Refinements of Generalized Aczél’s Inequality and Bellman

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013, Article ID 645263, 6 pageshttp://dx.doi.org/10.1155/2013/645263

Research ArticleRefinements of Generalized Aczél’s Inequality and Bellman’sInequality and Their Applications

Jing-Feng Tian1 and Shu-Yan Wang2

1 College of Science and Technology, North China Electric Power University, Baoding, Hebei Province 071051, China2Department of Mathematics and Computer, Baoding University, Baoding, Hebei Province 071000, China

Correspondence should be addressed to Jing-Feng Tian; tianjfhxm [email protected]

Received 13 October 2012; Accepted 30 December 2012

Academic Editor: Shanhe Wu

Copyright © 2013 J.-F. Tian and S.-Y. Wang. This is an open access article distributed under the Creative Commons AttributionLicense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properlycited.

We give some refinements of generalized Aczel’s inequality and Bellman’s inequality proposed by Tian. As applications, somerefinements of integral type of generalized Aczel’s inequality and Bellman’s inequality are given.

1. Introduction

The famous Aczel’s inequality [1] states that if 𝑎𝑖, 𝑏𝑖(𝑖 =

1, 2, . . . , 𝑛) are real numbers such that 𝑎21− ∑𝑛

𝑖=2𝑎2

𝑖> 0 and

𝑏2

1− ∑𝑛

𝑖=2𝑏2

𝑖> 0, then

(𝑎2

1−

𝑛

∑

𝑖=2

𝑎2

𝑖)(𝑏2

1−

𝑛

∑

𝑖=2

𝑏2

𝑖) ≤ (𝑎

1𝑏1−

𝑛

∑

𝑖=2

𝑎𝑖𝑏𝑖)

2

. (1)

It is well known that Aczel’s inequality plays an importantrole in the theory of functional equations in non-Euclideangeometry. In recent years, various attempts have been madeby many authors to improve and generalize the Aczel’sinequality (see [2–19] and references therein). We state heresome improvements of Aczel’s inequality.

One of the most important results in the referencesmentioned above is an exponential extension of (1), whichis stated in the following theorem.

Theorem A. Let 𝑝 and 𝑞 be real numbers such that 𝑝, 𝑞 = 0,and 1/𝑝 + 1/𝑞 = 1 and let 𝑎

𝑖, 𝑏𝑖(𝑖 = 1, 2, . . . , 𝑛) be positive

numbers such that 𝑎𝑝1− ∑𝑛

𝑖=2𝑎𝑝

𝑖> 0 and 𝑏𝑞

1− ∑𝑛

𝑖=2𝑏𝑞

𝑖> 0. If

𝑝 > 1, then

(𝑎𝑝

1−

𝑛

∑

𝑖=2

𝑎𝑝

𝑖)

1/𝑝

(𝑏𝑞

1−

𝑛

∑

𝑖=2

𝑏𝑞

𝑖)

1/𝑞

≤ 𝑎1𝑏1−

𝑛

∑

𝑖=2

𝑎𝑖𝑏𝑖. (2)

If 𝑝 < 1 (𝑝 = 0), then the reverse inequality in (2) holds.

Remark 1. The case 𝑝 > 1 of Theorem A was proved byPopoviciu [8]. The case 𝑝 < 1 was given in [15] by Vasic andPecaric.

Vasic and Pecaric [16] presented a further extension ofinequality (1).

Theorem B. Let 𝑎𝑟𝑗> 0, 𝜆

𝑗> 0, 𝑎𝜆𝑗

1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, 𝑟 =

1, 2, . . . , 𝑛, 𝑗 = 1, 2, . . . , 𝑚, and let ∑𝑚𝑗=1(1/𝜆𝑗) ≥ 1. Then

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≤

𝑚

∏

𝑗=1

𝑎1𝑗−

𝑛

∑

𝑟=2

𝑚

∏

𝑗=1

𝑎𝑟𝑗. (3)

In a recent paper [18], Wu and Debnath established aninteresting generalization of Aczel’s inequality [1] as follows.

Theorem C. Let 𝑎𝑟𝑗> 0, 𝜆

𝑗> 0, 𝑎𝜆𝑗

1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, 𝑟 =

1, 2, . . . , 𝑛, 𝑗 = 1, 2, . . . , 𝑚, and let 𝜌 = min{∑𝑚𝑗=1(1/𝜆𝑗), 1}.

Then

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≤ 𝑛1−𝜌

𝑚

∏

𝑗=1

𝑎1𝑗−

𝑛

∑

𝑟=2

𝑚

∏

𝑗=1

𝑎𝑟𝑗. (4)

In 2012, Tian [10] presented the following reversed ver-sion of inequality (4).

2 Journal of Applied Mathematics

Theorem D. Let 𝑎𝑟𝑗> 0, 𝜆

1= 0, 𝜆𝑗< 0 (𝑗 = 2, 3, . . . , 𝑚),

𝑎

𝜆𝑗

1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, 𝑟 = 1, 2, . . . , 𝑛, 𝑗 = 1, 2, . . . , 𝑚, and let

𝜏 = max{∑𝑚𝑗=1(1/𝜆𝑗), 1}. Then

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥ 𝑛1−𝜏

𝑚

∏

𝑗=1

𝑎1𝑗−

𝑛

∑

𝑟=2

𝑚

∏

𝑗=1

𝑎𝑟𝑗. (5)

Therefore, applying the above inequality, Tian gave the re-versed version of inequality (3) as follows.

Theorem E. Let 𝑎𝑟𝑗> 0, 𝜆

1= 0, 𝜆𝑗< 0 (𝑗 = 2, 3, . . . , 𝑚),

∑𝑚

𝑗=1(1/𝜆𝑗) ≤ 1, 𝑎𝜆𝑗

1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, and 𝑟 = 1, 2, . . . , 𝑛, 𝑗 =

1, 2, . . . , 𝑚. Then

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥

𝑚

∏

𝑗=1

𝑎1𝑗−

𝑛

∑

𝑟=2

𝑚

∏

𝑗=1

𝑎𝑟𝑗. (6)

Moreover, in [10], Tian obtained the following integralform of inequality (5).

TheoremF. Let 𝜆1> 0, 𝜆

𝑗< 0 (𝑗 = 2, 3, . . . , 𝑚),∑𝑚

𝑗=1𝜆𝑗= 1,

𝑃𝑗> 0 (𝑗 = 1, 2, . . . , 𝑚), and let 𝑓

𝑗(𝑥) (𝑗 = 1, 2, . . . , 𝑚) be

positive Riemann integrable functions on [𝑎, 𝑏] such that 𝑃𝜆𝑗𝑗−

∫

𝑏

𝑎𝑓

𝜆𝑗

𝑗(𝑥)d𝑥 > 0. Then

𝑚

∏

𝑗=1

(𝑃

𝜆𝑗

𝑗− ∫

𝑏

𝑎

𝑓

𝜆𝑗

𝑗(𝑥) d𝑥)

1/𝜆𝑗

≥

𝑚

∏

𝑗=1

𝑃𝑗− ∫

𝑏

𝑎

𝑚

∏

𝑗=1

𝑓𝑗 (𝑥) d𝑥.

(7)

Bellman inequality [20] related with Aczel’s inequality isstated as follows.

Theorem G. Let 𝑎𝑖, 𝑏𝑖(𝑖 = 1, 2, . . . , 𝑛) be positive numbers

such that 𝑎𝑝1− ∑𝑛

𝑖=2𝑎𝑝

𝑖> 0 and 𝑏𝑞

1− ∑𝑛

𝑖=2𝑏𝑞

𝑖> 0. If 𝑝 ≥ 1,

then

[

[

(𝑎𝑝

1−

𝑛

∑

𝑖=2

𝑎𝑝

𝑖)

1/𝑝

+ (𝑏𝑝

1−

𝑛

∑

𝑖=2

𝑏𝑝

𝑖)

1/𝑝

]

]

𝑝

≤ (𝑎1+ 𝑏1)𝑝−

𝑛

∑

𝑖=2

(𝑎𝑖+ 𝑏𝑖)𝑝.

(8)

If 0 < 𝑝 < 1, then the reverse inequality in (8) holds.

Remark 2. The case 𝑝 > 1 of Theorem G was proposed byBellman [20]. The case 0 < 𝑝 < 1 was proved in [15] by Vasicand Pecaric.

The main purpose of this work is to give refinements ofinequalities (5) and (8). As applications, some refinements ofintegral type of inequality (5) and (8) are given.

2. Refinements of Generalized Aczél’sInequality and Bellman’s Inequality

Theorem 3. Let 𝑎𝑟𝑗> 0, 𝜆

1= 0, 𝜆𝑗< 0 (𝑗 = 2, 3, . . . , 𝑚),

𝑎

𝜆𝑗

1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, 𝑟 = 1, 2, . . . , 𝑛, 𝑗 = 1, 2, . . . , 𝑚, and let

𝜏 = max{∑𝑚𝑗=1(1/𝜆𝑗), 1}. Then

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥ 𝑇 (𝑠, 𝑎𝑖𝑗) ≥ 𝑛1−𝜏

𝑚

∏

𝑗=1

𝑎1𝑗−

𝑛

∑

𝑟=2

𝑚

∏

𝑗=1

𝑎𝑟𝑗,

(9)

where

𝑇 (𝑠, 𝑎𝑖𝑗) = (𝑛 + 1 − 𝑠)

1−𝜏

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

−

𝑛

∑

𝑟=𝑠+1

𝑚

∏

𝑗=1

𝑎𝑟𝑗.

(10)

ProofCase (I). When∑𝑚

𝑗=1(1/𝜆𝑗) < 1, then 𝜏 = 1. On the one hand,

we split the left-hand side of inequality (9) as follows:

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

=

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗−

𝑛

∑

𝑟=𝑠+1

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

=

𝑚

∏

𝑗=1

(𝐴

𝜆𝑗

𝑗−

𝑛

∑

𝑟=𝑠+1

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

,

(11)

where

𝐴𝑗= (𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

, 𝑗 = 1, 2, . . . , 𝑚. (12)

From this hypothesis, it is immediate to obtain the inequality

𝐴

𝜆𝑗

𝑗≥

𝑛

∑

𝑟=𝑠+1

𝑎

𝜆𝑗

𝑟𝑗. (13)

Thus, by using inequality (6), we have

𝑚

∏

𝑗=1

(𝐴

𝜆𝑗

𝑗−

𝑛

∑

𝑟=𝑠+1

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥

𝑚

∏

𝑗=1

𝐴𝑗−

𝑛

∑

𝑟=𝑠+1

𝑚

∏

𝑗=1

𝑎𝑟𝑗. (14)

On the other hand, by using inequality (6) again, weobtain the inequality

𝑚

∏

𝑗=1

𝐴𝑗=

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥

𝑚

∏

𝑗=1

𝑎1𝑗−

𝑠

∑

𝑟=2

𝑚

∏

𝑗=1

𝑎𝑟𝑗. (15)

Combining inequalities (11), (14), and (15) we can get inequal-ity (9).Case (II). When ∑

𝑚

𝑗=1(1/𝜆𝑗) ≥ 1, using Case (I) with

∑𝑚+1

𝑗=1(1/𝜆𝑗) = 1, we have

𝑚+1

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥

𝑚+1

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

−

𝑛

∑

𝑟=𝑠+1

𝑚+1

∏

𝑗=1

𝑎𝑟𝑗,

(16)

Journal of Applied Mathematics 3

where 𝑎𝜆𝑗1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, 𝑎

𝑟𝑗> 0, 𝜆

1= 0, 𝜆𝑗< 0 (𝑗 =

2, 3, . . . , 𝑚 + 1).Hence, taking 𝑎𝜆𝑚+1

1(𝑚+1)= 1, 𝑎𝜆𝑚+1

2(𝑚+1)= 𝑎𝜆𝑚+1

3(𝑚+1)= ⋅ ⋅ ⋅ =

𝑎𝜆𝑚+1

𝑛(𝑚+1)= 1/𝑛 into (16), we obtain

(

1

𝑛

)

1−1/𝜆1−1/𝜆2−⋅⋅⋅−1/𝜆𝑚𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥ (

𝑛 + 1 − 𝑠

𝑛

)

1−1/𝜆1−1/𝜆2−⋅⋅⋅−1/𝜆𝑚

×

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

− (

1

𝑛

)

1−1/𝜆1−1/𝜆2−⋅⋅⋅−1/𝜆𝑚𝑛

∑

𝑟=𝑠+1

𝑚

∏

𝑗=1

𝑎𝑟𝑗,

(17)

that is,𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥ (𝑛 + 1 − 𝑠)1−1/𝜆1−1/𝜆2−⋅⋅⋅−1/𝜆𝑚

×

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

−

𝑛

∑

𝑟=𝑠+1

𝑚

∏

𝑗=1

𝑎𝑟𝑗.

(18)

Therefore, repeating the foregoing arguments, we get

(𝑛 + 1 − 𝑠)1−1/𝜆1−1/𝜆2−⋅⋅⋅−1/𝜆𝑚

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

−

𝑛

∑

𝑟=𝑠+1

𝑚

∏

𝑗=1

𝑎𝑟𝑗≥ 𝑛1−𝜏

𝑚

∏

𝑗=1

𝑎1𝑗−

𝑛

∑

𝑟=2

𝑚

∏

𝑗=1

𝑎𝑟𝑗.

(19)

Combining inequalities (18) and (19) leads to inequality (9)immediately. The proof of Theorem 3 is completed.

If we set∑𝑚𝑗=1(1/𝜆𝑗) ≤ 1, then fromTheorem 3, we obtain

the following refinement of inequality (6).

Corollary 4. Let 𝑎𝑟𝑗> 0, 𝜆

1= 0, 𝜆𝑗< 0 (𝑗 = 2, 3, . . . , 𝑚),

∑𝑚

𝑗=1(1/𝜆𝑗) ≤ 1, 𝑎𝜆𝑗

1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, 𝑟 = 1, 2, . . . , 𝑛, and 𝑗 =

1, 2, . . . , 𝑚. Then𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑛

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

≥

𝑚

∏

𝑗=1

(𝑎

𝜆𝑗

1𝑗−

𝑠

∑

𝑟=2

𝑎

𝜆𝑗

𝑟𝑗)

1/𝜆𝑗

−

𝑛

∑

𝑟=𝑠+1

𝑚

∏

𝑗=1

𝑎𝑟𝑗

≥

𝑚

∏

𝑗=1

𝑎1𝑗−

𝑛

∑

𝑟=2

𝑚

∏

𝑗=1

𝑎𝑟𝑗.

(20)

Putting𝑚 = 2, 𝜆1= 𝑝 = 0, 𝜆

2= 𝑞 < 0, 𝑎

𝑟1= 𝑎𝑟, and 𝑎

𝑟2=

𝑏𝑟(𝑟 = 1, 2, . . . , 𝑛) in Theorem 3, we obtain the refinement

and generalization of Theorem A for 𝑝 < 1.

Corollary 5. Let 𝑎𝑟> 0, 𝑏

𝑟> 0 (𝑟 = 1, 2, . . . , 𝑛), 𝑎𝑝

1−

∑𝑛

𝑟=2𝑎𝑝

𝑟> 0, 𝑏𝑞

1− ∑𝑛

𝑟=2𝑏𝑞

𝑟> 0, 𝑝 = 0, 𝑞 < 0, and 𝜌 =

max{1/𝑝 + 1/𝑞, 1}. Then, the following inequality holds:

(𝑎𝑝

1−

𝑛

∑

𝑟=2

𝑎𝑝

𝑟)

1/𝑝

(𝑏𝑞

1−

𝑛

∑

𝑟=2

𝑏𝑞

𝑟)

1/𝑞

≥ (𝑛 + 1 − 𝑠)1−𝜌(𝑎𝑝

1−

𝑠

∑

𝑟=2

𝑎𝑝

𝑟)

1/𝑝

× (𝑏𝑞

1−

𝑠

∑

𝑟=2

𝑏𝑞

𝑟)

1/𝑞

−

𝑛

∑

𝑟=𝑠+1

𝑎𝑟𝑏𝑟

≥ 𝑛1−𝜌𝑎1𝑏1−

𝑛

∑

𝑟=2

𝑎𝑟𝑏𝑟.

(21)

Based on themathematical induction, it is easy to see thatthe following generalized Bellman’s inequality is true.

Theorem 6. Let 𝑎𝑟𝑗> 0, 𝑎𝜆𝑗

1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, 𝑟 = 1, 2, . . . , 𝑛,

𝑗 = 1, 2, . . . , 𝑚, and let 0 < 𝑝 < 1. Then

[

[

𝑚

∏

𝑗=1

(𝑎𝑝

1𝑗−

𝑛

∑

𝑟=2

𝑎𝑝

𝑟𝑗)

1/𝑝

]

]

𝑝

≥ (

𝑚

∑

𝑗=1

𝑎1𝑗)

𝑝

−

𝑛

∑

𝑟=2

(

𝑚

∑

𝑗=1

𝑎𝑟𝑗)

𝑝

.

(22)

Next, we give a refinement of generalized Bellman’sinequality (22) as follows.

Theorem 7. Let 𝑎𝑟𝑗> 0, 𝑎𝜆𝑗

1𝑗− ∑𝑛

𝑟=2𝑎

𝜆𝑗

𝑟𝑗> 0, 𝑟 = 1, 2, . . . , 𝑛,

𝑗 = 1, 2, . . . , 𝑚, and let 0 < 𝑝 < 1. Then

[

[

𝑚

∏

𝑗=1

(𝑎𝑝

1𝑗−

𝑛

∑

𝑟=2

𝑎𝑝

𝑟𝑗)

1/𝑝

]

]

𝑝

≥[

[

𝑚

∏

𝑗=1

(𝑎𝑝

1𝑗−

𝑠

∑

𝑟=2

𝑎𝑝

𝑟𝑗)

1/𝑝

]

]

𝑝

−

𝑛

∑

𝑟=𝑠+1

(

𝑚

∑

𝑗=1

𝑎𝑟𝑗)

𝑝

≥ (

𝑚

∑

𝑗=1

𝑎1𝑗)

𝑝

−

𝑛

∑

𝑟=2

(

𝑚

∑

𝑗=1

𝑎𝑟𝑗)

𝑝

.

(23)

Proof. The proof of Theorem 7 is similar to the one ofTheorem 3. Applying generalized Bellman’s inequality (22)twice, we can deduce the inequality (23).

3. Application

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


Firstly, we give an improvement of inequality (7) by usingTheorem 3.

Theorem 8. Let 𝐵𝑗> 0 (𝑗 = 1, 2, . . . , 𝑚), let 𝜆

1> 0,

𝜆𝑗< 0 (𝑗 = 2, 3, . . . , 𝑚), ∑𝑚

𝑗=1𝜆𝑗= 1, and let 𝑓

𝑗(𝑥) (𝑗 =

1, 2, . . . , 𝑚) be positive integrable functions defined on [𝑎, 𝑏]with 𝐵𝜆𝑗

𝑗− ∫

𝑏

𝑎𝑓

𝜆𝑗

𝑗(𝑥)d𝑥 > 0. Then, for any 𝑡 ∈ [𝑎, 𝑏), one has

𝑚

∏

𝑗=1

(𝐵

𝜆𝑗

𝑗− ∫

𝑏

𝑎

𝑓

𝜆𝑗

𝑗(𝑥) d𝑥)

1/𝜆𝑗

≥

𝑚

∏

𝑗=1

(𝐵

𝜆𝑗

𝑗− ∫

𝑡

𝑎

𝑓

𝜆𝑗

𝑗(𝑥) d𝑥)

1/𝜆𝑗

− ∫

𝑏

𝑡

𝑚

∏

𝑗=1

𝑓𝑗 (𝑥) d𝑥

≥

𝑚

∏

𝑗=1

𝐵𝑗− ∫

𝑏

𝑎

𝑚

∏

𝑗=1

𝑓𝑗 (𝑥) d𝑥.

(24)

Proof. We need to prove only the left side of inequality (24).The proof of the right side of inequality (24) is similar. For anypositive integers 𝑛 and 𝑙, we choose an equidistant partitionof [𝑎, 𝑡] and [𝑡, 𝑏], respectively, as

𝑎 < 𝑎 +

𝑐 − 𝑎

𝑛

< ⋅ ⋅ ⋅ < 𝑎 +

𝑐 − 𝑎

𝑛

𝑘

< ⋅ ⋅ ⋅ < 𝑎 +

𝑐 − 𝑎

𝑛

(𝑛 − 1) < 𝑐,

𝑐 < 𝑐 +

𝑏 − 𝑐

𝑙

< ⋅ ⋅ ⋅ < 𝑐 +

𝑏 − 𝑐

𝑙

𝑖

< ⋅ ⋅ ⋅ < 𝑐 +

𝑏 − 𝑐

𝑙

(𝑙 − 1) < 𝑏,

𝑥𝑘= 𝑎 +

𝑐 − 𝑎

𝑛

𝑘, Δ𝑥𝑘=

𝑐 − 𝑎

𝑛

, 𝑘 = 1, 2, . . . , 𝑛,

𝑥𝑖= 𝑐 +

𝑏 − 𝑐

𝑙

𝑖, Δ𝑥𝑖=

𝑏 − 𝑐

𝑙

, 𝑖 = 1, 2, . . . , 𝑛.

(25)

Noting that 𝐵𝜆𝑗𝑗− ∫

𝑏

𝑎𝑓

𝜆𝑗

𝑗(𝑥)d𝑥 = 𝐵

𝜆𝑗

𝑗− (∫

𝑐

𝑎𝑓

𝜆𝑗

𝑗(𝑥)d𝑥 +

∫

𝑏

𝑐𝑓

𝜆𝑗

𝑗(𝑥)d𝑥) > 0 (𝑗 = 1, 2, . . . , 𝑚), we have

𝐵

𝜆𝑗

𝑗− { lim𝑛→∞

𝑛

∑

𝑘=1

𝑓

𝜆𝑗

𝑗[𝑎 +

𝑘 (𝑐 − 𝑎)

𝑛

]

𝑐 − 𝑎

𝑛

+ lim𝑙→∞

𝑙

∑

𝑖=1

𝑓

𝜆𝑗

𝑗[𝑐 +

𝑖 (𝑏 − 𝑐)

𝑙

]

𝑏 − 𝑐

𝑙

} > 0,

(𝑗 = 1, 2, . . . , 𝑚) .

(26)

Consequently, there exists a positive integer𝑁, such that

𝐵

𝜆𝑗

𝑗− [

𝑛

∑

𝑘=1

𝑓

𝜆𝑗

𝑗(𝑎 +

𝑘 (𝑐 − 𝑎)

𝑛

)

𝑐 − 𝑎

𝑛

+

𝑙

∑

𝑖=1

𝑓

𝜆𝑗

𝑗(𝑐 +

𝑖 (𝑏 − 𝑐)

𝑙

)

𝑏 − 𝑐

𝑙

] > 0

(27)

for all 𝑛, 𝑙 > 𝑁 and 𝑗 = 1, 2, . . . , 𝑚.By using Theorem 3, for any 𝑛, 𝑙 > 𝑁, the following

inequality holds:

𝑚

∏

𝑗=1

{𝐵

𝜆𝑗

𝑗− [

𝑛

∑

𝑘=1

𝑓

𝜆𝑗

𝑗(𝑎 +

𝑘 (𝑐 − 𝑎)

𝑛

)

𝑐 − 𝑎

𝑛

+

𝑙

∑

𝑖=1

𝑓

𝜆𝑗

𝑗(𝑐 +

𝑖 (𝑏 − 𝑐)

𝑙

)

𝑏 − 𝑐

𝑙

]}

1/𝜆𝑗

≥

𝑚

∏

𝑗=1

[𝐵

𝜆𝑗

𝑗−

𝑛

∑

𝑘=1

𝑓

𝜆𝑗

𝑗(𝑎 +

𝑘 (𝑐 − 𝑎)

𝑛

)

𝑐 − 𝑎

𝑛

]

1/𝜆𝑗

−

𝑙

∑

𝑖=1

[

[

𝑚

∏

𝑗=1

𝑓𝑗(𝑐 +

𝑖 (𝑏 − 𝑐)

𝑙

)]

]

(

𝑏 − 𝑐

𝑙

)

∑𝑚𝑗=1(1/𝜆𝑗)

.

(28)

Since𝑚

∑

𝑗=1

1

𝜆𝑗

= 1, (29)

we have𝑚

∏

𝑗=1

{𝐵

𝜆𝑗

𝑗− [

𝑛

∑

𝑘=1

𝑓

𝜆𝑗

𝑗(𝑎 +

𝑘 (𝑐 − 𝑎)

𝑛

)

𝑐 − 𝑎

𝑛

+

𝑙

∑

𝑖=1

𝑓

𝜆𝑗

𝑗(𝑐 +

𝑖 (𝑏 − 𝑐)

𝑙

)

𝑏 − 𝑐

𝑙

]}

1/𝜆𝑗

≥

𝑚

∏

𝑗=1

[𝐵

𝜆𝑗

𝑗−

𝑛

∑

𝑘=1

𝑓

𝜆𝑗

𝑗(𝑎 +

𝑘 (𝑐 − 𝑎)

𝑛

)

𝑐 − 𝑎

𝑛

]

1/𝜆𝑗

−

𝑙

∑

𝑖=1

[

[

𝑚

∏

𝑗=1

𝑓𝑗(𝑐 +

𝑖 (𝑏 − 𝑐)

𝑙

)]

]

(

𝑏 − 𝑐

𝑙

) .

(30)

Noting that 𝑓𝑗(𝑥) (𝑗 = 1, 2, . . . , 𝑚) are positive Riemann

integrable functions on [𝑎, 𝑏], we know that ∏𝑚𝑗=1𝑓𝑗(𝑥) and

𝑓

𝜆𝑗

𝑗(𝑥) are also integrable on [𝑎, 𝑏]. Letting 𝑛 → ∞ on both

sides of inequality (30), we get the left side of inequality (24).The proof of Theorem 8 is completed.

We give here a direct consequence from Theorem 8.Putting𝑚 = 2, 𝜆

1= 𝑝, 𝜆

2= 𝑞, 𝐵

1= 𝑎1, 𝐵2= 𝑏1, 𝑓1= 𝑓, and

𝑓2= 𝑔 in (24), we obtain a special important case as follows.

Journal of Applied Mathematics 5

Corollary 9. Let 𝑝 and 𝑞 be real numbers such that 𝑝 > 0, 𝑞 <0, and (1/𝑝) + (1/𝑞) = 1, let 𝑎

1, 𝑏1> 0, and let 𝑓, 𝑔 be positive

integrable functions defined on [𝑎, 𝑏]with 𝑎𝑝1−∫

𝑏

𝑎𝑓𝑝(𝑥)d𝑥 > 0

and 𝑏𝑞1− ∫

𝑏

𝑎𝑔𝑞(𝑥)d𝑥 > 0. Then, for any 𝑡 ∈ [𝑎, 𝑏), one has

(𝑎𝑝

1− ∫

𝑏

𝑎

𝑓𝑝(𝑥) d𝑥)

1/𝑝

(𝑏𝑞

1− ∫

𝑏

𝑎

𝑔𝑞(𝑥) d𝑥)

1/𝑞

≥ (𝑎𝑝

1− ∫

𝑐

𝑎


1/𝑝

(𝑏𝑞

1− ∫

𝑐

𝑎

𝑔𝑞(𝑥) d𝑥)

1/𝑞

≥ 𝑎1𝑏1− ∫

𝑏

𝑎

𝑓 (𝑥) 𝑔 (𝑥) d𝑥.

(31)

Finally, we present a refinement of integral type ofgeneralized Bellman’s inequality.

Theorem 10. Let 𝐵𝑗> 0 (𝑗 = 1, 2, . . . , 𝑚), let 0 < 𝑝 < 1,

and let 𝑓𝑗(𝑥) (𝑗 = 1, 2, . . . , 𝑚) be positive integrable functions

defined on [𝑎, 𝑏] with 𝐵𝜆𝑗𝑗− ∫

𝑏

𝑎𝑓

𝜆𝑗

𝑗(𝑥)d𝑥 > 0. Then, for any

𝑡 ∈ [𝑎, 𝑏), one has

[

[

𝑚

∑

𝑗=1

(𝐵𝑝

𝑗− ∫

𝑏

𝑎

𝑓𝑝

𝑗(𝑥) d𝑥)

1/𝑝

]

]

𝑝

≥[

[

𝑚

∑

𝑗=1

(𝐵𝑝

𝑗− ∫

𝑡

𝑎

𝑓𝑝

𝑗(𝑥) d𝑥)

1/𝑝

]

]

𝑝

− ∫

𝑏

𝑡

(

𝑚

∑

𝑗=1

𝑓𝑗 (𝑥))

𝑝

d𝑥

≥ (

𝑚

∑

𝑗=1

𝐵𝑗)

𝑝

− ∫

𝑏

𝑎

(

𝑚

∑

𝑗=1

𝑓𝑗 (𝑥))

𝑝

d𝑥.

(32)

Proof. The proof of Theorem 10 is similar to the proof ofTheorem 8.

A special case to the last theorem is as follows.

Corollary 11. Let 0 < 𝑝 < 1, let 𝑎1, 𝑏1> 0, and let 𝑓, 𝑔

be positive integrable functions defined on [𝑎, 𝑏] with 𝑎𝑝1−

∫

𝑏

𝑎𝑓𝑝(𝑥)d𝑥 > 0 and 𝑏𝑞

1− ∫

𝑏

𝑎𝑔𝑞(𝑥)d𝑥 > 0. Then, for any

𝑡 ∈ [𝑎, 𝑏), one has

[(𝑎𝑝

1− ∫

𝑏

𝑎


1/𝑝

+ (𝑏𝑝

1− ∫

𝑏

𝑎

𝑔𝑝(𝑥) d𝑥)

1/𝑝

]

𝑝

≥ [(𝑎𝑝

1− ∫

𝑡

𝑎


1/𝑝

+ (𝑏𝑝

1− ∫

𝑡

𝑎

𝑔𝑝(𝑥) d𝑥)

1/𝑝

]

𝑝

− ∫

𝑏

𝑡

(𝑓 (𝑥) + 𝑔 (𝑥))𝑝d𝑥

≥ (𝑎1+ 𝑏1)𝑝− ∫

𝑏

𝑎

(𝑓 (𝑥) + 𝑔 (𝑥))𝑝d𝑥.

(33)

Acknowledgments

The authors would like to express hearty thanks to theanonymous referees for their great efforts to improve thispaper. This work was supported by the NNSF of China (no.61073121), the Natural Science Foundation of Hebei Provinceof China (no. F2012402037), the Natural Science Foundationof Hebei Education Department (no. Q2012046), and theFundamental Research Funds for the Central Universities(no. 11ML65).

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