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

$$(a_1^2-\sum _{i=2}^n{a}_i^2)(b_1^2-\sum _{i=2}^n{b}_i^2)\le {(a_1{b}_1-\sum _{i=2}^n{a}_i{b}_i)}^2.$$

(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 [2–6] 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

$${(a_1^p-\sum _{i=2}^n{a}_i^p)}^{\frac{1}{p}}{(b_1^q-\sum _{i=2}^n{b}_i^q)}^{\frac{1}{q}}\le a_1{b}_1-\sum _{i=2}^n{a}_i{b}_i.$$

(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

$${(a_1^p-\sum _{i=2}^n{a}_i^p)}^{\frac{1}{p}}{(b_1^q-\sum _{i=2}^n{b}_i^q)}^{\frac{1}{q}}\ge a_1{b}_1-\sum _{i=2}^n{a}_i{b}_i.$$

(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

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

(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

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

(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 $f_j(x)$ ($j=1,2,\dots ,m$) be positive Riemann integrable functions on $[a,b]$ such that $A_j^{{\lambda }_j}-{\int }_a^b{f}_j^{{\lambda }_j}(x)\mathrm{d}x>0$. Then

$$\prod _{j=1}^m{(A_j^{{\lambda }_j}-{\int }_a^b{f}_j^{{\lambda }_j}(x)\mathrm{d}x)}^{\frac{1}{{\lambda }_j}}\ge \prod _{j=1}^m{A}_j-{\int }_a^b\prod _{j=1}^m{f}_j(x)\mathrm{d}x.$$

(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 $\beta =max\{{\sum }_{j=1}^m{\lambda }_j,1\}$. 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 $\gamma =min\{{\sum }_{j=1}^m{\lambda }_j,1\}$. 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

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

(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

$$\sum _{1\le i<j\le m}{(A_i-A_j)}^2=m\left(\sum _{i=1}^m{A}_i^2\right)-{\left(\sum _{i=1}^m{A}_i\right)}^2.$$

(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

$$\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^m{X}_j\ge {\{1-\frac{2}{m(m-1)}[m\left(\sum _{j=1}^m{X}_j^{2{\lambda }_j}\right)-{\left(\sum _{j=1}^m{X}_j^{{\lambda }_j}\right)}^2]\}}^{\frac{m}{2{\lambda }_1}}.$$

(11)

Proof From the assumptions in Lemma 2.5, we find

$$\frac{1}{(m-1){\lambda }_i}<0,\frac{1}{(m-1){\lambda }_j}-\frac{1}{(m-1){\lambda }_i}\le 0(1\le i<j\le m),$$

and

$$\begin{array}{r}\sum _{1\le i<j\le m}[\frac{1}{(m-1){\lambda }_i}+\frac{1}{(m-1){\lambda }_i}+\frac{1}{(m-1){\lambda }_j}-\frac{1}{(m-1){\lambda }_i}]\\ =\sum _{1\le i<j\le m}[\frac{1}{(m-1){\lambda }_i}+\frac{1}{(m-1){\lambda }_j}]=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}+\cdots +\frac{1}{{\lambda }_m}.\end{array}$$

(12)

Thus, by using inequality (7) we have

$$\begin{array}{r}\prod _{1\le i<j\le m}{[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{1}{(m-1){\lambda }_i}}\\ =\prod _{1\le i<j\le m}\{{[X_i^{{\lambda }_i}+(1-X_j^{{\lambda }_j})]}^{\frac{1}{(m-1){\lambda }_i}}{[X_j^{{\lambda }_j}+(1-X_i^{{\lambda }_i})]}^{\frac{1}{(m-1){\lambda }_i}}\\ \times {[X_j^{{\lambda }_j}+(1-X_j^{{\lambda }_j})]}^{\frac{1}{(m-1){\lambda }_j}-\frac{1}{(m-1){\lambda }_i}}\}\\ \le \prod _{1\le i<j\le m}\left[{\left(X_i^{{\lambda }_i}\right)}^{\frac{1}{(m-1){\lambda }_i}}{\left(X_j^{{\lambda }_j}\right)}^{\frac{1}{(m-1){\lambda }_i}}{\left(X_j^{{\lambda }_j}\right)}^{\frac{1}{(m-1){\lambda }_j}-\frac{1}{(m-1){\lambda }_i}}\right]\\ +\prod _{1\le i<j\le m}\left[{(1-X_j^{{\lambda }_j})}^{\frac{1}{(m-1){\lambda }_i}}{(1-X_i^{{\lambda }_i})}^{\frac{1}{(m-1){\lambda }_i}}{(1-X_j^{{\lambda }_j})}^{\frac{1}{(m-1){\lambda }_j}-\frac{1}{(m-1){\lambda }_i}}\right]\\ =\prod _{1\le i<j\le m}{X}_i^{\frac{1}{m-1}}{X}_j^{\frac{1}{m-1}}+\prod _{1\le i<j\le m}\left[{(1-X_i^{{\lambda }_i})}^{\frac{1}{(m-1){\lambda }_i}}{(1-X_j^{{\lambda }_j})}^{\frac{1}{(m-1){\lambda }_j}}\right]\\ =\prod _{j=1}^m{X}_j+\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}.\end{array}$$

(13)

Noting the fact that there are $\frac{m(m-1)}{2}$ product terms in the expression ${\prod }_{1\le i<j\le m}[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]$, and using the arithmetic-geometric mean’s inequality, we obtain

$$\begin{array}{rcl}\prod _{1\le i<j\le m}[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]& \le & {\left\{\frac{2}{m(m-1)}\sum _{1\le i<j\le m}[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]\right\}}^{\frac{m(m-1)}{2}}\\ =& {[1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m(m-1)}{2}}.\end{array}$$

(14)

Therefore, we have

$$\begin{array}{r}\prod _{1\le i<j\le m}{[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{1}{(m-1){\lambda }_i}}\\ \ge {\left\{\prod _{1\le i<j\le m}[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]\right\}}^{\frac{1}{(m-1){\lambda }_1}}\\ \ge {[1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m}{2{\lambda }_1}}.\end{array}$$

(15)

On the other hand, from Lemma 2.4 we have

$$\begin{array}{r}{[1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m}{2{\lambda }_1}}\\ ={\{1-\frac{2}{m(m-1)}[m\left(\sum _{j=1}^m{X}_j^{2{\lambda }_j}\right)-{\left(\sum _{j=1}^m{X}_j^{{\lambda }_j}\right)}^2]\}}^{\frac{m}{2{\lambda }_1}}.\end{array}$$

(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 $\alpha =max\{{\sum }_{j=1}^m\frac{1}{{\lambda }_j},1\}$. If $m>2$, then

$$\begin{array}{r}\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^m{X}_j\\ \ge n^{1-\alpha }{\{1-\frac{2}{(m-1)(m-2)}[(m-1)\left(\sum _{j=1}^{m-1}{X}_j^{2{\lambda }_j}\right)-{\left(\sum _{j=1}^{m-1}{X}_j^{{\lambda }_j}\right)}^2]\}}^{\frac{m-1}{2{\lambda }_1}}.\end{array}$$

(17)

If $m=2$, then

$$\prod _{j=1}^2{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^2{X}_j\ge n^{1-\alpha }{\{1-[2\left(\sum _{j=1}^2{X}_j^{2{\lambda }_j}\right)-{\left(\sum _{j=1}^2{X}_j^{{\lambda }_j}\right)}^2]\}}^{\frac{1}{{\lambda }_1}}.$$

(18)

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

$$\begin{array}{r}\prod _{1\le i<j\le m-1}\{{[X_i^{{\lambda }_i}+(1-X_j^{{\lambda }_j})]}^{\frac{1}{(m-2){\lambda }_i}}{[X_j^{{\lambda }_j}+(1-X_i^{{\lambda }_i})]}^{\frac{1}{(m-2){\lambda }_i}}\\ \times {[X_j^{{\lambda }_j}+(1-X_j^{{\lambda }_j})]}^{\frac{1}{(m-2){\lambda }_j}-\frac{1}{(m-2){\lambda }_i}}\}.\end{array}$$

(19)

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

$$\frac{1}{(m-2){\lambda }_i}<0,\frac{1}{(m-2){\lambda }_j}-\frac{1}{(m-2){\lambda }_i}\le 0(1\le i<j\le m-1),$$

and

$$\begin{array}{r}\sum _{1\le i<j\le m-1}[\frac{1}{(m-2){\lambda }_i}+\frac{1}{(m-2){\lambda }_i}+\frac{1}{(m-2){\lambda }_j}-\frac{1}{(m-2){\lambda }_i}]\\ =\sum _{1\le i<j\le m-1}[\frac{1}{(m-2){\lambda }_i}+\frac{1}{(m-2){\lambda }_j}]=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}+\cdots +\frac{1}{{\lambda }_{m-1}}.\end{array}$$

(20)

Then, applying inequality (7), we have

$$\begin{array}{r}\prod _{1\le i<j\le m-1}{[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{1}{(m-2){\lambda }_i}}\\ ={[X_m^{{\lambda }_m}+(1-X_m^{{\lambda }_m})]}^{\frac{1}{{\lambda }_m}}\prod _{1\le i<j\le m-1}\{{[X_i^{{\lambda }_i}+(1-X_j^{{\lambda }_j})]}^{\frac{1}{(m-2){\lambda }_i}}\\ \times {[X_j^{{\lambda }_j}+(1-X_i^{{\lambda }_i})]}^{\frac{1}{(m-2){\lambda }_i}}{[X_j^{{\lambda }_j}+(1-X_j^{{\lambda }_j})]}^{\frac{1}{(m-2){\lambda }_j}-\frac{1}{(m-2){\lambda }_i}}\}\\ \le n^{\alpha -1}\{X_m^{\frac{{\lambda }_m}{{\lambda }_m}}\prod _{1\le i<j\le m-1}\left[{\left(X_i^{{\lambda }_i}\right)}^{\frac{1}{(m-2){\lambda }_i}}{\left(X_j^{{\lambda }_j}\right)}^{\frac{1}{(m-2){\lambda }_i}}{\left(X_j^{{\lambda }_j}\right)}^{\frac{1}{(m-2){\lambda }_j}-\frac{1}{(m-2){\lambda }_i}}\right]\\ +{(1-X_m^{{\lambda }_m})}^{\frac{1}{{\lambda }_m}}\prod _{1\le i<j\le m-1}[{(1-X_j^{{\lambda }_j})}^{\frac{1}{(m-2){\lambda }_i}}{(1-X_i^{{\lambda }_i})}^{\frac{1}{(m-2){\lambda }_i}}\\ \times {(1-X_j^{{\lambda }_j})}^{\frac{1}{(m-2){\lambda }_j}-\frac{1}{(m-2){\lambda }_i}}\left]\right\}\\ =n^{\alpha -1}\{X_m\prod _{1\le i<j\le m-1}{X}_i^{\frac{1}{m-2}}{X}_j^{\frac{1}{m-2}}\\ +{(1-X_m^{{\lambda }_m})}^{\frac{1}{{\lambda }_m}}\prod _{1\le i<j\le m-1}\left[{(1-X_i^{{\lambda }_i})}^{\frac{1}{(m-2){\lambda }_i}}{(1-X_j^{{\lambda }_j})}^{\frac{1}{(m-2){\lambda }_j}}\right]\}\\ =n^{\alpha -1}[\prod _{j=1}^m{X}_j+\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}].\end{array}$$

(21)

There are $\frac{(m-1)(m-2)}{2}$ product terms in the expression ${\prod }_{1\le i<j\le m-1}[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]$, and then we derive from the arithmetic-geometric mean’s inequality that

$$\begin{array}{r}\prod _{1\le i<j\le m-1}[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]\\ \le {\left\{\frac{2}{(m-1)(m-2)}\sum _{1\le i<j\le m-1}[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]\right\}}^{\frac{(m-1)(m-2)}{2}}\\ ={[1-\frac{2}{(m-1)(m-2)}\sum _{1\le i<j\le m-1}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{(m-1)(m-2)}{2}}.\end{array}$$

(22)

Therefore, we have

$$\begin{array}{r}\prod _{1\le i<j\le m-1}{[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{1}{(m-2){\lambda }_i}}\\ \ge {\left\{\prod _{1\le i<j\le m-1}[1-{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]\right\}}^{\frac{1}{(m-2){\lambda }_1}}\\ \ge {[1-\frac{2}{(m-1)(m-2)}\sum _{1\le i<j\le m-1}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m-1}{2{\lambda }_1}}.\end{array}$$

(23)

On the other hand, from Lemma 2.4 we find

$$\begin{array}{r}{[1-\frac{2}{(m-1)(m-2)}\sum _{1\le i<j\le m-1}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m-1}{2{\lambda }_1}}\\ ={\{1-\frac{2}{(m-1)(m-2)}[(m-1)\left(\sum _{j=1}^{m-1}{X}_j^{2{\lambda }_j}\right)-{\left(\sum _{j=1}^{m-1}{X}_j^{{\lambda }_j}\right)}^2]\}}^{\frac{m-1}{2{\lambda }_1}}.\end{array}$$

(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 $m\ge 2$, $\rho =min\{{\sum }_{j=1}^m\frac{1}{{\lambda }_j},1\}$. Then

$$\begin{array}{r}\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^m{X}_j\\ \le n^{1-\rho }{\{1-\frac{2}{m(m-1)}[m\left(\sum _{j=1}^m{X}_j^{2{\lambda }_j}\right)-{\left(\sum _{j=1}^m{X}_j^{{\lambda }_j}\right)}^2]\}}^{\frac{m}{2{\lambda }_1}}.\end{array}$$

(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

$$\begin{array}{r}\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^m{X}_j\\ \ge {[1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}.\end{array}$$

(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

$$\begin{array}{r}\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^m{X}_j\\ ={(1-X_{j_1}^{{\lambda }_{j_1}})}^{\frac{1}{{\lambda }_{j_1}}}{(1-X_{j_2}^{{\lambda }_{j_2}})}^{\frac{1}{{\lambda }_{j_2}}}\cdots {(1-X_{j_m}^{{\lambda }_{j_m}})}^{\frac{1}{{\lambda }_{j_m}}}+X_{j_1}{X}_{j_2}\cdots X_{j_m}\\ \ge {\{1-\frac{2}{m(m-1)}[m\left(\sum _{k=1}^m{X}_{j_k}^{2{\lambda }_{j_k}}\right)-{\left(\sum _{k=1}^m{X}_{j_k}^{{\lambda }_{j_k}}\right)}^2]\}}^{\frac{m}{2{\lambda }_{j_1}}}\\ ={\{1-\frac{2}{m(m-1)}[m\left(\sum _{k=1}^m{X}_{j_k}^{2{\lambda }_{j_k}}\right)-{\left(\sum _{k=1}^m{X}_{j_k}^{{\lambda }_{j_k}}\right)}^2]\}}^{\frac{m}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}\\ ={[1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}.\end{array}$$

(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 $\alpha =max\{{\sum }_{j=1}^m\frac{1}{{\lambda }_j},1\}$. If $m>2$, then

$$\begin{array}{r}\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^m{X}_j\\ \ge n^{1-\alpha }{[1-\frac{2}{(m-1)(m-2)}\sum _{1\le i<j\le m-1}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m-1}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}.\end{array}$$

(28)

If $m=2$, then

$$\prod _{j=1}^2{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^2{X}_j\ge n^{1-\alpha }{[1-\sum _{1\le i<j\le 2}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{1}{{\lambda }_1}}.$$

(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 $m\ge 2$, $\rho =min\{{\sum }_{j=1}^m\frac{1}{{\lambda }_j},1\}$. Then

$$\begin{array}{r}\prod _{j=1}^m{(1-X_j^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^m{X}_j\\ \le n^{1-\rho }{[1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{(X_i^{{\lambda }_i}-X_j^{{\lambda }_j})}^2]}^{\frac{m}{2max\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}.\end{array}$$

(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

$$\begin{array}{r}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\\ \ge {\{1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2\}}^{\frac{m}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}\\ \ge \prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}.\end{array}$$

(31)

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

$$\frac{{(a_{1j}^{{\lambda }_j}-{\sum }_{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}}{{(a_{1j}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}}>1(j=1,2,\dots ,m).$$

(32)

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

$$\begin{array}{r}\prod _{j=1}^m{\left(\frac{{\sum }_{r=2}^n{a}_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}}\right)}^{\frac{1}{{\lambda }_j}}+\prod _{j=1}^m{\left(\frac{a_{1j}^{{\lambda }_j}-{\sum }_{r=2}^n{a}_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}}\right)}^{\frac{1}{{\lambda }_j}}\\ \ge {\{1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{[(1-\frac{{\sum }_{r=2}^n{a}_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}})-(1-\frac{{\sum }_{r=2}^n{a}_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})]}^2\}}^{\frac{m}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}\\ ={\{1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2\}}^{\frac{m}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}},\end{array}$$

(33)

which implies

$$\begin{array}{rl}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\ge & {\{1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2\}}^{\frac{m}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}\\ \times \prod _{j=1}^m{a}_{1j}-\prod _{j=1}^m{\left(\sum _{r=2}^n{a}_{rj}^{{\lambda }_j}\right)}^{\frac{1}{{\lambda }_j}}.\end{array}$$

(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 $\alpha =max\{{\sum }_{j=1}^m\frac{1}{{\lambda }_j},1\}$. If $m>2$, then

$$\begin{array}{r}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\\ \ge n^{1-\alpha }{\{1-\frac{2}{(m-1)(m-2)}\sum _{1\le i<j\le m-1}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2\}}^{\frac{m-1}{2min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}\\ \times \prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}\\ \ge n^{1-\alpha }\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}.\end{array}$$

(36)

If $m=2$, then

$$\begin{array}{rl}\prod _{j=1}^2{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}& \ge n^{1-\alpha }{\{1-\sum _{1\le i<j\le 2}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2\}}^{\frac{1}{{\lambda }_1}}\prod _{j=1}^2{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^2{a}_{rj}\\ \ge n^{1-\alpha }\prod _{j=1}^2{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^2{a}_{rj}.\end{array}$$

(37)

Proof From the hypotheses of Theorem 2.12, we find that

$$0<\frac{{(a_{1j}^{{\lambda }_j}-{\sum }_{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}}{{(a_{1j}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}}<1(j=1,2,\dots ,m-1),$$

and

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

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

$$\begin{array}{r}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\\ \le n^{1-\rho }{\{1-\frac{2}{m(m-1)}\sum _{1\le i<j\le m}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2\}}^{\frac{m}{2max\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}}\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}\\ \le n^{1-\rho }\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}.\end{array}$$

(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 $\rho =min\{{\sum }_{j=1}^m\frac{1}{{\lambda }_j},1\}$. If $max\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}\ge \frac{m}{2}$, then

$$\begin{array}{r}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\\ \le n^{1-\rho }\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}-\frac{n^{1-\rho }{\prod }_{j=1}^m{a}_{1j}}{(m-1)max\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}\sum _{1\le i<j\le m}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2\\ \le n^{1-\rho }\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}.\end{array}$$

(39)

If $max\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}<\frac{m}{2}$, then

$$\begin{array}{r}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\\ \le n^{1-\rho }\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}-\frac{2n^{1-\rho }{\prod }_{j=1}^m{a}_{1j}}{m(m-1)}\sum _{1\le i<j\le m}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2\\ \le n^{1-\rho }\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}.\end{array}$$

(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 $p,q>0$, $\rho =min\{\frac{1}{p}+\frac{1}{q},1\}$, 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

$$\begin{array}{r}{(a_1^p-\sum _{r=2}^n{a}_r^p)}^{\frac{1}{p}}{(b_1^q-\sum _{r=2}^n{b}_r^q)}^{\frac{1}{q}}\\ \le n^{1-\rho }{\{1-{\left[\sum _{r=2}^n(\frac{a_r^p}{a_1^p}-\frac{b_r^q}{b_1^q})\right]}^2\}}^{\frac{1}{max\{p,q\}}}{a}_1{b}_1-\sum _{r=2}^n{a}_r{b}_r.\end{array}$$

(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 $p<0$, $q\ne 0$, $\alpha =max\{\frac{1}{p}+\frac{1}{q},1\}$, 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

$$\begin{array}{r}{(a_1^p-\sum _{r=2}^n{a}_r^p)}^{\frac{1}{p}}{(b_1^q-\sum _{r=2}^n{b}_r^q)}^{\frac{1}{q}}\\ \ge n^{1-\alpha }{\{1-{\left[\sum _{r=2}^n(\frac{a_r^p}{a_1^p}-\frac{b_r^q}{b_1^q})\right]}^2\}}^{\frac{1}{min\{p,q\}}}{a}_1{b}_1-\sum _{r=2}^n{a}_r{b}_r.\end{array}$$

(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

$$\begin{array}{r}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\ge \prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}\\ -\frac{a_{11}{a}_{12}\cdots a_{1m}}{(m-1)min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}\sum _{1\le i<j\le m}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2.\end{array}$$

(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 $\alpha =max\{{\sum }_{j=1}^m\frac{1}{{\lambda }_j},1\}$, $m>2$. Then

$$\begin{array}{r}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\ge n^{1-\alpha }\prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}\\ -\frac{a_{11}{a}_{12}\cdots a_{1m}{n}^{1-\alpha }}{(m-2)min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}\sum _{1\le i<j\le m-1}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2.\end{array}$$

(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

$$\begin{array}{r}\prod _{j=1}^m{(a_{1j}^{{\lambda }_j}-\sum _{r=2}^n{a}_{rj}^{{\lambda }_j})}^{\frac{1}{{\lambda }_j}}\ge \prod _{j=1}^m{a}_{1j}-\sum _{r=2}^n\prod _{j=1}^m{a}_{rj}\\ -\frac{a_{11}{a}_{12}\cdots a_{1m}}{(m-1)min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}\sum _{1\le i<j\le m-1}{\left[\sum _{r=2}^n(\frac{a_{ri}^{{\lambda }_i}}{a_{1i}^{{\lambda }_i}}-\frac{a_{rj}^{{\lambda }_j}}{a_{1j}^{{\lambda }_j}})\right]}^2.\end{array}$$

(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 $f_j(x)$ ($j=1,2,\dots ,m$) be positive integrable functions defined on $[a,b]$ with $A_j^{{\lambda }_j}-{\int }_a^b{f}_j^{{\lambda }_j}(x)\mathrm{d}x>0$. Then

$$\begin{array}{r}\prod _{j=1}^m{(A_j^{{\lambda }_j}-{\int }_a^b{f}_j^{{\lambda }_j}(x)\mathrm{d}x)}^{\frac{1}{{\lambda }_j}}\ge \prod _{j=1}^m{A}_j-{\int }_a^b\prod _{j=1}^m{f}_j(x)\mathrm{d}x\\ -\frac{A_1{A}_2\cdots A_m}{(m-2)min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}\sum _{1\le i<j\le m-1}{\left[{\int }_a^b(\frac{f_i^{{\lambda }_i}(x)}{A_i^{{\lambda }_i}}-\frac{f_j^{{\lambda }_j}(x)}{A_j^{{\lambda }_j}})\mathrm{d}x\right]}^2.\end{array}$$

(46)

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

$$\begin{array}{c}a<a+\frac{b-a}{n}<\cdots <a+\frac{b-a}{n}k<\cdots <a+\frac{b-a}{n}(n-1)<b,\hfill \\ x_i=a+\frac{b-a}{n}i,i=0,1,\dots ,n,\mathrm{\Delta }{x}_k=\frac{b-a}{n},k=1,2,\dots ,n.\hfill \end{array}$$

Noting that $A_j^{{\lambda }_j}-{\int }_a^b{f}_j^{{\lambda }_j}(x)\mathrm{d}x>0$ ($j=1,2,\dots ,m$), we have

$$A_j^{{\lambda }_j}-\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n{f}_j^{{\lambda }_j}(a+\frac{k(b-a)}{n})\frac{b-a}{n}>0(j=1,2,\dots ,m).$$

Consequently, there exists a positive integer N, such that

$$A_j^{{\lambda }_j}-\sum _{k=1}^n{f}_j^{{\lambda }_j}(a+\frac{k(b-a)}{n})\frac{b-a}{n}>0,$$

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

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

$$\begin{array}{r}\prod _{j=1}^m{[A_j^{{\lambda }_j}-\sum _{k=1}^n{f}_j^{{\lambda }_j}(a+\frac{k(b-a)}{n})\frac{b-a}{n}]}^{\frac{1}{{\lambda }_j}}\\ \ge \prod _{j=1}^m{A}_j^{{\lambda }_j}-\sum _{k=1}^n\left[\prod _{j=1}^m{f}_j(a+\frac{k(b-a)}{n})\right]{\left(\frac{b-a}{n}\right)}^{\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}+\cdots +\frac{1}{{\lambda }_m}}\\ -\frac{A_1{A}_2\cdots A_m}{(m-2)min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}\sum _{1\le i<j\le m}\left\{\sum _{k=1}^n\right[\frac{1}{A_i^{{\lambda }_i}}{f}_i^{{\lambda }_i}(a+\frac{k(b-a)}{n})\frac{b-a}{n}\\ {-\frac{1}{A_j^{{\lambda }_j}}{f}_j^{{\lambda }_j}(a+\frac{k(b-a)}{n})\frac{b-a}{n}]\}}^2.\end{array}$$

(47)

Since

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

we have

$$\begin{array}{r}\prod _{j=1}^m{[A_j^{{\lambda }_j}-\sum _{k=1}^n{f}_j^{{\lambda }_j}(a+\frac{k(b-a)}{n})\frac{b-a}{n}]}^{\frac{1}{{\lambda }_j}}\\ \ge \prod _{j=1}^m{A}_j^{{\lambda }_j}-\sum _{k=1}^n\left[\prod _{j=1}^m{f}_j(a+\frac{k(b-a)}{n})\right]\left(\frac{b-a}{n}\right)\\ -\frac{A_1{A}_2\cdots A_m}{(m-2)min\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\}}\sum _{1\le i<j\le m}\left\{\sum _{k=1}^n\right[\frac{1}{A_i^{{\lambda }_i}}{f}_i^{{\lambda }_i}(a+\frac{k(b-a)}{n})\frac{b-a}{n}\\ {-\frac{1}{A_j^{{\lambda }_j}}{f}_j^{{\lambda }_j}(a+\frac{k(b-a)}{n})\frac{b-a}{n}]\}}^2.\end{array}$$

(48)

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