# 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 [1] 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 [26] and references therein) have given considerable attention to this inequality and its refinements.

In 1959, Popoviciu [3] 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 [7] 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 [8] 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 [5] 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 [5] 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 [5]

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 [9]

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 [10]

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 [10]

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 ([[11], 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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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Tian, J., Sun, Y. New refinements of generalized Aczél inequality. J Inequal Appl 2014, 239 (2014). https://doi.org/10.1186/1029-242X-2014-239