JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 10.1155/2014/581354 581354 Research Article Refinements of Aczél-Type Inequality and Their Applications http://orcid.org/0000-0002-0631-038X Tian Jingfeng 1 Wang Wen-Li 2 Wu Shanhe 1 College of Science and Technology, North China Electric Power University, Baoding, Hebei 071051 China ncepu.edu.cn 2 Department of Information Engineering China University of Geosciences Great Wall College Baoding 071000 China cug.edu.cn 2014 2662014 2014 22 04 2014 08 06 2014 09 06 2014 26 6 2014 2014 Copyright © 2014 Jingfeng Tian and Wen-Li Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We present some new sharpened versions of Aczél-type inequality. Moreover, as applications, some refinements of integral type of Aczél-type inequality are given.

1. Introduction

Let n be a positive integer, and let a i , b i    ( i = 1,2 , , n ) be real numbers such that a 1 2 - i = 2 n a i 2 > 0 or b 1 2 - i = 2 n b i 2 > 0 . Then, the famous Aczél inequality  can be stated as follows: (1) ( 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 .

Aczél’s inequality plays a very important role in the theory of functional equations in non-Euclidean geometry. Due to the importance of Aczél’s inequality (1), it has received considerable attention by many authors and has motivated a large number of research papers giving it various generalizations, improvements, and applications (see  and the references therein).

In 1959, Popoviciu  first obtained an exponential extension of the Aczél inequality as follows.

Theorem B.

Let p q > 1 , ( 1 / p ) + ( 1 / q ) = 1 , and let a i , b i    ( i = 1,2 , , n ) be positive numbers such that a 1 p - i = 2 n a i p > 0 and b 1 q - i = 2 n b i q > 0 . Then (2) ( 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 .

Later, in 1982, Vasić and Pečarić  established the following reversed version of inequality (2).

Theorem C.

Let q < 0 , p > 0 , ( 1 / p ) + ( 1 / q ) = 1 , and let a i , b i    ( i = 1,2 , , n ) be positive numbers such that a 1 p - i = 2 n a i p > 0 and b 1 q - i = 2 n b i q > 0 . Then (3) ( 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 .

In another paper, Vasić and Pečarić  generalized inequality (2) in the following form.

Theorem D.

Let a r j > 0 , β j > 0 , a 1 j β j - r = 2 n a r j β j > 0 , r = 1,2 , , n , j = 1,2 , , m , and let j = 1 m ( 1 / β j ) 1 . Then (4) j = 1 m ( a 1 j β j - r = 2 n a r j β j ) 1 / β j j = 1 m a 1 j - r = 2 n j = 1 m a r j .

In 2012, Tian  presented the reversed version of inequality (4) as follows.

Theorem E.

Let a r j > 0 , β 1 0 , β j < 0    ( j = 2,3 , , m ) , j = 1 m ( 1 / β j ) 1 , a 1 j β j - r = 2 n a r j β j > 0 , r = 1,2 , , n , j = 1,2 , , m . Then (5) j = 1 m ( a 1 j β j - r = 2 n a r j β j ) 1 / β j j = 1 m a 1 j - r = 2 n j = 1 m a r j .

Moreover, in  Tian established an integral type of inequality (5).

Theorem F.

Let β 1 > 0 , β j < 0    ( j = 2,3 , , m ) , j = 1 m ( 1 / β j ) = 1 , let t 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 t j β j - a b f j β j ( x ) d x > 0 . Then (6) j = 1 m ( t j β j - a b f j β j ( x ) d x ) 1 / β j j = 1 m t j - a b j = 1 m f j ( x ) d x .

Remark 1.

In fact, the integral form of inequality (4) is also valid; that is, one has the following.

Theorem G.

Let β j > 0    ( j = 1,2 , , m ) , j = 1 m ( 1 / β j ) = 1 , let t 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 t j β j - a b f j β j ( x ) d x > 0 . Then (7) j = 1 m ( t j β j - a b f j β j ( x ) d x ) 1 / β j j = 1 m t j - a b j = 1 m f j ( x ) d x .

The main purpose of this work is to give new refinements of inequalities (4) and (5). As applications, new refinements of inequalities (6) and (7) are also given.

2. Refinements of Aczél-Type Inequality

In order to present our main results, we need some lemmas as follows.

Lemma 2 (see [<xref ref-type="bibr" rid="B6">6</xref>]).

Let a i , x i    ( i = 1,2 , , n ) be real numbers such that a i 0 and x i > - 1 . If i = 1 n a i 1 , then (8) i = 1 n ( 1 + x i ) a i 1 + i = 1 n a i x i . If either a i 1    ( i = 1,2 , , n ) or a i 0    ( i = 1,2 , , n ) and if all x i are positive or negative with x i > - 1 , then the reverse inequality of (8) holds.

Lemma 3 (see [<xref ref-type="bibr" rid="B15">15</xref>]).

Let a i j > 0    ( i = 1,2 , , n , j = 1,2 , , m ) .

If λ j 0 and if j = 1 m λ j 1 , then (9) i = 1 n j = 1 m a i j λ j j = 1 m ( i = 1 n a i j ) λ j .

If λ j 0    ( j = 1,2 , , m ) , then (10) i = 1 n j = 1 m a i j λ j j = 1 m ( i = 1 n a i j ) λ j .

If λ 1 > 0 , λ j 0    ( j = 2,3 , , m ) , and j = 1 m λ j 1 , then (11) i = 1 n j = 1 m a i j λ j j = 1 m ( i = 1 n a i j ) λ j .

Lemma 4 (see [<xref ref-type="bibr" rid="B18">18</xref>]).

Let 0 x < 1 , α > 0 . Then (12) ( 1 - x ) 1 / α 1 - x max { α , 1 } .

Lemma 5.

Let 0 < β 1 β 2 β m , j = 1 m ( 1 / β j ) 1 , m 2 , let 0 < x j < 1    ( j = 1,2 , , m ) , and let ξ ( m ) = { m / 2 i f m e v e n ( m - 1 ) / 2 i f    m o d d .

Then (13) j = 1 m ( 1 - x j β j ) 1 / β j + j = 1 m x j 1 - 1 ξ ( m ) × j = 1 ξ ( m ) [ 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] .

Proof.

From the assumptions we have that (14) 1 β 1 1 β 2 1 β m - 1 1 β m > 0 , 1 β j - 1 β j + 1 0 ( j = 1,2 , , m - 1 ) .

Case (I) (let m be even). In view of ( 1 / β 1 - 1 / β 2 ) + 1 / β 2 + 1 / β 2 + ( 1 / β 3 - 1 / β 4 ) + 1 / β 4 + 1 / β 4 + + ( 1 / β m - 1 - 1 / β m ) + 1 / β m + 1 / β m = 1 / β 1 + 1 / β 2 + + 1 / β m 1 by using inequality (9),

we get (15) j = 1 m / 2 [ 1 - ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] 1 / β 2 j = j = 1 m / 2 { [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j - 1 β 2 j - 1 ] ( 1 / β 2 j - 1 ) - ( 1 / β 2 j ) [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j β 2 j ] 1 / β 2 j 00000000 × [ ( 1 - x 2 j β 2 j ) + x 2 j - 1 β 2 j - 1 ] 1 / β 2 j 00000000 × [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j - 1 β 2 j - 1 ] 1 / β 2 j - 1 - 1 / β 2 j [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j - 1 β 2 j - 1 ] ( 1 / β 2 j - 1 ) - ( 1 / β 2 j ) [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j β 2 j ] 1 / β 2 j } = [ ( 1 - x 1 β 1 ) + x 2 β 2 ] 1 / β 2 [ ( 1 - x 2 β 2 ) + x 1 β 1 ] 1 / β 2 × [ ( 1 - x 1 β 1 ) + x 1 β 1 ] 1 / β 1 - 1 / β 2 × [ ( 1 - x 3 β 3 ) + x 4 β 4 ] 1 / β 4 [ ( 1 - x 4 β 4 ) + x 3 β 3 ] 1 / β 4 × [ ( 1 - x 3 β 3 ) + x 3 β 3 ] 1 / β 3 - 1 / β 4 × [ ( 1 - x m - 1 β m - 1 ) + x m β m ] 1 / β m [ ( 1 - x m β m ) + x m - 1 β m - 1 ] 1 / β m × [ ( 1 - x m - 1 β m - 1 ) + x m - 1 β m - 1 ] 1 / β m - 1 - 1 / β m j = 1 m / 2 [ [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j - 1 β 2 j - 1 ] ( 1 / β 2 j - 1 ) - ( 1 / β 2 j ) ( 1 - x 2 j - 1 β 2 j - 1 ) 1 / β 2 j ( 1 - x 2 j β 2 j ) 1 / β 2 j 00000000 × ( 1 - x 2 j - 1 β 2 j - 1 ) 1 / β 2 j - 1 - 1 / β 2 j ] + j = 1 m / 2 [ [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j - 1 β 2 j - 1 ] ( 1 / β 2 j - 1 ) - ( 1 / β 2 j ) ( x 2 j β 2 j ) 1 / β 2 j ( x 2 j - 1 β 2 j - 1 ) 1 / β 2 j 0000000000 × ( x 2 j - 1 β 2 j - 1 ) 1 / β 2 j - 1 - 1 / β 2 j ] = j = 1 m / 2 ( 1 - x j β j ) 1 / β j + j = 1 m x j .

On the other hand, applying Lemma 4 and the arithmetic-geometric means inequality we obtain (16) j = 1 m / 2 [ 1 - ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] 1 / β 2 j j = 1 m / 2 [ 1 - 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] { 2 m j = 1 m / 2 [ 1 - 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] } m / 2 = { 1 - 2 m j = 1 m / 2 [ 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] } m / 2 . Applying Lemma 4 again, we get (17) { 1 - 2 m j = 1 m / 2 [ 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] } m / 2 1 - 2 m j = 1 m / 2 [ 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] .

Combining (15), (16), and (17) yields immediately inequality (13).

Case (II) (let m be odd). In view of ( 1 / β 1 - 1 / β 2 ) + 1 / β 2 + 1 / β 2 + ( 1 / β 3 - 1 / β 4 ) + 1 / β 4 + 1 / β 4 + + ( 1 / β m - 2 - 1 / β m - 1 ) + 1 / β m - 1 + 1 / β m - 1 + 1 / β m = 1 / β 1 + 1 / β 2 + + 1 / β m 1 , by using inequality (9), we have (18) j = 1 ( m - 1 ) / 2 [ 1 - ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] 1 / β 2 j = { j = 1 ( m - 1 ) / 2 [ 1 - ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] 1 / β 2 j } × [ ( 1 - x m β m ) + x m β m ] 1 / β m = { j = 1 ( m - 1 ) / 2 { [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j β 2 j ] 1 / β 2 j 0000000000000 × [ ( 1 - x 2 j β 2 j ) + x 2 j - 1 β 2 j - 1 ] 1 / β 2 j 0000000000000 × [ ( 1 - x 2 j - 1 β 2 j - 1 ) + x 2 j - 1 β 2 j - 1 ] 1 / β 2 j - 1 - 1 / β 2 j } j = 1 ( m - 1 ) / 2 } × [ ( 1 - x m β m ) + x m β m ] 1 / β m { j = 1 ( m - 1 ) / 2 [ ( 1 - x 2 j - 1 β 2 j - 1 ) 1 / β 2 j ( 1 - x 2 j β 2 j ) 1 / β 2 j 0000000000000 ( 1 - x 2 j - 1 β 2 j - 1 ) 1 / β 2 j - 1 - 1 / β 2 j ] j = 1 ( m - 1 ) / 2 } ( 1 - x m β m ) 1 / β m + { j = 1 ( m - 1 ) / 2 [ ( x 2 j β 2 j ) 1 / β 2 j ( x 2 j - 1 β 2 j - 1 ) 1 / β 2 j 00000000000000 × ( x 2 j - 1 β 2 j - 1 ) 1 / β 2 j - 1 - 1 / β 2 j ] j = 1 ( m - 1 ) / 2 } ( x m β m ) 1 / β m = j = 1 m ( 1 - x j β j ) 1 / β j + j = 1 m x j . On the other hand, applying Lemma 4 and the arithmetic-geometric means inequality we obtain (19) j = 1 ( m - 1 ) / 2 [ 1 - ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] 1 / β 2 j j = 1 ( m - 1 ) / 2 [ 1 - 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] { 2 m - 1 j = 1 ( m - 1 ) / 2 [ 1 - 1 max { β 2 j , 1 } 000000000000000000 × ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 1 max { β 2 j , 1 } ] j = 1 ( m - 1 ) / 2 } ( m - 1 ) / 2 = { 1 - 2 m - 1 j = 1 ( m - 1 ) / 2 [ 1 max { β 2 j , 1 } 000000000000000000000 × ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 1 max { β 2 j , 1 } ] j = 1 ( m - 1 ) / 2 } ( m - 1 ) / 2 . Applying Lemma 4 again, we have (20) { 1 - 2 m - 1 j = 1 ( m - 1 ) / 2 [ 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] } ( m - 1 ) / 2 1 - 2 m - 1 j = 1 ( m - 1 ) / 2 [ 1 max { β 2 j , 1 } ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 ] . Hence, combining (18), (19), and (20) yields immediately inequality (13).

Similar to the proof of Lemma 5 but using Lemma 2 in place of Lemma 4, we immediately obtain the following result.

Lemma 6.

Let β 1 > 0 , 0 > β 2 β 3 β m , j = 1 m ( 1 / β j ) 1 , m 2 , let 0 < x 1 < 1 , x j > 1    ( j = 2,3 , , m ) , and let ξ ( m ) = { m / 2 i f m e v e n ( m - 1 ) / 2 i f    m o d d .

Then (21) j = 1 m ( 1 - x j β j ) 1 / β j + j = 1 m x j 1 - j = 1 ξ ( m ) ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 β 2 j .

Using the same methods as in Lemma 6, we get the following Lemma.

Lemma 7.

Let 0 > β 1 β 2 β m , m 2 , let x j > 1    ( j = 1,2 , , m ) , and let ξ ( m ) = { m / 2 i f    m e v e n ( m - 1 ) / 2 i f    m o d d .

Then (22) j = 1 m ( 1 - x j β j ) 1 / β j + j = 1 m x j 1 - j = 1 ξ ( m ) ( x 2 j β 2 j - x 2 j - 1 β 2 j - 1 ) 2 β 2 j .

Now, we present some new refinements of inequalities (4) and (5).

Theorem 8.

Let a r j > 0 , r = 1,2 , , n , j = 1,2 , , m , m 2 , n 2 , let 0 < β 1 β 2 β m , j = 1 m ( 1 / β j ) 1 , a 1 j β j - r = 2 n a r j β j > 0 , and let ξ ( m ) = { m / 2 i f    m e v e n ( m - 1 ) / 2 i f    m o d d .

Then (23) j = 1 m ( a 1 j β j - r = 2 n a r j β j ) 1 / β j j = 1 m a 1 j - r = 2 n j = 1 m a r j - a 11 a 12 a 1 m ξ ( m ) × j = 1 ξ ( m ) { [ r = 2 n ( a r ( 2 j ) β 2 j a 1 ( 2 j ) β 2 j - a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 1 max { β 2 j , 1 } 000000000000 × [ r = 2 n ( a r ( 2 j ) β 2 j a 1 ( 2 j ) β 2 j - a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 } .

Proof.

From the assumptions we find that (24) 0 < ( a 1 j β j - r = 2 n a r j β j ) 1 / β j ( a 1 j β j ) 1 / β j < 1 ( j = 1,2 , , m ) .

Thus, by using Lemma 5 with a substitution x j ( ( a 1 j β j - r = 2 n a r j β j ) / a 1 j β j ) 1 / β j    ( j = 1,2 , , m ) in (13), we obtain (25) j = 1 m ( r = 2 n a r j β j a 1 j β j ) 1 / β j + j = 1 m ( a 1 j β j - r = 2 n a r j β j a 1 j β j ) 1 / β j 1 - 1 ξ ( m ) j = 1 ξ ( m ) { - ( 1 - r = 2 n a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 1 max { β 2 j , 1 } 00000000000000000 × [ ( 1 - r = 2 n a r ( 2 j ) β 2 j a 1 ( 2 j ) β 2 j ) 00000000000000000000 - ( 1 - r = 2 n a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 } = 1 - 1 ξ ( m ) j = 1 ξ ( m ) { [ r = 2 n ( a r ( 2 j ) β 2 j a 1 ( 2 j ) β 2 j - a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 1 max { β 2 j , 1 } 0000000000000000 × [ r = 2 n ( a r ( 2 j ) β 2 j a 1 ( 2 j ) β 2 j - a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 } , which implies (26) j = 1 m ( a 1 j β j - r = 2 n a r j β j ) 1 / β j j = 1 m a 1 j - j = 1 m ( r = 2 n a r j β j ) 1 / β j - a 11 a 12 a 1 m ξ ( m ) × j = 1 ξ ( m ) { [ r = 2 n ( a r ( 2 j ) β 2 j a 1 ( 2 j ) β 2 j - a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 1 max { β 2 j , 1 } 000000000000 × [ r = 2 n ( a r ( 2 j ) β 2 j a 1 ( 2 j ) β 2 j - a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 } .

On the other hand, we get from Lemma 3 that (27) j = 1 m ( r = 2 n a r j β j ) 1 / β j r = 2 n j = 1 m a r j . Combining (26) and (27) yields immediately the desired inequality (23).

Theorem 9.

Let a r j > 0 , 0 > β 1 β 2 β m , a 1 j β j - r = 2 n a r j β j > 0 , r = 1,2 , , n , j = 1,2 , , m , let m 2 , n 2 , and let ξ ( m ) = { m / 2 i f    m e v e n ( m - 1 ) / 2 i f    m o d d .

Then (28) j = 1 m ( a 1 j β j - r = 2 n a r j β j ) 1 / β j j = 1 m a 1 j - r = 2 n j = 1 m a r j - a 11 a 12 , , a 1 m × j = 1 ξ ( m ) { 1 β 2 j [ r = 2 n ( a r ( 2 j ) β 2 j a 1 ( 2 j ) β 2 j - a r ( 2 j - 1 ) β 2 j - 1 a 1 ( 2 j - 1 ) β 2 j - 1 ) ] 2 } . Inequality (28) is also valid for β 1 > 0 , 0 > β 2 β 3 β m , j = 1 m ( 1 / β j ) 1 .

Proof.

The proof of Theorem 9 is similar to the one of Theorem 8, and we omit it.

3. Applications

In this section, we show two applications of the inequalities newly obtained in Section 2.

Firstly, we present a new refinement of inequality (6) by using Theorem 9.

Theorem 10.

Let t j > 0    ( j = 1,2 , , m ) , β 1 > 0 , 0 > β 2 β 3 β m , j = 1 m ( 1 / β j ) = 1 , let f j ( x )    ( j = 1,2 , , m ) be positive integrable functions defined on [ a , b ] with t j β j - a b f j β j ( x ) d x > 0 , and let ξ ( m ) = { m / 2 if    m even ( m - 1 ) / 2 if    m odd .

Then (29) j = 1 m ( t j β j - a b f j β j ( x ) d x ) 1 / β j j = 1 m t j - a b j = 1 m f j ( x ) d x - t 1 t 2 , , t m × j = 1 ξ ( m ) [ 1 β 2 j a b ( f 2 j β 2 j ( x ) t 2 j β 2 j - f 2 j - 1 β 2 j - 1 ( x ) t 2 j - 1 β 2 j - 1 ) d x ] 2 .

Proof.

For any positive integer n , we choose an equidistant partition of [ a , b ] as (30) 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 , (31) Δ x k = b - a n , k = 1,2 , , n .

Since t j β j - a b f j β j ( x ) d x > 0    ( j = 1,2 , , m ) , it follows that (32) t j β j - lim n k = 1 n f j β j ( a + k ( b - a ) n ) b - a n > 0 ( j = 1 , 2 , , m ) . Therefore, there exists a positive integer N such that (33) t j β j - k = 1 n f j β j ( a + k ( b - a ) n ) b - a n > 0 , for all n > N and j = 1,2 , , m .

Moreover, for any n > N , it follows from Theorem 9 that (34) j = 1 m [ t j β j - k = 1 n f j β j ( a + k ( b - a ) n ) b - a n ] 1 / β j j = 1 m t j β j - k = 1 n ( j = 1 m f j ( a + k ( b - a ) n ) ) × ( b - a n ) 1 / β 1 + 1 / β 2 + + 1 / β m - t 1 t 2 , , t m j = 1 ξ ( m ) 1 β 2 j × [ ( a + k ( b - a ) n ) b - a n k = 1 n ( 1 t 2 j β 2 j f 2 j β 2 j ( a + k ( b - a ) n ) b - a n 00000000000 - 1 t 2 j - 1 β 2 j - 1 f 2 j - 1 β 2 j - 1 ( a + k ( b - a ) n ) b - a n ) ] 2 .

Noting that (35) j = 1 m 1 β j = 1 , we get (36) j = 1 m [ t j β j - k = 1 n f j β j ( a + k ( b - a ) n ) b - a n ] 1 / β j j = 1 m t j β j - k = 1 n ( j = 1 m f j ( a + k ( b - a ) n ) ) ( b - a n ) - t 1 t 2 , , t m j = 1 ξ ( m ) 1 β 2 j × [ ( a + k ( b - a ) n ) b - a n k = 1 n ( 1 t 2 j β 2 j f 2 j β 2 j ( a + k ( b - a ) n ) b - a n 000000000000 - 1 t 2 j - 1 β 2 j - 1 f 2 j - 1 β 2 j - 1 ( a + k ( b - a ) n ) b - a n ) ] 2 . In view of the assumption that f j ( x )    ( j = 1,2 , , m ) are positive Riemann integrable functions on [ a , b ] , we find 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 (36), we get the desired inequality (29).

Next, we give a new refinement of inequality (7) by using Theorem 8.

Theorem 11.

Let t j > 0    ( j = 1,2 , , m ) , 0 < β 1 β 2 β m , j = 1 m ( 1 / β j ) = 1 , m 2 , and let f j ( x )    ( j = 1,2 , , m ) be positive integrable functions defined on [ a , b ] with t j β j - a b f j β j ( x ) d x > 0 , and let ξ ( m ) = { m / 2 if    m even ( m - 1 ) / 2 if    m odd .

Then (37) j = 1 m ( t j β j - a b f j β j ( x ) d x ) 1 / β j j = 1 m t j - a b j = 1 m f j ( x ) d x - t 1 t 2 , , t m ξ ( m ) × j = 1 ξ ( m ) { [ a b ( f 2 j β 2 j ( x ) t 2 j β 2 j - f 2 j - 1 β 2 j - 1 ( x ) t 2 j - 1 β 2 j - 1 ) d x ] 2 1 β 2 j 00000000000 × [ a b ( f 2 j β 2 j ( x ) t 2 j β 2 j - f 2 j - 1 β 2 j - 1 ( x ) t 2 j - 1 β 2 j - 1 ) d x ] 2 } .

Proof.

The proof of Theorem 11 is similar to the one of Theorem 10, and we omit it.

Acknowledgments

The authors would like to thank the reviewers and the editors for their valuable suggestions and comments. This work was supported by the Fundamental Research Funds for the Central Universities (Grant no. 13ZD19).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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