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# A sharpened and generalized version of Aczél-Vasić-Pečarić inequality and its application

Journal of Inequalities and Applications20132013:497

https://doi.org/10.1186/1029-242X-2013-497

• Received: 27 June 2013
• Accepted: 3 October 2013
• Published:

## Abstract

In this paper, we present a sharpened and generalized version of Aczél-Vasić-Pečarić inequality. As an application, an integral type of Aczél-Vasić-Pečarić inequality is obtained.

MSC:26D15, 26D10.

## Keywords

• Aczél’s inequality
• Aczél-Vasić-Pečarić inequality
• Hölder’s inequality

## 1 Introduction

In 1956, Aczél  established the following inequality, which is of wide application in the theory of functional equations in non-Euclidean geometry.

Theorem A If ${a}_{i}$, ${b}_{i}$ ($i=1,2,\dots ,n$) are positive numbers such that ${a}_{1}^{2}-{\sum }_{i=2}^{n}{a}_{i}^{2}>0$ or ${b}_{1}^{2}-{\sum }_{i=2}^{n}{b}_{i}^{2}>0$, then
$\left({a}_{1}^{2}-\sum _{i=2}^{n}{a}_{i}^{2}\right)\left({b}_{1}^{2}-\sum _{i=2}^{n}{b}_{i}^{2}\right)\le {\left({a}_{1}{b}_{1}-\sum _{i=2}^{n}{a}_{i}{b}_{i}\right)}^{2}.$
(1)

Later, in 1959, Popoviciu  gave a generalization of the above inequality.

Theorem B Let $p>1$, $q>1$, $\frac{1}{p}+\frac{1}{q}=1$, and let ${a}_{i}$, ${b}_{i}$ ($i=1,2,\dots ,n$) be positive numbers such that ${a}_{1}^{p}-{\sum }_{i=2}^{n}{a}_{i}^{p}>0$ and ${b}_{1}^{q}-{\sum }_{i=2}^{n}{b}_{i}^{q}>0$. Then
${\left({a}_{1}^{p}-\sum _{i=2}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left({b}_{1}^{q}-\sum _{i=2}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\le {a}_{1}{b}_{1}-\sum _{i=2}^{n}{a}_{i}{b}_{i}.$
(2)

In 1982, Vasić and Pečarić  presented the following reversed version of inequality (2).

Theorem C Let $p<1$ ($p\ne 0$), $\frac{1}{p}+\frac{1}{q}=1$, and let ${a}_{i}$, ${b}_{i}$ ($i=1,2,\dots ,n$) be positive numbers such that ${a}_{1}^{p}-{\sum }_{i=2}^{n}{a}_{i}^{p}>0$ and ${b}_{1}^{q}-{\sum }_{i=2}^{n}{b}_{i}^{q}>0$. Then
${\left({a}_{1}^{p}-\sum _{i=2}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left({b}_{1}^{q}-\sum _{i=2}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\ge {a}_{1}{b}_{1}-\sum _{i=2}^{n}{a}_{i}{b}_{i}.$
(3)

Recently inequalities (2) and (3) were generalized and refined in many different ways; see, for example,  and . In , Wu established an interesting generalization of Aczél-Popoviciu inequality (2) as follows.

Theorem D Let $p,q>0$, ${a}_{i},{b}_{i}>0$ ($i=1,2,\dots ,n$), let k ($1\le k) be a positive integer such that ${\sum }_{i=1}^{k}{a}_{i}^{p}-{\sum }_{i=k+1}^{n}{a}_{i}^{p}>0$ and ${\sum }_{i=1}^{k}{b}_{i}^{q}-{\sum }_{i=k+1}^{n}{b}_{i}^{q}>0$. Then
$\begin{array}{c}{\left(\sum _{i=1}^{k}{a}_{i}^{p}-\sum _{i=k+1}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}-\sum _{i=k+1}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\le {2}^{max\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left(\sum _{i=1}^{k}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}\right)}^{\frac{1}{q}}-{\left(\sum _{i=k+1}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=k+1}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\frac{{2}^{max\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}}{max\left\{p,q,1\right\}}{\left(\sum _{i=1}^{k}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}\right)}^{\frac{1}{q}}{\left(\frac{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}-\frac{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{2},\hfill \end{array}$
(4)
and equality holds if and only if
$\frac{{\sum }_{i=1}^{k}{a}_{i}^{p}}{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}=\frac{{\sum }_{i=1}^{k}{b}_{i}^{q}}{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}=2$
for $\frac{1}{p}+\frac{1}{q}<1$, or
$\frac{{\sum }_{i=1}^{k}{a}_{i}^{p}}{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}=\frac{{\sum }_{i=1}^{k}{b}_{i}^{q}}{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}$

for $\frac{1}{p}+\frac{1}{q}=1$.

The main purpose of this work is to give a sharpened and generalized version of Aczél-Vasić-Pečarić inequality (3). Moreover, a new Aczél-Vasić-Pečarić type integral inequality is established.

## 2 A sharpened and generalized version of Aczél-Vasić-Pečarić inequality

We begin this section with some lemmas, which will be used in the sequel.

Lemma 2.1 

If $x>-1$, $\alpha \ge 1$ or $\alpha <0$, then
${\left(1+x\right)}^{\alpha }\ge 1+\alpha x.$
(5)

The inequality is reversed for $0<\alpha <1$. The sign of equality holds if and only if $x=0$ or $\alpha =1$.

Lemma 2.2 

Let ${a}_{rj}>0$ ($r=1,2,\dots ,n$, $j=1,2,\dots ,m$), let ${\lambda }_{1}\ne 0$, ${\lambda }_{j}<0$ ($j=2,3,\dots ,m$), and let $\tau =max\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$. Then
$\sum _{r=1}^{n}\prod _{j=1}^{m}{a}_{rj}\ge {n}^{1-\tau }\prod _{j=1}^{m}{\left(\sum _{r=1}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}.$
(6)

The sign of equality holds if and only if the m sets $\left({a}_{r1}\right),\left({a}_{r2}\right),\dots ,\left({a}_{rm}\right)$ are proportional for ${\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}}\le 1$, or ${a}_{1j}={a}_{2j}=\cdots ={a}_{nj}$, $j=1,2,\dots ,m$, for ${\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}}>1$.

Lemma 2.3 Let $x>1$, $y>1$, and let $p<0$, $q<0$. Then
$xy+{\left(1-{x}^{p}\right)}^{\frac{1}{p}}{\left(1-{y}^{q}\right)}^{\frac{1}{q}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(1-min\left\{{p}^{-1},{q}^{-1}\right\}{\left({x}^{p}-{y}^{q}\right)}^{2}\right),$
(7)

and equality holds if and only if ${x}^{p}={y}^{q}=\frac{1}{2}$.

Proof Case (I). When $p, it implies that $\frac{1}{q}<0$, $\frac{1}{p}-\frac{1}{q}>0$. By applying Lemma 2.2 and Lemma 2.1, we have
$\begin{array}{c}xy+{\left(1-{x}^{p}\right)}^{\frac{1}{p}}{\left(1-{y}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({x}^{p}\right)}^{\frac{1}{q}}{\left({y}^{q}\right)}^{\frac{1}{q}}{\left({x}^{p}\right)}^{\frac{1}{p}-\frac{1}{q}}+{\left(1-{y}^{q}\right)}^{\frac{1}{q}}{\left(1-{x}^{p}\right)}^{\frac{1}{q}}{\left(1-{x}^{p}\right)}^{\frac{1}{p}-\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left({x}^{p}+\left(1-{y}^{q}\right)\right)}^{\frac{1}{q}}{\left({y}^{q}+\left(1-{x}^{p}\right)\right)}^{\frac{1}{q}}{\left({x}^{p}+\left(1-{x}^{p}\right)\right)}^{\frac{1}{p}-\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}={2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left(1-{\left({x}^{p}-{y}^{q}\right)}^{2}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(1-{q}^{-1}{\left({x}^{p}-{y}^{q}\right)}^{2}\right).\hfill \end{array}$
(8)
Case (II). When $q, it implies that $\frac{1}{p}<0$, $\frac{1}{q}-\frac{1}{p}>0$. By using Lemma 2.2 and Lemma 2.1, we obtain
$\begin{array}{c}xy+{\left(1-{x}^{p}\right)}^{\frac{1}{p}}{\left(1-{y}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({y}^{q}\right)}^{\frac{1}{p}}{\left({x}^{p}\right)}^{\frac{1}{p}}{\left({y}^{q}\right)}^{\frac{1}{q}-\frac{1}{p}}+{\left(1-{x}^{p}\right)}^{\frac{1}{p}}{\left(1-{y}^{q}\right)}^{\frac{1}{p}}{\left(1-{y}^{q}\right)}^{\frac{1}{q}-\frac{1}{p}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left({y}^{q}+\left(1-{x}^{p}\right)\right)}^{\frac{1}{p}}{\left({x}^{p}+\left(1-{y}^{q}\right)\right)}^{\frac{1}{p}}{\left({y}^{q}+\left(1-{y}^{q}\right)\right)}^{\frac{1}{q}-\frac{1}{p}}\hfill \\ \phantom{\rule{1em}{0ex}}={2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left(1-{\left({x}^{p}-{y}^{q}\right)}^{2}\right)}^{\frac{1}{p}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(1-{p}^{-1}{\left({x}^{p}-{y}^{q}\right)}^{2}\right).\hfill \end{array}$
(9)
Case (III). When $p=q$, $p<0$, $q<0$. From Lemma 2.2 and Lemma 2.1 we have
$\begin{array}{c}xy+{\left(1-{x}^{p}\right)}^{\frac{1}{p}}{\left(1-{y}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({y}^{p}\right)}^{\frac{1}{p}}{\left({x}^{p}\right)}^{\frac{1}{p}}+{\left(1-{x}^{p}\right)}^{\frac{1}{p}}{\left(1-{y}^{p}\right)}^{\frac{1}{p}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{2}{p},0\right\}}{\left({y}^{p}+\left(1-{x}^{p}\right)\right)}^{\frac{1}{p}}{\left({x}^{p}+\left(1-{y}^{p}\right)\right)}^{\frac{1}{p}}\hfill \\ \phantom{\rule{1em}{0ex}}={2}^{min\left\{1-\frac{2}{p},0\right\}}{\left(1-{\left({x}^{p}-{y}^{p}\right)}^{2}\right)}^{\frac{1}{p}}\hfill \\ \phantom{\rule{1em}{0ex}}={2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left(1-{\left({x}^{p}-{y}^{q}\right)}^{2}\right)}^{\frac{1}{p}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(1-{p}^{-1}{\left({x}^{p}-{y}^{q}\right)}^{2}\right).\hfill \end{array}$
(10)

Combining inequalities (8)-(10) yields inequality (7). The condition of equality in (7) follows immediately from Lemma 2.2 and Lemma 2.1. The proof of Lemma 2.3 is completed. □

Lemma 2.4 Let $0, $y>1$, and let $p>0$, $q<0$. Then
$xy+{\left(1-{x}^{p}\right)}^{\frac{1}{p}}{\left(1-{y}^{q}\right)}^{\frac{1}{q}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(1-min\left\{{p}^{-1},{q}^{-1}\right\}{\left({x}^{p}-{y}^{q}\right)}^{2}\right),$
(11)

and equality holds if and only if ${x}^{p}={y}^{q}=\frac{1}{2}$ for $\frac{1}{p}+\frac{1}{q}<1$, or ${x}^{p}={y}^{q}$ for $\frac{1}{p}+\frac{1}{q}=1$.

Proof By using Lemma 2.2 and Lemma 2.1, we have
$\begin{array}{c}xy+{\left(1-{x}^{p}\right)}^{\frac{1}{p}}{\left(1-{y}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({x}^{p}\right)}^{\frac{1}{q}}{\left({y}^{q}\right)}^{\frac{1}{q}}{\left({x}^{p}\right)}^{\frac{1}{p}-\frac{1}{q}}+{\left(1-{y}^{q}\right)}^{\frac{1}{q}}{\left(1-{x}^{p}\right)}^{\frac{1}{q}}{\left(1-{x}^{p}\right)}^{\frac{1}{p}-\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left({x}^{p}+\left(1-{y}^{q}\right)\right)}^{\frac{1}{q}}{\left({y}^{q}+\left(1-{x}^{p}\right)\right)}^{\frac{1}{q}}{\left({x}^{p}+\left(1-{x}^{p}\right)\right)}^{\frac{1}{p}-\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}={2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left(1-{\left({x}^{p}-{y}^{q}\right)}^{2}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(1-{q}^{-1}{\left({x}^{p}-{y}^{q}\right)}^{2}\right)\hfill \\ \phantom{\rule{1em}{0ex}}={2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(1-min\left\{{p}^{-1},{q}^{-1}\right\}{\left({x}^{p}-{y}^{q}\right)}^{2}\right).\hfill \end{array}$
(12)

In addition, the condition of equality for inequality (11) can easily be obtained by Lemma 2.1 and Lemma 2.2. The proof of Lemma 2.4 is completed. □

Theorem 2.5 Let ${a}_{i}>0$, ${b}_{i}>0$ ($i=1,2,\dots ,n$), let $p\ne 0$, $q<0$, and let k ($1\le k) be a positive integer such that ${\sum }_{i=1}^{k}{a}_{i}^{p}-{\sum }_{i=k+1}^{n}{a}_{i}^{p}>0$ and ${\sum }_{i=1}^{k}{b}_{i}^{q}-{\sum }_{i=k+1}^{n}{b}_{i}^{q}>0$. Then
$\begin{array}{c}{\left(\sum _{i=1}^{k}{a}_{i}^{p}-\sum _{i=k+1}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}-\sum _{i=k+1}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left(\sum _{i=1}^{k}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}\right)}^{\frac{1}{q}}-{\left(\frac{1}{n-k}\right)}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(\sum _{i=k+1}^{n}{a}_{i}{b}_{i}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-{2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}min\left\{{p}^{-1},{q}^{-1}\right\}{\left(\sum _{i=1}^{k}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\left(\frac{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}-\frac{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{2},\hfill \end{array}$
(13)

and equality holds if and only if ${\left(2n-2k\right)}^{-1}{\sum }_{i=1}^{k}{a}_{i}^{p}={a}_{k+1}^{p}={a}_{k+2}^{p}=\cdots ={a}_{n}^{p}$ and ${\left(2n-2k\right)}^{-1}{\sum }_{i=1}^{k}{b}_{i}^{q}={b}_{k+1}^{q}={b}_{k+2}^{q}=\cdots ={b}_{n}^{q}$ for $\frac{1}{p}+\frac{1}{q}<1$, or $\frac{{\sum }_{i=1}^{k}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}=\frac{{a}_{k+1}^{p}}{{b}_{k+1}^{q}}=\frac{{a}_{k+2}^{p}}{{b}_{k+2}^{q}}=\cdots =\frac{{a}_{n}^{p}}{{b}_{n}^{q}}$ for $\frac{1}{p}+\frac{1}{q}=1$.

Proof Case (I). When $p>0$, $q<0$. From the hypotheses of Theorem 2.5, we find that
$\begin{array}{c}0<{\left(\frac{{\sum }_{i=1}^{k}{a}_{i}^{p}-{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}\right)}^{\frac{1}{p}}<1,\hfill \\ {\left(\frac{{\sum }_{i=1}^{k}{b}_{i}^{q}-{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{\frac{1}{q}}>1.\hfill \end{array}$
Thus, by using Lemma 2.4 with a substitution
$x={\left(\frac{{\sum }_{i=1}^{k}{a}_{i}^{p}-{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}\right)}^{\frac{1}{p}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}y={\left(\frac{{\sum }_{i=1}^{k}{b}_{i}^{q}-{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{\frac{1}{q}}$
in (11), we have
$\begin{array}{c}{\left(\frac{{\sum }_{i=1}^{k}{a}_{i}^{p}-{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}\right)}^{\frac{1}{p}}{\left(\frac{{\sum }_{i=1}^{k}{b}_{i}^{q}-{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{\frac{1}{q}}+{\left(\frac{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}\right)}^{\frac{1}{p}}{\left(\frac{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×\left[1-min\left\{{p}^{-1},{q}^{-1}\right\}{\left(\frac{{\sum }_{i=1}^{k}{a}_{i}^{p}-{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}-\frac{{\sum }_{i=1}^{k}{b}_{i}^{q}-{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{2}\right],\hfill \end{array}$
(14)
which implies
$\begin{array}{c}{\left(\sum _{i=1}^{k}{a}_{i}^{p}-\sum _{i=k+1}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}-\sum _{i=k+1}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}+{\left(\sum _{i=k+1}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=k+1}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left(\sum _{i=1}^{k}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×\left[1-min\left\{{p}^{-1},{q}^{-1}\right\}{\left(\frac{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}-\frac{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{2}\right].\hfill \end{array}$
(15)
Hence, we obtain
$\begin{array}{c}{\left(\sum _{i=1}^{k}{a}_{i}^{p}-\sum _{i=k+1}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}-\sum _{i=k+1}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{\left(\sum _{i=1}^{k}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}\right)}^{\frac{1}{q}}-{\left(\sum _{i=k+1}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=k+1}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-{2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}min\left\{{p}^{-1},{q}^{-1}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\left(\sum _{i=1}^{k}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=1}^{k}{b}_{i}^{q}\right)}^{\frac{1}{q}}{\left(\frac{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}-\frac{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}\right)}^{2},\hfill \end{array}$
(16)

where the equality holds if and only if $\frac{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}=\frac{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}=\frac{1}{2}$ for $\frac{1}{p}+\frac{1}{q}<1$, or $\frac{{\sum }_{i=k+1}^{n}{a}_{i}^{p}}{{\sum }_{i=1}^{k}{a}_{i}^{p}}=\frac{{\sum }_{i=k+1}^{n}{b}_{i}^{q}}{{\sum }_{i=1}^{k}{b}_{i}^{q}}$ for $\frac{1}{p}+\frac{1}{q}=1$.

On the other hand, by using Lemma 2.2, we obtain
${\left(\sum _{i=k+1}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left(\sum _{i=k+1}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\le {\left(\frac{1}{n-k}\right)}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(\sum _{i=k+1}^{n}{a}_{i}{b}_{i}\right),$
(17)

where the equality holds if and only if ${a}_{k+1}={a}_{k+2}=\cdots ={a}_{n}$ and ${b}_{k+1}={b}_{k+2}=\cdots ={b}_{n}$ for $\frac{1}{p}+\frac{1}{q}<1$, or $\frac{{a}_{k+1}^{p}}{{b}_{k+1}^{q}}=\frac{{a}_{k+2}^{p}}{{b}_{k+2}^{q}}=\cdots =\frac{{a}_{n}^{p}}{{b}_{n}^{q}}$ for $\frac{1}{p}+\frac{1}{q}=1$.

Combining the above two inequalities gives the desired result.

Case (II). When $p<0$, $q<0$. By the same method as in the above case (I) and using Lemma 2.3 and Lemma 2.2, we get that inequality (13) is also valid. The proof of Theorem 2.5 is completed. □

Remark 2.6 If we set $k=1$, $\frac{1}{p}+\frac{1}{q}=1$, and $\frac{{\sum }_{i=2}^{n}{a}_{i}^{p}}{{a}_{1}^{p}}=\frac{{\sum }_{i=2}^{n}{b}_{i}^{q}}{{b}_{1}^{q}}$ in Theorem 2.5, then inequality (13) reduces to inequality (3).

If we set $k=1$, then from Theorem 2.5 we obtain the following sharpened and generalized version of Aczél-Vasić-Pečarić inequality (3).

Corollary 2.7 Let $p\ne 0$, $q<0$, and let ${a}_{i}>0$, ${b}_{i}>0$, ${a}_{1}^{p}-{\sum }_{i=2}^{n}{a}_{i}^{p}>0$, ${b}_{1}^{q}-{\sum }_{i=2}^{n}{b}_{i}^{q}>0$ $\left(i=1,2,\dots ,n\right)$. Then
$\begin{array}{c}{\left({a}_{1}^{p}-\sum _{i=2}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left({b}_{1}^{q}-\sum _{i=2}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}{a}_{1}{b}_{1}-{\left(\frac{1}{n-1}\right)}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}\left(\sum _{i=2}^{n}{a}_{i}{b}_{i}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-{2}^{min\left\{1-\frac{1}{p}-\frac{1}{q},0\right\}}min\left\{{p}^{-1},{q}^{-1}\right\}{a}_{1}{b}_{1}{\left[\sum _{i=2}^{n}\left(\frac{{a}_{i}^{p}}{{a}_{1}^{p}}-\frac{{b}_{i}^{q}}{{b}_{1}^{q}}\right)\right]}^{2},\hfill \end{array}$
(18)

and equality holds if and only if ${\left(2n-2\right)}^{-\frac{1}{p}}{a}_{1}={a}_{2}=\cdots ={a}_{n}$ and ${\left(2n-2\right)}^{-\frac{1}{p}}{b}_{1}={b}_{2}=\cdots ={b}_{n}$ for $\frac{1}{p}+\frac{1}{q}<1$, or $\frac{{a}_{1}^{p}}{{b}_{1}^{q}}=\frac{{a}_{2}^{p}}{{b}_{2}^{q}}=\cdots =\frac{{a}_{n}^{p}}{{b}_{n}^{q}}$ for $\frac{1}{p}+\frac{1}{q}=1$.

In particular, if we set $\frac{1}{p}+\frac{1}{q}\le 1$, then from Corollary 2.7 we get the sharpened version of Aczél-Vasić-Pečarić inequality (3) as follows.

Corollary 2.8 Let $p>0$, $q<0$, $\frac{1}{p}+\frac{1}{q}\le 1$, and let ${a}_{i}>0$, ${b}_{i}>0$, ${a}_{1}^{p}-{\sum }_{i=2}^{n}{a}_{i}^{p}>0$, ${b}_{1}^{q}-{\sum }_{i=2}^{n}{b}_{i}^{q}>0$ ($i=1,2,\dots ,n$). Then
$\begin{array}{c}{\left({a}_{1}^{p}-\sum _{i=2}^{n}{a}_{i}^{p}\right)}^{\frac{1}{p}}{\left({b}_{1}^{q}-\sum _{i=2}^{n}{b}_{i}^{q}\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {a}_{1}{b}_{1}-\left(\sum _{i=2}^{n}{a}_{i}{b}_{i}\right)-\frac{{a}_{1}{b}_{1}}{q}{\left[\sum _{i=2}^{n}\left(\frac{{a}_{i}^{p}}{{a}_{1}^{p}}-\frac{{b}_{i}^{q}}{{b}_{1}^{q}}\right)\right]}^{2},\hfill \end{array}$
(19)

and equality holds if and only if $\frac{{a}_{2}^{p}}{{b}_{2}^{q}}=\frac{{a}_{3}^{p}}{{b}_{3}^{q}}=\cdots =\frac{{a}_{n}^{p}}{{b}_{n}^{q}}$ and $\frac{1}{p}+\frac{1}{q}=1$.

## 3 Application

As application of the above results, we establish here an integral type of Aczél-Vasić-Pečarić inequality.

Theorem 3.1 Let $p>0$, $q<0$, $\frac{1}{p}+\frac{1}{q}=1$, let $A>0$, $B>0$, and let $f\left(x\right)$, $g\left(x\right)$ be positive Riemann integrable functions on $\left[a,b\right]$ such that ${A}^{p}-{\int }_{a}^{b}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x>0$ and ${B}^{q}-{\int }_{a}^{b}{g}^{q}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x>0$. Then
$\begin{array}{c}{\left({A}^{p}-{\int }_{a}^{b}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\right)}^{\frac{1}{p}}{\left({B}^{q}-{\int }_{a}^{b}{g}^{q}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\right)}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge AB-{\int }_{a}^{b}f\left(x\right)g\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x-\frac{AB}{q}{\left[{\int }_{a}^{b}\left(\frac{{f}^{p}\left(x\right)}{{A}^{p}}-\frac{{g}^{q}\left(x\right)}{{B}^{q}}\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\right]}^{2}.\hfill \end{array}$
(20)
Proof For any positive integer n, we choose an equidistant partition of $\left[a,b\right]$ as
$\begin{array}{c}a
Since
${A}^{p}-{\int }_{a}^{b}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x>0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{B}^{q}-{\int }_{a}^{b}{g}^{q}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x>0,$
we have
${A}^{p}-\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}{f}^{p}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}>0,$
and
${B}^{q}-\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}{g}^{q}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}>0.$
Hence, there exists a positive integer N such that
${A}^{p}-\sum _{k=1}^{n}{f}^{p}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}>0,$
and
By using Corollary 2.8, we obtain that for any $n>N$, the following inequality holds:
$\begin{array}{c}{\left[{A}^{p}-\sum _{k=1}^{n}{f}^{p}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}\right]}^{\frac{1}{p}}{\left[{B}^{q}-\sum _{k=1}^{n}{g}^{q}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}\right]}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge AB-\sum _{k=1}^{n}{f}^{p}\left(a+\frac{k\left(b-a\right)}{n}\right){g}^{q}\left(a+\frac{k\left(b-a\right)}{n}\right){\left(\frac{b-a}{n}\right)}^{\frac{1}{p}+\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\frac{AB}{q}{\left\{\sum _{k=1}^{n}\left[\frac{1}{{A}^{p}}{f}^{p}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}-\frac{1}{{B}^{q}}{g}^{q}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}\right]\right\}}^{2}.\hfill \end{array}$
(21)
Since
$\frac{1}{p}+\frac{1}{q}=1,$
we have
$\begin{array}{c}{\left[{A}^{p}-\sum _{k=1}^{n}{f}^{p}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}\right]}^{\frac{1}{p}}{\left[{B}^{q}-\sum _{k=1}^{n}{g}^{q}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}\right]}^{\frac{1}{q}}\hfill \\ \phantom{\rule{1em}{0ex}}\ge AB-\sum _{k=1}^{n}{f}^{p}\left(a+\frac{k\left(b-a\right)}{n}\right){g}^{q}\left(a+\frac{k\left(b-a\right)}{n}\right)\left(\frac{b-a}{n}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\frac{AB}{q}{\left\{\sum _{k=1}^{n}\left[\frac{1}{{A}^{p}}{f}^{p}\left(a+\frac{k\left(b-a\right)}{n}\right)-\frac{1}{{B}^{q}}{g}^{q}\left(a+\frac{k\left(b-a\right)}{n}\right)\right]\left(\frac{b-a}{n}\right)\right\}}^{2}.\hfill \end{array}$
(22)

In view of the hypotheses that $f\left(x\right)$, $g\left(x\right)$ are positive Riemann integrable functions on $\left[a,b\right]$, we conclude that $f\left(x\right)g\left(x\right)$, ${f}^{p}\left(x\right)$ and ${g}^{q}\left(x\right)$ are also integrable on $\left[a,b\right]$. Passing the limit as $n\to \mathrm{\infty }$ in both sides of inequality (22), we obtain inequality (20). The proof of Theorem 3.1 is completed. □

## Declarations

### Acknowledgements

The author would like to express his sincere thanks to the anonymous referees for making great efforts to improve this paper. This work was supported by the NNSF of China (Grant No. 61073121), and the Fundamental Research Funds for the Central Universities (No. 13ZD19).

## Authors’ Affiliations

(1)
College of Science and Technology, North China Electric Power University, Baoding, Hebei Province, 071051, P.R. China

## References 