# Generalizations of Refined Hölder’s Inequalities and Their Applications

Authors: Jingfeng Tian; Wen-Li Wang
Journal: Mathematical Problems in Engineering
 (2014)

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## Abstract

We present several new generalized versions of refined Hölder’s inequalities proposed by Tian and Hu. And then we obtained some new generalized and sharp versions of Hölder’s inequalities. As the applications, the obtained results are used to improve Aczél-Popoviciu type inequality and Aczél-Vasić-Pečarić inequality.


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## Body

## 1. Introduction

The famous Hölder’s inequality asserts that(1)∑i=1n‍aibi≤∑i=1n‍aip1/p∑i=1n‍biq1/q,
where ai≥0,bi≥0(i=1,2,…,n),p≥q>0, and 1/p+1/q=1. The sign of inequality is reversed for p<1,p≠0. For p<0, we assume that ai,bi>0. The above inequality plays an important role in many areas of pure and applied mathematics. In recent years, considerable attention has been given to this inequality involving its generalizations, refinements, variations, and applications (see [1–5] and references therein). Among various generalizations of (1), Vasić and Pečarić in [6] presented the following important theorems.Theorem A.
LetAij≥0(i=1,2,…,n,j=1,2,…,m). If βj are positive numbers such that ∑j=1m1/βj≥1, then
(2)∑i=1n‍∏j=1m‍Aij≤∏j=1m‍∑i=1n‍Aijβj1/βj.Theorem B.
LetAij>0(i=1,2,…,n,j=1,2,…,m).(a)
Ifβ1>0,βj<0(j=2,3,…,m), and if ∑j=1m‍1/βj≤1, then
(3)∑i=1n‍∏j=1m‍Aij≥∏j=1m‍∑i=1n‍Aijβj1/βj.(b)
Ifβj<0(j=1,2,…,m),
(4)∑i=1n‍∏j=1m‍Aij≥∏j=1m‍∑i=1n‍Aijβj1/βj.In 2013, Tian and Hu [7] established some interesting refinements of inequalities (2), (3), and (4) as follows.Theorem C.
LetAij≥0,(i=1,2,…,n,j=1,2,…,m), let β1≥β2≥⋯≥βm>0, ∑j=1m‍1/βj≥1, and let α(m)=m/2ifmevenm-1/2ifmodd. Then
(5)∑i=1n‍∏j=1m‍Aij≤∏j=1m‍∑i=1n‍Aijβj1/βj×∏j=1αm‍1-∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j2-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j21/2β2jkkkkkkkkkkkkkkkkk-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j21/2β2j≤∏j=1m‍∑i=1n‍Aijβj1/βj.Theorem D.
LetAij>0(i=1,2,…,n,j=1,2,…,m).(a)
Ifβ1>0,βj<0(j=2,3,…,m), and if 0<∑j=1m1/βj≤1, then
(6)∑i=1n‍∏j=1m‍Aij≥∏j=1m‍∑i=1n‍Aijβj1/βj×∏j=2m‍1-12βj∑i=1n1/iAi1β1∑i=1nAi1β1∑i=1n1/iAijβj∑i=1nAijβj-∑i=1n1/iAijβj∑i=1nAijβj2kkkkkkkkkkkkkkkkk-∑i=1n1/iAijβj∑i=1nAijβj2.(b)
Ifβ1>0,βj<0(j=2,3,…,m), and if ∑j=1m1/βj<0, then
(7)∑i=1n‍∏j=1m‍Aij≥∏j=1m‍∑i=1n‍Aijβj1/βj×∏j=2m‍1-12βj∑i=1n1/iAi1β1*∑i=1nAi1β1*∑i=1n1/iAijβj∑i=1nAijβj-∑i=1n1/iAijβj∑i=1nAijβj2-∑i=1n1/iAijβj∑i=1nAijβj2.(c)
Ifβj<0(j=1,2,…,m), then
(8)∑i=1n‍∏j=1m‍Aij≥∏j=1m‍∑i=1n‍Aijβj1/βj×∏j=2m‍1-12βj∑i=1n1/iAi1β1*∑i=1nAi1β1*∑i=1n1/iAijβj∑i=1nAijβj-∑i=1n1/iAijβj∑i=1nAijβj2-∑i=1n1/iAijβj∑i=1nAijβj2,where 1/β1*=1-∑j=2m1/βj.The main goal of this paper is to give generalizations of inequality (5), (6), (7), and (8), and then some new generalized and sharp versions of Hölder’s inequalities are given. Moreover, the obtained results will be applied to improve Aczél-Popoviciu type inequality and Aczél-Vasić-Pečarić inequality.


## 2. Generalizations of Refined Hölder’s Inequalities

In order to prove our main results, we need the following lemmas.Lemma 1 (see [8]).
Ifx>-1,α>1 or α<0, then
(9)1+xα≥1+αx.
The sign of inequality is reversed for 0<α<1.Lemma 2 (see [9]).
Ifxi>0,λi>0,i=1,2,…,n,0≤p≤1, then
(10)∑i=1n‍λixip≤∑i=1n‍λi1-p∑i=1n‍λixip.
The sign of inequality is reversed for p≥1 or p<0.Lemma 3 (see [10]).
Let0≤x<1,α>0. Then
(11)1-x1/α≤1-xmax⁡⁡α,1.Next we provide the generalization of inequality (5).Theorem 4.
LetAij>0,(i=1,2,…,n,j=1,2,…,m), let β1≥β2≥⋯≥βm>0, and let τ=min⁡⁡{∑j=1k1/βj,1}, α(m)=m/2ifmevenm-1/2ifmodd. Then
(12)∑i=1n‍∏j=1m‍Aij≤n1-τ∏j=1m‍∑i=1n‍Aijβj1/βj×-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/2β2j∏j=1αm‍-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21-∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/2β2j≤n1-τ∏j=1m‍∑i=1n‍Aijβj1/βj.Proof.
After some simple calculations, we obtain(13)∑i=1n‍∏j=1m‍Aij∑s=1n‍∏t=1m‍Ast1-1i+1s=∑i=1n‍∑s=1n‍∏j=1m‍Aij∏t=1m‍Ast-∑i=1n‍∑s=1n‍1i∏j=1m‍Aij∏t=1m‍Ast+∑i=1n‍∑s=1n‍1s∏j=1m‍Aij∏t=1m‍Ast=∑i=1n‍∏j=1m‍Aij2.Case (I). When ∑j=1m1/βj≥1, then τ=1. Obviously, inequality (12) is equivalent to inequality (5).
Case (II). Let m be even and let ∑j=1k1/βj<1. Write ∑j=1k‍1/βj=τ(0<τ<1), which implies ∑j=1k1/τβj=1. By inequality (2), we have
(14)∑i=1n‍∏j=1m‍Aij∑s=1n‍∏t=1m‍Ast1-1i+1s=∑i=1n‍∏j=1m‍Aij∑s=1n‍∏t=1m‍Ast1-1i+1s1/τβt≤∑i=1n‍∏j=1m‍Aij∏t=1m‍∑s=1n‍Astτβt1-1i+1s1/τβt=∑i=1n‍∏j=1m/2‍Ai2j-1τβ2j-1∑s=1n‍As(2j-1)τβ2j-1kk∑s=1n‍×1-1i+1s1/τβ2j-1-1/τβ2ji×Ai2j-1τβ2j-1∑s=1n‍As2jτβ2j1-1i+1s1/τβ2jki∏j=1m/2‍×Ai2jτβ2j∑s=1n‍As2j-1τβ2j-11-1i+1s1/τβ2j.
Consequently, according to (1/τβ1-1/τβ2)+ 1/τβ2+1/τβ2+ (1/τβ3-1/τβ4)+ 1/τβ4+1/τβ4+⋯+ (1/τβm-1-1/τβm)+ 1/τβm+1/τβm=1, by using inequality (2) on the right side of (14), we observe that
(15)∑i=1n‍∏j=1m‍Aij∑s=1n‍∏t=1m‍Ast1-1i+1s≤∏j=1m/2‍∑i=1n‍Ai2j-1τβ2j-1∑s=1n‍As2j-1τβ2j-1∑i=1n‍Ai2jτβ2j∑s=1n‍As2j-1τβ2j-11-1i+1s1/τβ2jkkkkkkkk∑i=1n‍kk×1-1i+1s1/τβ2j-1-1/τβ2jkkkk×∑i=1n‍Ai2j-1τβ2j-1∑s=1n‍As2jτβ2j1-1i+1s1/τβ2jkkkk×∑i=1n‍Ai2jτβ2j∑s=1n‍As2j-1τβ2j-11-1i+1s1/τβ2j=∏j=1m/2‍∑i=1n‍∑s=1n‍Ai(2j-1)τβ2j-1As(2j-1)τβ2j-1∑i=1n‍∑s=1n‍Ai2jτβ2jAs2j-1τβ2j-11-1i+1s1/τβ2jkkkkkkkkkk∑i=1n‍×1-1i+1s1/τβ2j-1-1/τβ2jkkkk×∑i=1n‍∑s=1n‍Ai2j-1τβ2j-1As2jτβ2j1-1i+1s1/τβ2jkkkkkkki×∑i=1n‍∑s=1n‍Ai2jτβ2jAs2j-1τβ2j-11-1i+1s1/τβ2j.
Additionally, by using Lemma 2 on the right side of (15), we have
(16)∏j=1m/2‍∑i=1n‍∑s=1n‍Ai2j-1τβ2j-1As2j-1τβ2j-1∑i=1n‍∑s=1n‍Ai2jτβ2jAs2j-1τβ2j-11-1i+1s1/τβ2jkkkkkk∑i=1n‍×1-1i+1s1/τβ2j-1-1/τβ2jkkkkk×∑i=1n‍∑s=1n‍Ai2j-1τβ2j-1As2jτβ2j1-1i+1s1/τβ2jkkkkk×∑i=1n‍∑s=1n‍Ai2jτβ2jAs2j-1τβ2j-11-1i+1s1/τβ2j≤∏j=1m/2‍∑i=1n‍∑s=1n‍1-1i+1s1-τ1/τβ2j-1-1/τβ2jkkkkkkkkk×∑i=1n‍∑s=1n‍Ai2j-1β2j-1As2j-1β2j-1kkkkkkkkkkkkkk∑i=1n‍×1-1i+1s1/β2j-1-1/β2jkkkkkkkkk×∑i=1n‍∑s=1n‍1-1i+1s1-τ/τβ2j∑i=1n‍∑s=1n‍Ai2j-1β2j-1As2jβ2j1-1i+1s1/β2jkkkkkkkki×∑i=1n‍∑s=1n‍Ai2j-1β2j-1As2jβ2j1-1i+1s1/β2jkkkkkkkki×∑i=1n‍∑s=1n‍1-1i+1s1-τ/τβ2jkkkkkkkki∑i=1n‍∑s=1n‍1-1i+1s1-τ1/τβ2j-1-1/τβ2j×∑i=1n‍∑s=1n‍Ai2jβ2jAs2j-1β2j-11-1i+1s1/β2j=∑i=1n‍∑s=1n‍1-1i+1s1-τ×∏j=1m/2‍∑i=1n‍∑s=1n‍Ai(2j-1)β2j-1As(2j-1)β2j-1∑i=1n‍∑s=1n‍Ai2jβ2jAs2j-1β2j-11-1i+1s1/β2jkkkkkkkkkkkkk∑i=1n‍×1-1i+1s1/β2j-1-1/β2jkkkkkkkkkk×∑i=1n‍∑s=1n‍Ai2j-1β2j-1As2jβ2j1-1i+1s1/β2jkkkkkkkki×∑i=1n‍∑s=1n‍Ai2jβ2jAs2j-1β2j-11-1i+1s1/β2j=n2-2τ∏j=1m/2‍∑i=1n‍Ai2j-1β2j-12/β2j-1-2/β2jkkkkkkkkkkkkk×∑i=1n‍∑s=1n‍Ai2j-1β2j-1As2jβ2j1-1i+1skkkkkkkkkkkkkkkkk×∑i=1n‍∑s=1n‍Ai(2j)β2jAs(2j-1)β2j-1kkkkkkkkkkkkkkkkkkk∑i=1n‍×1-1i+1s1/β2j∑i=1n‍Ai2j-1β2j-12/β2j-1-2/β2j=n2-2τ∏j=1m/2‍∑i=1n‍Ai2j-1β2j-12/β2j-1-2/β2jkkkkkkkkkkkik×∑i=1n‍Ai(2j-1)β2j-1∑s=1n‍As(2j)β2jkkkkkkkkkkkkkkkki-∑i=1n‍1iAi(2j-1)β2j-1∑s=1n‍As(2j)β2jkkkkkkkkkkkkkkkki+∑i=1n‍Ai(2j-1)β2j-1∑s=1n‍1sAs(2j)β2jkkkkkkkkkkkkkkkii×∑i=1n‍Ai(2j)β2j∑s=1n‍As(2j-1)β2j-1kkkkkkkkkkkkkkkkkkki-∑i=1n‍1iAi(2j)β2j∑s=1n‍As(2j-1)β2j-1kkkkkkkkkkkkkkkkkiii+∑i=1n‍Ai(2j)β2j∑s=1n‍1sAs(2j-1)β2j-11/β2j∑i=1n‍Ai2j-1β2j-12/β2j-1-2/β2j=n2-2τ∏j=1m/2‍∑i=1n‍Ai2j-1β2j-12/β2j-1-2/β2j×∑i=1n‍1iAi2jβ2j21/β2jkkkkkkkkkkki×∑i=1n‍Ai2j-1β2j-1∑i=1n‍Ai2jβ2j2kkkkkkkkkkkkkkk-∑i=1n‍1iAi2j-1β2j-1∑i=1n‍Ai2jβ2jkkkkkkkkkkkkkkkkkk-∑i=1n‍Ai2j-1β2j-1kkkkkkkkkkkkkiki×∑i=1n‍1iAi2jβ2j21/β2j=n2-2τ∏j=1m/2‍∑i=1n‍Ai2j-1β2j-12/β2j-1-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/β2jkkkkkkkkkkkk×∑i=1n‍Ai2jβ2j2/β2jkkkkkkkkkkkk×1-∑i=1n1/iAi(2j-1)β2j-1∑i=1nAi(2j-1)β2j-1-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j2kkkkkkkkkkkkikkkkkkk-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/β2j=n2-2τ∏j=1m‍∑i=1n‍Aijβj2/βj×∏j=1m/2‍-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21-∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/β2jkkkkkkkkkkkkkkkk-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/β2j.
Combining inequalities (13), (15), and (16) we can get
(17)∑i=1n‍∏j=1m‍Aij≤n1-τ∏j=1m‍∑i=1n‍Aijβj1/βj×∏j=1m/2‍1-∑i=1n1/iAi(2j-1)β2j-1∑i=1nAi(2j-1)β2j-1-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j2-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/2β2jkkkkkkkkkkkkkkkk-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/2β2j.
Furthermore, noting that(18)0<∑i=1n1/iAi(2j-1)β2j-1∑i=1nAi(2j-1)β2j-1,∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j≤1,
we have
(19)∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j<1.
Consequently, from (17) and (19), we obtain the desired inequality (12) when m is even.
Case (III). Let m be odd and let ∑j=1m1/βj<1. By the same method as in the above Case (II), we have
(20)∑i=1n‍∏j=1m‍Aij≤n1-τ∏j=1m‍∑i=1n‍Aijβj1/βj×-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/2β2j∏j=1m-1/2‍-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21-∑i=1n1/iAi(2j-1)β2j-1∑i=1nAi(2j-1)β2j-1kkkkk-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j21/2β2j≤n1-τ∏j=1m‍∑i=1n‍Aijβj1/βj.If we setm=2,Ai1=ai,Ai2=bi,β1=p,β2=q, then from Theorem 4 we obtain a new generalized and sharp version of Hölder’s inequality (1) as follows.Corollary 5.
Letai>0,bi>0(i=1,2,…,n),p≥q>0, and let τ=min⁡⁡{(1/p)+(1/q),1}. Then
(21)∑i=1n‍aibi≤n1-τ∑i=1n‍aip1/p∑i=1n‍biq1/q×1-∑i=1naip/i∑i=1naip-∑i=1nbiq/i∑i=1nbiq21/2q=n1-τ∑i=1n‍aip1/p∑i=1n‍biq1/q×1-∑i=1n‍1iaip∑k=1nakp-biq∑k=1nbkq21/2q≤n1-τ∑i=1n‍aip1/p∑i=1n‍biq1/q.Moreover, by using Theorem4 and Lemma 3 we obtain the following refinement and generalization of the Hölder’s inequality (2).Corollary 6.
LetAij>0(i=1,2,…,n,j=1,2,…,m), let β1≥β2≥⋯≥βm>0, and let τ=min⁡⁡{∑j=1m1/βj,1},α(m)=m/2ifmevenm-1/2ifmodd. Then
(22)∑i=1n‍∏j=1m‍Aij≤n1-τ∏j=1m‍∑i=1n‍Aijβj1/βj×∏j=1α(m)‍1-1max⁡⁡{2β2j,1}-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j2kkkkkkkkkkki×∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1kkkkkkkkkkkkkki∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j2≤n1-τ∏j=1m‍∑i=1n‍Aijβj1/βj.Similarly, if we setm=2, Ai1=ai, Ai2=bi, β1=p, β2=q, then from Corollary 6 we obtain the following refinement and generalization of the Hölder’s inequality (1).Corollary 7.
Letai>0, bi>0(i=1,2,…,n), p≥q>0 and let τ=min⁡⁡{(1/p)+(1/q),1}. Then
(23)∑i=1n‍aibi≤n1-τ∑i=1n‍aip1/p∑i=1n‍biq1/q×1-1max⁡⁡2q,1∑i=1naip/i∑i=1naip-∑i=1nbiq/i∑i=1nbiq2=n1-τ∑i=1n‍aip1/p∑i=1n‍biq1/q×1-1max⁡⁡2q,1∑i=1n‍1iaip∑k=1nakp-biq∑k=1nbkq2≤n1-τ∑i=1n‍aip1/p∑i=1n‍biq1/q.Next, we will give the generalizations of inequalities (6), (7), and (8).Theorem 8.
LetAij>0(i=1,2,…,n,j=1,2,…,m), let β1≠0, βj<0(j=2,3,…,m), and let λ=max⁡⁡{∑j=1m1/βj,1}. Then
(24)∑i=1n‍∏j=1m‍Aij≥n1-λ∏j=1m‍∑i=1n‍Aijβj1/βj×∑i=1n1/iAijβj∑i=1nAijβj2∏j=2m‍∑i=1n1/iAijβj∑i=1nAijβj21-12βj∑i=1n1/iAijβj∑i=1nAijβj∑i=1n1/iAi1β1*∑i=1nAi1β1kkkkkkkkkkkkkkkkkiii-∑i=1n1/iAijβj∑i=1nAijβj2≥n1-λ∏j=1m‍∑i=1n‍Aijβj1/βj,
where 1/β1*=1/β1if∑j=1m1/βj>01-∑j=2m1/βjif∑j=1m1/βj<0.Proof.
Case (I). When β1>0, βj<0(j=2,3,…,m) with ∑j=1m1/βj≥1. Obviously, λ=∑j=1m1/βj, which implies ∑j=1m1/λβj=1. From inequality (3), we have
(25)∑i=1n‍∏j=1m‍Aij∑s=1n‍∏t=1m‍Ast1-1s+1i=∑i=1n‍∏j=1m‍Aij∑s=1n‍∏t=1m‍Ast1-1s+1i1/λβt≥∑i=1n‍∏j=1m‍Aij∏t=1m‍∑s=1n‍Astλβt1-1s+1i1/λβt=∑i=1n‍Ai1λβ1∑s=1n‍As1λβ11-1s+1i1/λβ1-∑j=2m1/λβjk×∏j=2m‍Ai1λβ1∑s=1n‍Asjλβj1-1s+1i1/λβjkkkkkkkkAi1λβ1∑s=1n‍As1λβ11-1s+1i1/λβ1-∑j=2m1/λβj×∏j=2m‍Aijλβj∑s=1n‍As1λβ11-1s+1i1/λβj.
Consequently, according to(1/λβ1-∑j=2m1/λβj)+ 1/λβ2+1/λβ3+⋯+ 1/λβm+1/λβ2+ 1/λβ3+⋯+ 1/λβm=1, by using the inequality (3) on the right side of (25), we observe that
(26)∑i=1n‍∏j=1m‍Aij∑s=1n‍∏t=1m‍Ast1-1s+1i≥∑i=1n‍∑s=1n‍Ai1λβ1As1λβ11-1s+1i1/λβ1-∑j=2m1/λβj×∏j=2m‍∑i=1n‍∑s=1n‍Ai1λβ1Asjλβj1-1s+1i1/λβj×∏j=2m‍∑i=1n‍∑s=1n‍AijλβjAs1λβ11-1s+1i1/λβj.
Moreover, using Lemma 2 together with λ≥1, we find
(27)∑i=1n‍∑s=1n‍Ai1λβ1As1λβ11-1s+1i1/λβ1-∑j=2m1/λβj×∏j=2m‍∑i=1n‍∑s=1n‍Ai1λβ1Asjλβj1-1s+1i1/λβj×∏j=2m‍∑i=1n‍∑s=1n‍AijλβjAs1λβ11-1s+1i1/λβj≥∑i=1n‍∑s=1n‍1-1s+1i(1-λ)(1/λβ1-∑j=2m1/λβj)×∑i=1n‍∑s=1n‍Ai1β1As1β11-1s+1i1/β1-∑j=2m1/βj×∏j=2m‍∑i=1n‍∑s=1n‍1-1s+1i1-λ/λβjkkkkkkkk×∑i=1n‍∑s=1n‍Ai1β1Asjβj1-1s+1i1/βj×∏j=2m‍∑i=1n‍∑s=1n‍1-1s+1i1-λ/λβjkkkkkkk×∑i=1n‍∑s=1n‍AijβjAs1β11-1s+1i1/βj=∑i=1n‍∑s=1n‍1-1s+1i1-λ×∑i=1n‍∑s=1n‍Ai1β1As1β11-1s+1i1/β1-∑j=2m1/βj×∏j=2m‍∑i=1n‍∑s=1n‍Ai1β1Asjβj1-1s+1i1/βj×∏j=2m‍∑i=1n‍∑s=1n‍AijβjAs1β11-1s+1i1/βj=n2-2λ∑i=1n‍Ai1β12/β1-∑j=2m2/βj×∏j=2m‍∑i=1n‍∑s=1n‍Ai1β1Asjβj1-1s+1i×∑i=1n‍∑s=1n‍AijβjAs1β11-1s+1i1/βjkkkkkkkkkkk×∑i=1n‍∑s=1n‍AijβjAs1β11-1s+1i1/βj=n2-2λ∑i=1n‍Ai1β12/β1-∑j=2m2/βj×∏j=2m‍∑i=1n‍Ai1β1∑s=1n‍Asjβj+∑i=1n‍1iAijβj∑s=1n‍As1β11/βjkkkkkkkkkkkkki-∑i=1n‍Ai1β1∑s=1n‍1sAsjβjkkkkkkkkkkkkkki+∑i=1n‍1iAi1β1∑s=1n‍Asjβjkkkkkkkkkk×∑i=1n‍Aijβj∑s=1n‍As1β1-∑i=1n‍Aijβj∑s=1n‍1sAs1β1kkkkkkkkkkkkkkkkk+∑i=1n‍1iAijβj∑s=1n‍As1β11/βj=n2-2λ∑i=1n‍Ai1β12/β1-∑j=2m2/βj×∏j=2m‍∑i=1n‍Ai1β1∑s=1n‍Asjβj2-(∑i=1n‍1iAi1β1)(∑s=1n‍Asjβj)21/βjkkkkkkkkkkk-∑i=1n‍Ai1β1∑s=1n‍1sAsjβjkkkkkkkkkkki-∑i=1n‍1iAi1β1∑s=1n‍Asjβj21/βj=n2-2λ∑i=1n‍Ai1β12/β1-∑j=2m2/βj×-∑i=1n1/iAi1β1∑i=1nAi1β1∑s=1n1/sAsjβj∑s=1nAsjβj2-∑i=1n1/iAi1β1∑i=1nAi1β1∑s=1n1/sAsjβj∑s=1nAsjβj2sy1/βj∏j=2m‍∑i=1n‍Ai1β1∑s=1n‍Asjβj2/βj-∑i=1n1/iAi1β1∑i=1nAi1β121/βjkkkkkkki×-∑i=1n1/iAi1β1∑i=1nAi1β1∑s=1n1/sAsjβj∑s=1nAsjβj21/βj1-∑s=1n1/sAsjβj∑s=1nAsjβj-∑i=1n1/iAi1β1∑i=1nAi1β12kkkkkkkkkkkkkk-∑i=1n1/iAi1β1∑i=1nAi1β1∑s=1n1/sAsjβj∑s=1nAsjβj2-∑i=1n1/iAi1β1∑i=1nAi1β1∑s=1n1/sAsjβj∑s=1nAsjβj2sy1/βj=n2-2λ∏j=1m‍∑i=1n‍Aijβj2/βj×-∑i=1n1/iAi1β1∑i=1nAi1β121/βj∏j=2m‍1-∑i=1n1/iAijβj∑i=1nAijβj-∑i=1n1/iAi1β1∑i=1nAi1β12-∑i=1n1/iAi1β1∑i=1nAi1β121/βjkkkkkkkkikkkkkk-∑i=1n1/iAi1β1∑i=1nAi1β121/βj.
Combining inequalities (13), (26), and (27) we can get
(28)∑i=1n‍∏j=1m‍Aij≥n1-λ∏j=1m‍∑i=1n‍Aijβj1/βj×∏j=2m‍1-∑i=1n1/iAijβj∑i=1nAijβj-∑i=1n1/iAi1β1∑i=1nAi1β12-∑i=1n1/iAi1β1∑i=1nAi1β121/2βjkk-∑i=1n1/iAi1β1∑i=1nAi1β121/2βj.
Furthermore, noting that(29)0<∑i=1n1/iAijβj∑i=1nAijβj,∑i=1n1/iAi1β1∑i=1nAi1β1≤1,
we have
(30)∑i=1n1/iAijβj∑i=1nAijβj-∑i=1n1/iAi1β1∑i=1nAi1β1<1.
Consequently, from Lemma 1 and the inequalities (28) and (30), we have the desired inequality (24).
Case (II). When β1>0,βj<0(j=2,3,…,m) with ∑j=1m1/βj<1, then λ=1. The inequality (24) is equivalent to inequalities (6) and (7).
Case (III). When βj<0(j=1,2,…,m), then λ=1. Inequality (24) is equivalent to inequality (8).If we setm=2,Ai1=ai,Ai2=bi,β1=p,β2=q, then from Theorem 8 we obtain the following refinement and generalization of the Hölder’s inequality (1).Corollary 9.
Letai>0,bi>0(i=1,2,…,n),p≠0,q<0, and let λ=max⁡⁡{1/p+1/q,1}. Then
(31)∑i=1n‍aibi≥n1-λ∑i=1n‍aip1/p∑i=1n‍biq1/q×1-12q∑i=1naip*/i∑i=1naip-∑i=1nbiq/i∑i=1nbiq2=n1-λ∑i=1n‍aip1/p∑i=1n‍biq1/q×1-12q∑i=1n‍1iaip*∑k=1nakp-biq∑k=1nbkq2≥n1-λ∑i=1n‍aip1/p∑i=1n‍biq1/q,
where 1/p*=1/pif(1/p)+(1/q)>01-(1/q)if(1/p)+(1/q)<0.


## 3. Applications

In this section, we show two applications of our new inequalities. Firstly, we provide an application of the obtained results to improve the Aczél-Popoviciu type inequality, which is due to Wu and Debnath.The classical Aczél [11] inequality states that if ai>0, bi>0(i=1,2,…,n), a12-∑i=2nai2>0, b12-∑i=2nbi2>0, then
(32)a12-∑i=2n‍ai2b12-∑i=2n‍bi2≤a1b1-∑i=2n‍aibi2.As is well-known, Aczél’s inequality plays an important role in the theory of functional equations in non-Euclidean geometry, and many researchers (see [12–17] and references therein) have given considerable attention to this inequality and its generalizations and refinements.In 1959, Popoviciu [18] presented the following extension of inequality (32).Theorem E.
Letp>1,q>1,1/p+1/q=1, and let ai>0,  bi>0(i=1,2,…,n),  a1p-∑i=2naip>0,  b1q-∑i=2nbiq>0. Then
(33)a1p-∑i=2n‍aip1/pb1q-∑i=2n‍biq1/q≤a1b1-∑i=2n‍aibi.The inequality (33) is called Aczél-Popoviciu inequality.Later, in 1982, Vasić and Pečarić [5] obtained the reversed version of inequality (33). The inequality is called Aczél-Vasić-Pečarić inequality.Theorem F.
Letp<1(p≠0),1/p+1/q=1, and let ai>0, bi>0(i=1,2,…,n), a1p-∑i=2naip>0, b1q-∑i=2nbiq>0. Then
(34)a1p-∑i=2n‍aip1/pb1q-∑i=2n‍biq1/q≥a1b1-∑i=2n‍aibi.In a recent paper [19], Wu and Debnath established an interesting generalization of Aczél-Popoviciu inequality (33) as follows.Theorem G.
LetAij>0, tj>0, βj>0, (∑i=1nAij)βj-tjβj∑i=1nAijβj>0, i=1,2,…,n, j=1,2,…,m, and let τ=min⁡⁡{∑j=1m1/βj,1}. Then
(35)∏j=1m‍∑i=1n‍Aijβj-tjβj∑i=1n‍Aijβj1/βj≤n+11-τ∏j=1m‍∑i=1n‍Aij-t1t2⋯tm∑i=1n‍∏j=1m‍Aij.Next, by using Corollary6 we present the following refinement of inequality (35).Theorem 10.
Letβ1≥β2≥⋯≥βm>0, let Aij>0, tj>0,  (∑i=1nAij)βj- tjβj∑i=1nAijβj>0, i=1,2,…,n, j=1,2,…,m, and let τ=min⁡⁡{∑j=1m1/βj,1}. Then
(36)∏j=1m‍∑i=1n‍Aijβj-tjβj∑i=1n‍Aijβj1/βj≤n+11-τ∏j=1m‍∑i=1n‍Aij×∏j=1α(m)‍1-1max⁡⁡2β2j,1-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j2k×∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1kkkkkkkkkkkk∏j=1α(m)‍-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j2-t1t2⋯tm∑i=1n‍∏j=1m‍Aij,
where α(m)=m/2ifmevenm-1/2ifmodd.Proof.
Denote(37)tjβjA(n+1)jβj=∑i=1n‍Aijβj-tjβj∑i=1n‍Aijβj.
Since (∑i=1nAij)βj-tjβj∑i=1nAijβj>0, we obtain A(n+1)j>0,j=2,3,…,m. From Corollary 6, we have
(38)∏j=1m‍∑i=1n‍Aij=t1t2⋯tm∏j=1m‍∑i=1n+1‍Aijβj1/βj≥t1t2⋯tmn+1τ-1∑i=1n+1‍∏j=1m‍Aij×∏j=1α(m)‍1-1max⁡⁡2β2j,1-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j2kkkkkkkik×∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1kkkkkkkkkkkkki∏j=1α(m)‍-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j2-1=t1t2⋯tmn+1τ-1×∏j=1m‍An+1j+∑i=1n‍∏j=1m‍Aij×∏j=1α(m)‍1-1max⁡⁡2β2j,1-∑i=1n1/iAi(2j)β2j∑i=1nAi(2j)β2j2kkkkkkkkk×∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1kkkkkkkkkkkkk∏j=1αm‍∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j2-1;
therefore,
(39)t1t2⋯tm∏j=1m‍A(n+1)j≤n+11-τ∏j=1m‍∑i=1n‍Aij×∏j=1α(m)‍1-1max⁡⁡2β2j,1-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j2kkkkkkkkkki×∑i=1n1/iAi2j-1β2j-1∑i=1nAi2j-1β2j-1kkkkkkkkkikkkkk∏j=1α(m)‍-∑i=1n1/iAi2jβ2j∑i=1nAi2jβ2j2-t1t2⋯tm∑i=1n‍∏j=1m‍Aij.
Substituting
(40)tjA(n+1)j=∑i=1n‍Aijβj-tjβj∑i=1n‍Aijβj1/βj
into the inequality (39), we have the desired inequality (36).Finally, we give the following Aczél-Vasić-Pečarić type inequality, which is the refinement of Aczél-Vasić-Pečarić inequality (34).Theorem 11.
Letβ1≠0,βj<0(j=2,3,…,m), let Aij>0, tj>0, (∑i=1nAij)βj-tjβj∑i=1nAijβj>0,  i=1,2,⋯,n,j=1,2,⋯,m, and let λ=max⁡⁡{∑j=1m1/βj,1}. Then
(41)∏j=1m‍∑i=1n‍Aijβj-tjβj∑i=1n‍Aijβj1/βj≥n+11-λ∏j=1m‍∑i=1n‍Aij×∏j=2m‍1-12βj-∑i=1n1/iAijβj∑i=1nAijβj2kkkkkkkkkkk×∑i=1n1/iAijβj∑i=1nAijβj∑i=1n1/iAi1β1*∑i=1nAi1β1kkkkkkkkkkkkkkkk-∑i=1n1/iAijβj∑i=1nAijβj2-t1t2⋯tm∑i=1n‍∏j=1m‍Aij,
where 1/β1*=1/β1if∑j=1m1/βj>01-∑j=2m1/βjif∑j=1m1/βj<0.Proof.
By the same method as in Theorem10, from Theorem 8, we get the desired inequality (41).



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*Source: 581962-2014-10-30.xml*