 ) 1 i < j m [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 } m 2 min { λ 1 , λ 2 , , λ m } × j = 1 m a 1 j j = 1 m ( r = 2 n a r j λ j ) 1 λ j .
(34)

On the other hand, it follows from Lemma 2.1 that

j = 1 m ( r = 2 n a r j λ j ) 1 λ j r = 2 n j = 1 m a r j .
(35)

Combining inequalities (34) and (35) yields inequality (31).

The proof of Theorem 2.11 is completed. □

Theorem 2.12 Let λ m >0, λ j <0 (j=1,2,,m1), let a r j >0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m, and let α=max{ j = 1 m 1 λ j ,1}. If m>2, then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j n 1 α { 1 2 ( m 1 ) ( m 2 ) 1 i < j m 1 [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 } m 1 2 min { λ 1 , λ 2 , , λ m } × j = 1 m a 1 j r = 2 n j = 1 m a r j n 1 α j = 1 m a 1 j r = 2 n j = 1 m a r j .
(36)

If m=2, then

j = 1 2 ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j n 1 α { 1 1 i < j 2 [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 } 1 λ 1 j = 1 2 a 1 j r = 2 n j = 1 2 a r j n 1 α j = 1 2 a 1 j r = 2 n j = 1 2 a r j .
(37)

Proof From the hypotheses of Theorem 2.12, we find that

0< ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j ( a 1 j λ j ) 1 λ j <1(j=1,2,,m1),

and

( a 1 m λ m r = 2 n a r m λ m ) 1 λ m ( a 1 m λ m ) 1 λ m >1.

Consequently, by the same method as in Theorem 2.11, and using Lemma 2.9 with a substitution X j ( a 1 j λ j r = 2 n a r j λ j a 1 j λ j ) 1 λ j (j=1,2,,m) in (28) and (29), respectively, we obtain the desired inequalities (36) and (37). □

By the same method as in Theorem 2.11, and using Lemma 2.10, we obtain the following sharpened and generalized version of inequality (4).

Theorem 2.13 Let a r j >0, λ j >0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m, let m2, and let ρ=min{ j = 1 m 1 λ j ,1}. Then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j n 1 ρ { 1 2 m ( m 1 ) 1 i < j m [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 } m 2 max { λ 1 , λ 2 , , λ m } j = 1 m a 1 j r = 2 n j = 1 m a r j n 1 ρ j = 1 m a 1 j r = 2 n j = 1 m a r j .
(38)

Therefore, from Lemma 2.3 and Theorem 2.13 we get a new refinement and generalization of inequality (4).

Corollary 2.14 Let a r j >0, λ j >0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m, let m2, and let ρ=min{ j = 1 m 1 λ j ,1}. If max{ λ 1 , λ 2 ,, λ m } m 2 , then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j n 1 ρ j = 1 m a 1 j r = 2 n j = 1 m a r j n 1 ρ j = 1 m a 1 j ( m 1 ) max { λ 1 , λ 2 , , λ m } 1 i < j m [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 n 1 ρ j = 1 m a 1 j r = 2 n j = 1 m a r j .
(39)

If max{ λ 1 , λ 2 ,, λ m }< m 2 , then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j n 1 ρ j = 1 m a 1 j r = 2 n j = 1 m a r j 2 n 1 ρ j = 1 m a 1 j m ( m 1 ) 1 i < j m [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 n 1 ρ j = 1 m a 1 j r = 2 n j = 1 m a r j .
(40)

Remark 2.15 If we set j = 1 m 1 λ j 1 in Corollary 2.14, then inequalities (39) and (40) reduce to Wu’s inequality ([[11], Theorem 1]).

In particular, putting m=2, λ 1 =p, λ 2 =q, a r 1 = a r , a r 2 = b r (r=1,2,,n) in Theorem 2.13, we obtain a new refinement and generalization of inequality (2).

Corollary 2.16 Let a r >0, b r >0 (r=1,2,,n), let p,q>0, ρ=min{ 1 p + 1 q ,1}, and let a 1 p r = 2 n a r p >0, b 1 q r = 2 n b r q >0. Then

( a 1 p r = 2 n a r p ) 1 p ( b 1 q r = 2 n b r q ) 1 q n 1 ρ { 1 [ r = 2 n ( a r p a 1 p b r q b 1 q ) ] 2 } 1 max { p , q } a 1 b 1 r = 2 n a r b r .
(41)

Similarly, putting m=2, λ 1 =p, λ 2 =q, a r 1 = a r , a r 2 = b r (r=1,2,,n) in Theorem 2.12 and Theorem 2.11, respectively, we obtain a new refinement and generalization of inequality (3).

Corollary 2.17 Let a r >0, b r >0 (r=1,2,,n), let p<0, q0, α=max{ 1 p + 1 q ,1}, and let a 1 p r = 2 n a r p >0, b 1 q r = 2 n b r q >0. Then

( a 1 p r = 2 n a r p ) 1 p ( b 1 q r = 2 n b r q ) 1 q n 1 α { 1 [ r = 2 n ( a r p a 1 p b r q b 1 q ) ] 2 } 1 min { p , q } a 1 b 1 r = 2 n a r b r .
(42)

From Lemma 2.3 and Theorem 2.11 we obtain the following refinement of inequality (5).

Corollary 2.18 Let a r j >0, λ j <0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m, and let m2. Then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j j = 1 m a 1 j r = 2 n j = 1 m a r j a 11 a 12 a 1 m ( m 1 ) min { λ 1 , λ 2 , , λ m } 1 i < j m [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 .
(43)

Similarly, from Lemma 2.3 and Theorem 2.12 we obtain the following refinement and generalization of inequality (5).

Corollary 2.19 Let λ m >0, λ j <0 (j=1,2,,m1), let a r j >0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m, and let α=max{ j = 1 m 1 λ j ,1}, m>2. Then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j n 1 α j = 1 m a 1 j r = 2 n j = 1 m a r j a 11 a 12 a 1 m n 1 α ( m 2 ) min { λ 1 , λ 2 , , λ m } 1 i < j m 1 [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 .
(44)

If we set j = 1 m 1 λ j 1, then from Corollary 2.18 and Corollary 2.19 we obtain the following refinement of inequality (5).

Corollary 2.20 Let λ 1 0, λ j <0 (j=2,3,,m), j = 1 m 1 λ j 1, let a r j >0, a 1 j λ j r = 2 n a r j λ j >0, r=1,2,,n, j=1,2,,m, and let m>2. Then

j = 1 m ( a 1 j λ j r = 2 n a r j λ j ) 1 λ j j = 1 m a 1 j r = 2 n j = 1 m a r j a 11 a 12 a 1 m ( m 1 ) min { λ 1 , λ 2 , , λ m } 1 i < j m 1 [ r = 2 n ( a r i λ i a 1 i λ i a r j λ j a 1 j λ j ) ] 2 .
(45)

3 Application

In this section, we show an application of the inequality newly obtained in Section 2.

Theorem 3.1 Let A j >0 (j=1,2,,m), let λ 1 >0, λ j <0 (j=2,3,,m), j = 1 m λ j =1, m>2, and let f j (x) (j=1,2,,m) be positive integrable functions defined on [a,b] with A j λ j a b f j λ j (x)dx>0. Then

j = 1 m ( A j λ j a b f j λ j ( x ) d x ) 1 λ j j = 1 m A j a b j = 1 m f j ( x ) d x A 1 A 2 A m ( m 2 ) min { λ 1 , λ 2 , , λ m } 1 i < j m 1 [ a b ( f i λ i ( x ) A i λ i f j λ j ( x ) A j λ j ) d x ] 2 .
(46)

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

a < a + b a n < < a + b a n k < < a + b a n ( n 1 ) < b , x i = a + b a n i , i = 0 , 1 , , n , Δ x k = b a n , k = 1 , 2 , , n .

Noting that A j λ j a b f j λ j (x)dx>0 (j=1,2,,m), we have

A j λ j lim n k = 1 n f j λ j ( a + k ( b a ) n ) b a n >0(j=1,2,,m).

Consequently, there exists a positive integer N, such that

A j λ j k = 1 n f j λ j ( a + k ( b a ) n ) b a n >0,

for all n,l>N and j=1,2,,m.

By using Theorem 2.12, for any n>N, the following inequality holds:

j = 1 m [ A j λ j k = 1 n f j λ j ( a + k ( b a ) n ) b a n ] 1 λ j j = 1 m A j λ j k = 1 n [ j = 1 m f j ( a + k ( b a ) n ) ] ( b a n ) 1 λ 1 + 1 λ 2 + + 1 λ m A 1 A 2 A m ( m 2 ) min { λ 1 , λ 2 , , λ m } 1 i < j m { k = 1 n [ 1 A i λ i f i λ i ( a + k ( b a ) n ) b a n 1 A j λ j f j λ j ( a + k ( b a ) n ) b a n ] } 2 .
(47)

Since

j = 1 m 1 λ j =1,

we have

j = 1 m [ A j λ j k = 1 n f j λ j ( a + k ( b a ) n ) b a n ] 1 λ j j = 1 m A j λ j k = 1 n [ j = 1 m f j ( a + k ( b a ) n ) ] ( b a n ) A 1 A 2 A m ( m 2 ) min { λ 1 , λ 2 , , λ m } 1 i < j m { k = 1 n [ 1 A i λ i f i λ i ( a + k ( b a ) n ) b a n 1 A j λ j f j λ j ( a + k ( b a ) n ) b a n ] } 2 .
(48)

Noting that f j (x) (j=1,2,,m) are positive Riemann integrable functions on [a,b], we know that j = 1 m f j (x) and f j λ j (x) are also integrable on [a,b]. Letting n on both sides of inequality (48), we get the desired inequality (46). The proof of Theorem 3.1 is completed. □

Remark 3.2 Obviously, inequality (46) is sharper than inequality (6).

References

  1. Aczél J: Some general methods in the theory of functional equations in one variable. New applications of functional equations. Usp. Mat. Nauk 1956,11(3):3–68. (in Russian)

    MathSciNet  Google Scholar 

  2. Farid G, Pečarić J, Ur Rehman A: On refinements of Aczél’s, Popoviciu, Bellman’s inequalities and related results. J. Inequal. Appl. 2010., 2010: Article ID 579567

    Google Scholar 

  3. Popoviciu T: On an inequality. Gaz. Mat. Fiz., Ser. A 1959,11(64):451–461. (in Romanian)

    MathSciNet  Google Scholar 

  4. Tian J: A sharpened and generalized version of Aczél-Vasić-Pečarić inequality and its application. J. Inequal. Appl. 2013., 2013: Article ID 497

    Google Scholar 

  5. Tian J-F: Reversed version of a generalized Aczél’s inequality and its application. J. Inequal. Appl. 2012., 2012: Article ID 202

    Google Scholar 

  6. Tian J-F, Wang S: Refinements of generalized Aczél’s inequality and Bellman’s inequality and their applications. J. Appl. Math. 2013., 2013: Article ID 645263

    Google Scholar 

  7. Vasić PM, Pečarić JE: On Hölder and some related inequalities. Rev. Anal. Numér. Théor. Approx. 1982, 25: 95–103.

    MATH  Google Scholar 

  8. Vasić PM, Pečarić JE: On the Jensen inequality for monotone functions. An. Univ. Timiş., Ser. ştiinţe Mat. 1979,17(1):95–104.

    MathSciNet  MATH  Google Scholar 

  9. Wu S, Debnath L: Generalizations of Aczél’s inequality and Popoviciu’s inequality. Indian J. Pure Appl. Math. 2005,36(2):49–62.

    MathSciNet  MATH  Google Scholar 

  10. Beckenbach EF, Bellman R: Inequalities. Springer, Berlin; 1983.

    MATH  Google Scholar 

  11. Wu S: Some improvements of Aczél’s inequality and Popoviciu’s inequality. Comput. Math. Appl. 2008,56(5):1196–1205. 10.1016/j.camwa.2008.02.021

    MathSciNet  Article  MATH  Google Scholar 

Download references

Acknowledgements

The authors would like to express their gratitude to the referee for his/her very valuable comments and suggestions. This work was supported by the Fundamental Research Funds for the Central Universities (Grant No. 13ZD19).

Author information

Affiliations

Authors

Corresponding author

Correspondence to Jingfeng Tian.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Tian, J., Sun, Y. New refinements of generalized Aczél inequality. J Inequal Appl 2014, 239 (2014). https://doi.org/10.1186/1029-242X-2014-239

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1029-242X-2014-239

Keywords

  • Aczél’s inequality
  • Aczél-Vasić-Pečarić inequality
  • refinement
  • generalization