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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).

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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).

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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

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