# 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  established the following inequality, which is called the Aczél inequality.

Theorem A Let ${a}_{i}>0$, ${b}_{i}>0$ ($i=1,2,\dots ,n$), ${a}_{1}^{2}-{\sum }_{i=2}^{n}{a}_{i}^{2}>0$, ${b}_{1}^{2}-{\sum }_{i=2}^{n}{b}_{i}^{2}>0$. 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)

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  and references therein) have given considerable attention to this inequality and its refinements.

In 1959, Popoviciu  generalized the Aczél inequality (1) in the form asserted by Theorem B below.

Theorem B Let $p>1$, $q>1$, $\frac{1}{p}+\frac{1}{q}=1$, let ${a}_{i}>0$, ${b}_{i}>0$ ($i=1,2,\dots ,n$), ${a}_{1}^{p}-{\sum }_{i=2}^{n}{a}_{i}^{p}>0$, ${b}_{1}^{q}-{\sum }_{i=2}^{n}{b}_{i}^{q}>0$. 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)

Later, in 1982, Vasić and Pečarić  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 $p<1$ ($p\ne 0$), $\frac{1}{p}+\frac{1}{q}=1$, and let ${a}_{i}>0$, ${b}_{i}>0$ ($i=1,2,\dots ,n$), ${a}_{1}^{p}-{\sum }_{i=2}^{n}{a}_{i}^{p}>0$, ${b}_{1}^{q}-{\sum }_{i=2}^{n}{b}_{i}^{q}>0$. 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)

In another paper, Vasić and Pečarić  presented an interesting generalization of inequality (2). The inequality is called the generalized Aczél-Vasić-Pečarić inequality.

Theorem D Let ${a}_{rj}>0$, ${\lambda }_{j}>0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$, and let ${\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}}\ge 1$. Then

$\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\le \prod _{j=1}^{m}{a}_{1j}-\sum _{r=2}^{n}\prod _{j=1}^{m}{a}_{rj}.$
(4)

In 2012, Tian  gave the reversed version of inequality (4) in the following form.

Theorem E Let ${\lambda }_{1}\ne 0$, ${\lambda }_{j}<0$ ($j=2,3,\dots ,m$), ${\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}}\le 1$, and let ${a}_{rj}>0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$. Then

$\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\ge \prod _{j=1}^{m}{a}_{1j}-\sum _{r=2}^{n}\prod _{j=1}^{m}{a}_{rj}.$
(5)

Moreover, in  Tian established an integral type of generalized Aczél-Vasić-Pečarić inequality.

Theorem F Let ${\lambda }_{1}>0$, ${\lambda }_{j}<0$ ($j=2,3,\dots ,m$), ${\sum }_{j=1}^{m}{\lambda }_{j}=1$, let ${A}_{j}>0$ ($j=1,2,\dots ,m$), and let ${f}_{j}\left(x\right)$ ($j=1,2,\dots ,m$) be positive Riemann integrable functions on $\left[a,b\right]$ such that ${A}_{j}^{{\lambda }_{j}}-{\int }_{a}^{b}{f}_{j}^{{\lambda }_{j}}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x>0$. Then

$\prod _{j=1}^{m}{\left({A}_{j}^{{\lambda }_{j}}-{\int }_{a}^{b}{f}_{j}^{{\lambda }_{j}}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\right)}^{\frac{1}{{\lambda }_{j}}}\ge \prod _{j=1}^{m}{A}_{j}-{\int }_{a}^{b}\prod _{j=1}^{m}{f}_{j}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x.$
(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 

Let ${a}_{rj}>0$ ($r=1,2,\dots ,n$, $j=1,2,\dots ,m$), let ${\lambda }_{1}$ be a real number, ${\lambda }_{j}\le 0$ ($j=2,3,\dots ,m$), and let $\beta =max\left\{{\sum }_{j=1}^{m}{\lambda }_{j},1\right\}$. Then

$\sum _{r=1}^{n}\prod _{j=1}^{m}{a}_{rj}^{{\lambda }_{j}}\ge {n}^{1-\beta }\prod _{j=1}^{m}{\left(\sum _{r=1}^{n}{a}_{rj}\right)}^{{\lambda }_{j}}.$
(7)

Lemma 2.2 

Let ${a}_{rj}>0$ ($r=1,2,\dots ,n$, $j=1,2,\dots ,m$), let ${\lambda }_{j}\ge 0$ ($j=1,2,\dots ,m$), and let $\gamma =min\left\{{\sum }_{j=1}^{m}{\lambda }_{j},1\right\}$. Then

$\sum _{r=1}^{n}\prod _{j=1}^{m}{a}_{rj}^{{\lambda }_{j}}\le {n}^{1-\gamma }\prod _{j=1}^{m}{\left(\sum _{r=1}^{n}{a}_{rj}\right)}^{{\lambda }_{j}}.$
(8)

Lemma 2.3 

If $x>-1$, $\alpha >1$ or $\alpha <0$, then

${\left(1+x\right)}^{\alpha }\ge 1+\alpha x.$
(9)

The inequality is reversed for $0<\alpha <1$.

Lemma 2.4 

Let ${A}_{1},{A}_{2},\dots ,{A}_{m}$ be real numbers, let m be a natural number, and let $m\ge 2$. Then

$\sum _{1\le i
(10)

Lemma 2.5 Let ${\lambda }_{1}\le {\lambda }_{2}\le \cdots \le {\lambda }_{m}<0$, let ${X}_{j}>1$ ($j=1,2,\dots ,m$), and let $m\ge 2$. Then

$\prod _{j=1}^{m}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{m}{X}_{j}\ge {\left\{1-\frac{2}{m\left(m-1\right)}\left[m\left(\sum _{j=1}^{m}{X}_{j}^{2{\lambda }_{j}}\right)-{\left(\sum _{j=1}^{m}{X}_{j}^{{\lambda }_{j}}\right)}^{2}\right]\right\}}^{\frac{m}{2{\lambda }_{1}}}.$
(11)

Proof From the assumptions in Lemma 2.5, we find

$\frac{1}{\left(m-1\right){\lambda }_{i}}<0,\phantom{\rule{2em}{0ex}}\frac{1}{\left(m-1\right){\lambda }_{j}}-\frac{1}{\left(m-1\right){\lambda }_{i}}\le 0\phantom{\rule{1em}{0ex}}\left(1\le i

and

$\begin{array}{r}\sum _{1\le i
(12)

Thus, by using inequality (7) we have

$\begin{array}{r}\prod _{1\le i
(13)

Noting the fact that there are $\frac{m\left(m-1\right)}{2}$ product terms in the expression ${\prod }_{1\le i, and using the arithmetic-geometric mean’s inequality, we obtain

$\begin{array}{rcl}\prod _{1\le i
(14)

Therefore, we have

$\begin{array}{r}\prod _{1\le i
(15)

On the other hand, from Lemma 2.4 we have

$\begin{array}{r}{\left[1-\frac{2}{m\left(m-1\right)}\sum _{1\le i
(16)

Consequently, from (13), (15), and (16), we obtain the desired inequality (11). □

Lemma 2.6 Let ${\lambda }_{m}>0$, ${\lambda }_{1}\le {\lambda }_{2}\le \cdots \le {\lambda }_{m-1}<0$, let $0<{X}_{m}<1$, ${X}_{j}>1$ ($j=1,2,\dots ,m-1$), and let $\alpha =max\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$. If $m>2$, then

$\begin{array}{r}\prod _{j=1}^{m}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{m}{X}_{j}\\ \phantom{\rule{1em}{0ex}}\ge {n}^{1-\alpha }{\left\{1-\frac{2}{\left(m-1\right)\left(m-2\right)}\left[\left(m-1\right)\left(\sum _{j=1}^{m-1}{X}_{j}^{2{\lambda }_{j}}\right)-{\left(\sum _{j=1}^{m-1}{X}_{j}^{{\lambda }_{j}}\right)}^{2}\right]\right\}}^{\frac{m-1}{2{\lambda }_{1}}}.\end{array}$
(17)

If $m=2$, then

$\prod _{j=1}^{2}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{2}{X}_{j}\ge {n}^{1-\alpha }{\left\{1-\left[2\left(\sum _{j=1}^{2}{X}_{j}^{2{\lambda }_{j}}\right)-{\left(\sum _{j=1}^{2}{X}_{j}^{{\lambda }_{j}}\right)}^{2}\right]\right\}}^{\frac{1}{{\lambda }_{1}}}.$
(18)

Proof Case I. When $m>2$. Let us consider the following product:

$\begin{array}{r}\prod _{1\le i
(19)

From the hypotheses of Lemma 2.6, it is easy to see that

$\frac{1}{\left(m-2\right){\lambda }_{i}}<0,\phantom{\rule{2em}{0ex}}\frac{1}{\left(m-2\right){\lambda }_{j}}-\frac{1}{\left(m-2\right){\lambda }_{i}}\le 0\phantom{\rule{1em}{0ex}}\left(1\le i

and

$\begin{array}{r}\sum _{1\le i
(20)

Then, applying inequality (7), we have

$\begin{array}{r}\prod _{1\le i
(21)

There are $\frac{\left(m-1\right)\left(m-2\right)}{2}$ product terms in the expression ${\prod }_{1\le i, and then we derive from the arithmetic-geometric mean’s inequality that

$\begin{array}{r}\prod _{1\le i
(22)

Therefore, we have

$\begin{array}{r}\prod _{1\le i
(23)

On the other hand, from Lemma 2.4 we find

$\begin{array}{r}{\left[1-\frac{2}{\left(m-1\right)\left(m-2\right)}\sum _{1\le i
(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 ${\lambda }_{1}\ge {\lambda }_{2}\ge \cdots \ge {\lambda }_{m}>0$, let $0<{X}_{j}<1$ ($j=1,2,\dots ,m$), and let $m\ge 2$, $\rho =min\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{m}{X}_{j}\\ \phantom{\rule{1em}{0ex}}\le {n}^{1-\rho }{\left\{1-\frac{2}{m\left(m-1\right)}\left[m\left(\sum _{j=1}^{m}{X}_{j}^{2{\lambda }_{j}}\right)-{\left(\sum _{j=1}^{m}{X}_{j}^{{\lambda }_{j}}\right)}^{2}\right]\right\}}^{\frac{m}{2{\lambda }_{1}}}.\end{array}$
(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 ${\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}<0$, let ${X}_{j}>1$ ($j=1,2,\dots ,m$), and let $m\ge 2$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{m}{X}_{j}\\ \phantom{\rule{1em}{0ex}}\ge {\left[1-\frac{2}{m\left(m-1\right)}\sum _{1\le i
(26)

Proof After simply rearranging, we write by ${\lambda }_{{j}_{1}}\le {\lambda }_{{j}_{2}}\le \cdots \le {\lambda }_{{j}_{m}}$ the component of ${\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}$ in increasing order, where ${j}_{1},{j}_{2},\dots ,{j}_{m}$ is a permutation of $1,2,\dots ,m$.

Then from Lemma 2.5 and Lemma 2.4 we get

$\begin{array}{r}\prod _{j=1}^{m}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{m}{X}_{j}\\ \phantom{\rule{1em}{0ex}}={\left(1-{X}_{{j}_{1}}^{{\lambda }_{{j}_{1}}}\right)}^{\frac{1}{{\lambda }_{{j}_{1}}}}{\left(1-{X}_{{j}_{2}}^{{\lambda }_{{j}_{2}}}\right)}^{\frac{1}{{\lambda }_{{j}_{2}}}}\cdots {\left(1-{X}_{{j}_{m}}^{{\lambda }_{{j}_{m}}}\right)}^{\frac{1}{{\lambda }_{{j}_{m}}}}+{X}_{{j}_{1}}{X}_{{j}_{2}}\cdots {X}_{{j}_{m}}\\ \phantom{\rule{1em}{0ex}}\ge {\left\{1-\frac{2}{m\left(m-1\right)}\left[m\left(\sum _{k=1}^{m}{X}_{{j}_{k}}^{2{\lambda }_{{j}_{k}}}\right)-{\left(\sum _{k=1}^{m}{X}_{{j}_{k}}^{{\lambda }_{{j}_{k}}}\right)}^{2}\right]\right\}}^{\frac{m}{2{\lambda }_{{j}_{1}}}}\\ \phantom{\rule{1em}{0ex}}={\left\{1-\frac{2}{m\left(m-1\right)}\left[m\left(\sum _{k=1}^{m}{X}_{{j}_{k}}^{2{\lambda }_{{j}_{k}}}\right)-{\left(\sum _{k=1}^{m}{X}_{{j}_{k}}^{{\lambda }_{{j}_{k}}}\right)}^{2}\right]\right\}}^{\frac{m}{2min\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}}}\\ \phantom{\rule{1em}{0ex}}={\left[1-\frac{2}{m\left(m-1\right)}\sum _{1\le i
(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 ${\lambda }_{m}>0$, ${\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m-1}<0$, let $0<{X}_{m}<1$, ${X}_{j}>1$ ($j=1,2,\dots ,m-1$), and let $\alpha =max\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$. If $m>2$, then

$\begin{array}{r}\prod _{j=1}^{m}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{m}{X}_{j}\\ \phantom{\rule{1em}{0ex}}\ge {n}^{1-\alpha }{\left[1-\frac{2}{\left(m-1\right)\left(m-2\right)}\sum _{1\le i
(28)

If $m=2$, then

$\prod _{j=1}^{2}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{2}{X}_{j}\ge {n}^{1-\alpha }{\left[1-\sum _{1\le i
(29)

Lemma 2.10 Let ${\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}>0$, let $0<{X}_{j}<1$ ($j=1,2,\dots ,m$), and let $m\ge 2$, $\rho =min\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left(1-{X}_{j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{m}{X}_{j}\\ \phantom{\rule{1em}{0ex}}\le {n}^{1-\rho }{\left[1-\frac{2}{m\left(m-1\right)}\sum _{1\le i
(30)

Now, we give the refinement and generalization of inequality (5).

Theorem 2.11 Let ${a}_{rj}>0$, ${\lambda }_{j}<0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$, and let $m\ge 2$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\\ \phantom{\rule{1em}{0ex}}\ge {\left\{1-\frac{2}{m\left(m-1\right)}\sum _{1\le i
(31)

Proof From the assumptions in Theorem 2.11, it is easy to verify that

$\frac{{\left({a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}}{{\left({a}_{1j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}}>1\phantom{\rule{1em}{0ex}}\left(j=1,2,\dots ,m\right).$
(32)

It thus follows from Lemma 2.8 with the substitution ${X}_{j}={\left(\frac{{a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}}{{a}_{1j}^{{\lambda }_{j}}}\right)}^{\frac{1}{{\lambda }_{j}}}$ in (26) that

$\begin{array}{r}\prod _{j=1}^{m}{\left(\frac{{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}}{{a}_{1j}^{{\lambda }_{j}}}\right)}^{\frac{1}{{\lambda }_{j}}}+\prod _{j=1}^{m}{\left(\frac{{a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}}{{a}_{1j}^{{\lambda }_{j}}}\right)}^{\frac{1}{{\lambda }_{j}}}\\ \phantom{\rule{1em}{0ex}}\ge {\left\{1-\frac{2}{m\left(m-1\right)}\sum _{1\le i
(33)

which implies

$\begin{array}{rl}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\ge & {\left\{1-\frac{2}{m\left(m-1\right)}\sum _{1\le i
(34)

On the other hand, it follows from Lemma 2.1 that

$\prod _{j=1}^{m}{\left(\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\le \sum _{r=2}^{n}\prod _{j=1}^{m}{a}_{rj}.$
(35)

Combining inequalities (34) and (35) yields inequality (31).

The proof of Theorem 2.11 is completed. □

Theorem 2.12 Let ${\lambda }_{m}>0$, ${\lambda }_{j}<0$ ($j=1,2,\dots ,m-1$), let ${a}_{rj}>0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$, and let $\alpha =max\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$. If $m>2$, then

$\begin{array}{r}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\\ \phantom{\rule{1em}{0ex}}\ge {n}^{1-\alpha }{\left\{1-\frac{2}{\left(m-1\right)\left(m-2\right)}\sum _{1\le i
(36)

If $m=2$, then

$\begin{array}{rl}\prod _{j=1}^{2}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}& \ge {n}^{1-\alpha }{\left\{1-\sum _{1\le i
(37)

Proof From the hypotheses of Theorem 2.12, we find that

$0<\frac{{\left({a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}}{{\left({a}_{1j}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}}<1\phantom{\rule{1em}{0ex}}\left(j=1,2,\dots ,m-1\right),$

and

$\frac{{\left({a}_{1m}^{{\lambda }_{m}}-{\sum }_{r=2}^{n}{a}_{rm}^{{\lambda }_{m}}\right)}^{\frac{1}{{\lambda }_{m}}}}{{\left({a}_{1m}^{{\lambda }_{m}}\right)}^{\frac{1}{{\lambda }_{m}}}}>1.$

Consequently, by the same method as in Theorem 2.11, and using Lemma 2.9 with a substitution ${X}_{j}\to {\left(\frac{{a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}}{{a}_{1j}^{{\lambda }_{j}}}\right)}^{\frac{1}{{\lambda }_{j}}}$ ($j=1,2,\dots ,m$) 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 ${a}_{rj}>0$, ${\lambda }_{j}>0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$, let $m\ge 2$, and let $\rho =min\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\\ \phantom{\rule{1em}{0ex}}\le {n}^{1-\rho }{\left\{1-\frac{2}{m\left(m-1\right)}\sum _{1\le i
(38)

Therefore, from Lemma 2.3 and Theorem 2.13 we get a new refinement and generalization of inequality (4).

Corollary 2.14 Let ${a}_{rj}>0$, ${\lambda }_{j}>0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$, let $m\ge 2$, and let $\rho =min\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$. If $max\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}\ge \frac{m}{2}$, then

$\begin{array}{r}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\\ \phantom{\rule{1em}{0ex}}\le {n}^{1-\rho }\prod _{j=1}^{m}{a}_{1j}-\sum _{r=2}^{n}\prod _{j=1}^{m}{a}_{rj}-\frac{{n}^{1-\rho }{\prod }_{j=1}^{m}{a}_{1j}}{\left(m-1\right)max\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}}\sum _{1\le i
(39)

If $max\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}<\frac{m}{2}$, then

$\begin{array}{r}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\\ \phantom{\rule{1em}{0ex}}\le {n}^{1-\rho }\prod _{j=1}^{m}{a}_{1j}-\sum _{r=2}^{n}\prod _{j=1}^{m}{a}_{rj}-\frac{2{n}^{1-\rho }{\prod }_{j=1}^{m}{a}_{1j}}{m\left(m-1\right)}\sum _{1\le i
(40)

Remark 2.15 If we set ${\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}}\ge 1$ in Corollary 2.14, then inequalities (39) and (40) reduce to Wu’s inequality ([, Theorem 1]).

In particular, putting $m=2$, ${\lambda }_{1}=p$, ${\lambda }_{2}=q$, ${a}_{r1}={a}_{r}$, ${a}_{r2}={b}_{r}$ ($r=1,2,\dots ,n$) in Theorem 2.13, we obtain a new refinement and generalization of inequality (2).

Corollary 2.16 Let ${a}_{r}>0$, ${b}_{r}>0$ ($r=1,2,\dots ,n$), let $p,q>0$, $\rho =min\left\{\frac{1}{p}+\frac{1}{q},1\right\}$, and let ${a}_{1}^{p}-{\sum }_{r=2}^{n}{a}_{r}^{p}>0$, ${b}_{1}^{q}-{\sum }_{r=2}^{n}{b}_{r}^{q}>0$. Then

$\begin{array}{r}{\left({a}_{1}^{p}-\sum _{r=2}^{n}{a}_{r}^{p}\right)}^{\frac{1}{p}}{\left({b}_{1}^{q}-\sum _{r=2}^{n}{b}_{r}^{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{1em}{0ex}}\le {n}^{1-\rho }{\left\{1-{\left[\sum _{r=2}^{n}\left(\frac{{a}_{r}^{p}}{{a}_{1}^{p}}-\frac{{b}_{r}^{q}}{{b}_{1}^{q}}\right)\right]}^{2}\right\}}^{\frac{1}{max\left\{p,q\right\}}}{a}_{1}{b}_{1}-\sum _{r=2}^{n}{a}_{r}{b}_{r}.\end{array}$
(41)

Similarly, putting $m=2$, ${\lambda }_{1}=p$, ${\lambda }_{2}=q$, ${a}_{r1}={a}_{r}$, ${a}_{r2}={b}_{r}$ ($r=1,2,\dots ,n$) in Theorem 2.12 and Theorem 2.11, respectively, we obtain a new refinement and generalization of inequality (3).

Corollary 2.17 Let ${a}_{r}>0$, ${b}_{r}>0$ ($r=1,2,\dots ,n$), let $p<0$, $q\ne 0$, $\alpha =max\left\{\frac{1}{p}+\frac{1}{q},1\right\}$, and let ${a}_{1}^{p}-{\sum }_{r=2}^{n}{a}_{r}^{p}>0$, ${b}_{1}^{q}-{\sum }_{r=2}^{n}{b}_{r}^{q}>0$. Then

$\begin{array}{r}{\left({a}_{1}^{p}-\sum _{r=2}^{n}{a}_{r}^{p}\right)}^{\frac{1}{p}}{\left({b}_{1}^{q}-\sum _{r=2}^{n}{b}_{r}^{q}\right)}^{\frac{1}{q}}\\ \phantom{\rule{1em}{0ex}}\ge {n}^{1-\alpha }{\left\{1-{\left[\sum _{r=2}^{n}\left(\frac{{a}_{r}^{p}}{{a}_{1}^{p}}-\frac{{b}_{r}^{q}}{{b}_{1}^{q}}\right)\right]}^{2}\right\}}^{\frac{1}{min\left\{p,q\right\}}}{a}_{1}{b}_{1}-\sum _{r=2}^{n}{a}_{r}{b}_{r}.\end{array}$
(42)

From Lemma 2.3 and Theorem 2.11 we obtain the following refinement of inequality (5).

Corollary 2.18 Let ${a}_{rj}>0$, ${\lambda }_{j}<0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$, and let $m\ge 2$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\ge \prod _{j=1}^{m}{a}_{1j}-\sum _{r=2}^{n}\prod _{j=1}^{m}{a}_{rj}\\ \phantom{\rule{1em}{0ex}}-\frac{{a}_{11}{a}_{12}\cdots {a}_{1m}}{\left(m-1\right)min\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}}\sum _{1\le i
(43)

Similarly, from Lemma 2.3 and Theorem 2.12 we obtain the following refinement and generalization of inequality (5).

Corollary 2.19 Let ${\lambda }_{m}>0$, ${\lambda }_{j}<0$ ($j=1,2,\dots ,m-1$), let ${a}_{rj}>0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$, and let $\alpha =max\left\{{\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}},1\right\}$, $m>2$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\ge {n}^{1-\alpha }\prod _{j=1}^{m}{a}_{1j}-\sum _{r=2}^{n}\prod _{j=1}^{m}{a}_{rj}\\ \phantom{\rule{1em}{0ex}}-\frac{{a}_{11}{a}_{12}\cdots {a}_{1m}{n}^{1-\alpha }}{\left(m-2\right)min\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}}\sum _{1\le i
(44)

If we set ${\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}}\le 1$, then from Corollary 2.18 and Corollary 2.19 we obtain the following refinement of inequality (5).

Corollary 2.20 Let ${\lambda }_{1}\ne 0$, ${\lambda }_{j}<0$ ($j=2,3,\dots ,m$), ${\sum }_{j=1}^{m}\frac{1}{{\lambda }_{j}}\le 1$, let ${a}_{rj}>0$, ${a}_{1j}^{{\lambda }_{j}}-{\sum }_{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}>0$, $r=1,2,\dots ,n$, $j=1,2,\dots ,m$, and let $m>2$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left({a}_{1j}^{{\lambda }_{j}}-\sum _{r=2}^{n}{a}_{rj}^{{\lambda }_{j}}\right)}^{\frac{1}{{\lambda }_{j}}}\ge \prod _{j=1}^{m}{a}_{1j}-\sum _{r=2}^{n}\prod _{j=1}^{m}{a}_{rj}\\ \phantom{\rule{1em}{0ex}}-\frac{{a}_{11}{a}_{12}\cdots {a}_{1m}}{\left(m-1\right)min\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}}\sum _{1\le i
(45)

## 3 Application

In this section, we show an application of the inequality newly obtained in Section 2.

Theorem 3.1 Let ${A}_{j}>0$ ($j=1,2,\dots ,m$), let ${\lambda }_{1}>0$, ${\lambda }_{j}<0$ ($j=2,3,\dots ,m$), ${\sum }_{j=1}^{m}{\lambda }_{j}=1$, $m>2$, and let ${f}_{j}\left(x\right)$ ($j=1,2,\dots ,m$) be positive integrable functions defined on $\left[a,b\right]$ with ${A}_{j}^{{\lambda }_{j}}-{\int }_{a}^{b}{f}_{j}^{{\lambda }_{j}}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x>0$. Then

$\begin{array}{r}\prod _{j=1}^{m}{\left({A}_{j}^{{\lambda }_{j}}-{\int }_{a}^{b}{f}_{j}^{{\lambda }_{j}}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\right)}^{\frac{1}{{\lambda }_{j}}}\ge \prod _{j=1}^{m}{A}_{j}-{\int }_{a}^{b}\prod _{j=1}^{m}{f}_{j}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\\ \phantom{\rule{1em}{0ex}}-\frac{{A}_{1}{A}_{2}\cdots {A}_{m}}{\left(m-2\right)min\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}}\sum _{1\le i
(46)

Proof For any positive integers n, we choose an equidistant partition of $\left[a,b\right]$ as

$\begin{array}{c}a

Noting that ${A}_{j}^{{\lambda }_{j}}-{\int }_{a}^{b}{f}_{j}^{{\lambda }_{j}}\left(x\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}x>0$ ($j=1,2,\dots ,m$), we have

${A}_{j}^{{\lambda }_{j}}-\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}{f}_{j}^{{\lambda }_{j}}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}>0\phantom{\rule{1em}{0ex}}\left(j=1,2,\dots ,m\right).$

Consequently, there exists a positive integer N, such that

${A}_{j}^{{\lambda }_{j}}-\sum _{k=1}^{n}{f}_{j}^{{\lambda }_{j}}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}>0,$

for all $n,l>N$ and $j=1,2,\dots ,m$.

By using Theorem 2.12, for any $n>N$, the following inequality holds:

$\begin{array}{r}\prod _{j=1}^{m}{\left[{A}_{j}^{{\lambda }_{j}}-\sum _{k=1}^{n}{f}_{j}^{{\lambda }_{j}}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}\right]}^{\frac{1}{{\lambda }_{j}}}\\ \phantom{\rule{1em}{0ex}}\ge \prod _{j=1}^{m}{A}_{j}^{{\lambda }_{j}}-\sum _{k=1}^{n}\left[\prod _{j=1}^{m}{f}_{j}\left(a+\frac{k\left(b-a\right)}{n}\right)\right]{\left(\frac{b-a}{n}\right)}^{\frac{1}{{\lambda }_{1}}+\frac{1}{{\lambda }_{2}}+\cdots +\frac{1}{{\lambda }_{m}}}\\ \phantom{\rule{2em}{0ex}}-\frac{{A}_{1}{A}_{2}\cdots {A}_{m}}{\left(m-2\right)min\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}}\sum _{1\le i
(47)

Since

$\sum _{j=1}^{m}\frac{1}{{\lambda }_{j}}=1,$

we have

$\begin{array}{r}\prod _{j=1}^{m}{\left[{A}_{j}^{{\lambda }_{j}}-\sum _{k=1}^{n}{f}_{j}^{{\lambda }_{j}}\left(a+\frac{k\left(b-a\right)}{n}\right)\frac{b-a}{n}\right]}^{\frac{1}{{\lambda }_{j}}}\\ \phantom{\rule{1em}{0ex}}\ge \prod _{j=1}^{m}{A}_{j}^{{\lambda }_{j}}-\sum _{k=1}^{n}\left[\prod _{j=1}^{m}{f}_{j}\left(a+\frac{k\left(b-a\right)}{n}\right)\right]\left(\frac{b-a}{n}\right)\\ \phantom{\rule{2em}{0ex}}-\frac{{A}_{1}{A}_{2}\cdots {A}_{m}}{\left(m-2\right)min\left\{{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{m}\right\}}\sum _{1\le i
(48)

Noting that ${f}_{j}\left(x\right)$ ($j=1,2,\dots ,m$) are positive Riemann integrable functions on $\left[a,b\right]$, we know that ${\prod }_{j=1}^{m}{f}_{j}\left(x\right)$ and ${f}_{j}^{{\lambda }_{j}}\left(x\right)$ are also integrable on $\left[a,b\right]$. Letting $n\to \mathrm{\infty }$ 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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