Extensions and Applications of Power-Type Aczél-Vasić-Pečarić’s Inequalities

Abstract

In this paper, we enrich and develop power-type Aczél-Vasić-Pečarić’s inequalities. First of all, we give some new versions of theorems and corollaries about Aczél-Vasić-Pečarić’s inequalities by quoting some lemmas. Moreover, in combination with Hölder’s inequality, we give some applications of the new version of Aczél-Vasić-Pečarić’s inequality and give its proof process.

This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

In 1956, Aczél [1] discovered the Aczél’s inequality.

Theorem 1.

Let n be a positive integer and let ${\lambda }_s$, ${\nu }_s\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n)$, $n\ge \mathrm{2}$, be positive numbers such that ${\lambda }_{\mathrm{1}}^{\mathrm{2}}-{\sum }_{s=\mathrm{2}}^n{\lambda }_s^{\mathrm{2}}\ge \mathrm{0}$; ${\nu }_{\mathrm{1}}^{\mathrm{2}}-{\sum }_{s=\mathrm{2}}^n{\nu }_s^{\mathrm{2}}\ge \mathrm{0}$. Then$$\begin{array}{}\text{(1)}& \left({\lambda }_{\mathrm{1}}^{\mathrm{2}}-{\displaystyle \sum }_{s=\mathrm{2}}^n{\lambda }_s^{\mathrm{2}}\right)\left({\nu }_{\mathrm{1}}^{\mathrm{2}}-{\displaystyle \sum }_{s=\mathrm{2}}^n{\nu }_s^{\mathrm{2}}\right)\le {\left({\lambda }_{\mathrm{1}}{\nu }_{\mathrm{1}}-\sum _{s=\mathrm{2}}^n{\lambda }_s{\nu }_s\right)}^{\mathrm{2}}.\end{array}$$

It is matter of common observation that Aczél’s inequality (1) is of great significance in the theory of functional equations in non-Euclidean geometry; meanwhile, many authors including Bellman [2], Hu et al.[3], Tian [4, 5], Tian and Ha [6], Tian and Sun [7], Tian and Wu [8], Tian and Wang [9], Tian and Zhou [10], Wu [11], and Wu and Debnath [12, 13] pay more attention to this inequality and its refinements.

In 1959, Popoviciu [14] gave a generalization of Aczél’s inequality, as follows.

Theorem 2.

Let $p>\mathrm{1},q>\mathrm{1}$, and $\mathrm{1}/p+\mathrm{1}/q=\mathrm{1}$, and let ${\lambda }_s,{\nu }_s\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n)$ be positive numbers such that ${\lambda }_{\mathrm{1}}^p-{\sum }_{s=\mathrm{2}}^n{\lambda }_s^p>\mathrm{0}$ and ${\nu }_{\mathrm{1}}^p-{\sum }_{s=\mathrm{2}}^n{\nu }_s^p>\mathrm{0}$. Then$$\begin{array}{}\text{(2)}& {\left({\lambda }_{\mathrm{1}}^p-\sum _{s=\mathrm{2}}^n{\lambda }_s^p\right)}^{\mathrm{1}/p}{\left({\nu }_{\mathrm{1}}^q-\sum _{s=\mathrm{2}}^n{\nu }_s^q\right)}^{\mathrm{1}/q}\le {\lambda }_{\mathrm{1}}{\nu }_{\mathrm{1}}-{\displaystyle \sum }_{s=\mathrm{2}}^n{\lambda }_s{\nu }_s.\end{array}$$

In 1979, Vasić and Pečarić [15] proved the following extension of inequality (2).

Theorem 3.

Let $p_t>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(t=\mathrm{1,2},\dots ,m)$ and ${\sum }_{t=\mathrm{1}}^m\mathrm{1}/p_t\ge \mathrm{1}$, and let ${\lambda }_{st}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$ be positive integers such that ${\lambda }_{\mathrm{1}t}^{p_t}-{\sum }_{s=2}^n{\lambda }_{st}^{p_t}>\mathrm{0}$. Then$$\begin{array}{}\text{(3)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\lambda }_{\mathrm{1}t}^{p_t}-\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}\right)}^{\mathrm{1}/p_t}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$

Inequality (3) is known as Aczél- Vasić-Pečarić’s inequality.

In 2005, Wu and Debnath [13] generalized inequality (3) in the following form.

Theorem 4.

Let $p_t$ and ${\lambda }_{st}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$ be positive numbers such that ${\lambda }_{\mathrm{1}t}^{p_t}-{\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}>\mathrm{0}$ for $t=\mathrm{1,2},\dots ,m$, and let $\tau =\min \{{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right),\mathrm{1}\}$. Then$$\begin{array}{}\text{(4)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}\right]}^{\mathrm{1}/p_t}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_t-{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_t.\end{array}$$

Later, Wu in [11] established the Aczél-Vasić-Pečarić inequality (3).

Theorem 5.

Let $p_{\mathrm{1}},p_{\mathrm{2}},\dots ,p_m>\mathrm{0}$ and $\rho =\mathrm{1}/p_{\mathrm{1}}+\mathrm{1}/p_{\mathrm{2}}+\cdots +\mathrm{1}/p_m$, and let ${\lambda }_{st}>\mathrm{0}$, ${\lambda }_{\mathrm{1}t}^{p_t}-{\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and $m\ge \mathrm{2}$, $n\ge \mathrm{2}$. Then we have $$\begin{array}{}\text{(5)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\lambda }_{\mathrm{1}t}^{p_t}-\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}\right)}^{\mathrm{1}/p_t}\le n^{\mathrm{1}-\mathrm{m}\mathrm{i}\mathrm{n}\left\{\rho ,\mathrm{1}\right\}}{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_t-{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_t-a{\displaystyle \sum }_{\mathrm{1}\le t\le k\le m}{\left(\sum _{s=\mathrm{2}}^n\left(\frac{{\lambda }_{st}^{p_t}}{{\lambda }_{\mathrm{1}t}^{p_t}}-\frac{{\lambda }_{sk}^{p_k}}{{\lambda }_{\mathrm{1}k}^{p_k}}\right)\right)}^{\mathrm{2}},\end{array}$$ where $$\begin{array}{}\text{(6)}& a=\frac{{\prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}}{\eta },\hspace{1em}\eta =\left\{\begin{array}{cc}\left(m-\mathrm{1}\right)\max \left\{p_{\mathrm{1}},p_{\mathrm{2}},\dots ,p_m,\frac{m}{\mathrm{2}}\right\},\hfill & \rho \ge \mathrm{1},\hfill \\ n^{\rho -\mathrm{1}}{m}^{\mathrm{2}}{\left(m-\mathrm{1}\right)}^{-\mathrm{1}}\max \left\{p_{\mathrm{1}},p_{\mathrm{2}},\dots ,p_m,{\left(\mathrm{1}-\rho \right)}^{-\mathrm{1}}\right\},\hfill & \rho <\mathrm{1}.\hfill \end{array}\right.\end{array}$$

Moreover, in 2014 Tian [7] also presented an improvement of the Aczél-Vasić-Pečarić inequality (3).

Theorem 6.

Let ${\lambda }_{rt}>\mathrm{0}$, $p_t>\mathrm{0}$, and ${\lambda }_{\mathrm{1}t}^{p_t}-{\sum }_{r=\mathrm{2}}^n{\lambda }_{rt}^{p_t}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(r=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$, let $m\ge \mathrm{2}$, and let $\rho =\mathrm{m}\mathrm{i}\mathrm{n}\{{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right),\mathrm{1}\}$. Then $$\begin{array}{}\text{(7)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\lambda }_{\mathrm{1}t}^{p_t}-\sum _{r=\mathrm{2}}^n{\lambda }_{rt}^{p_t}\right)}^{\mathrm{1}/p_t}\le n^{\mathrm{1}-\rho }{\left\{\mathrm{1}-\frac{\mathrm{2}}{m\left(m-\mathrm{1}\right)}\sum _{\mathrm{1}\le i\le t\le m}{\left[\sum _{r=\mathrm{2}}^n\left(\frac{{\lambda }_{ri}^{p_i}}{{\lambda }_{\mathrm{1}i}^{p_i}}-\frac{{\lambda }_{rt}^{p_t}}{{\lambda }_{\mathrm{1}t}^{p_t}}\right)\right]}^{\mathrm{2}}\right\}}^{\gamma }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\displaystyle \sum }_{r=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{rt}\le n^{\mathrm{1}-\rho }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\displaystyle \sum }_{r=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{rt},\end{array}$$ where $\gamma =m/\left(\mathrm{2}\max \{p_{\mathrm{1}},p_{\mathrm{2}},\dots ,p_m\}\right)$.

Recently, the power mean has attracted of many researcher ([16–24]), and many remarkable inequalities for the power mean including Hölder-type inequalities can be found in the literature [25–30]. The purpose of the article is to establish some distinctive versions of Aczél-Vasić-Pečarić’s inequality (3) for power mean type. As consequences, several integral inequalities of the obtained results are given.

2. Some Power Mean Types of Aczél-Vasić-Pečarić’s Inequality

Lemma 7 (see [13]).

Let ${\lambda }_{st}>\mathrm{0}$ and $\mathrm{1}/p_t>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let $\tau =\min \{{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right),\mathrm{1}\}$. Then $$\begin{array}{}\text{(8)}& {\displaystyle \sum }_{s=\mathrm{1}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\left(\sum _{s=\mathrm{1}}^n{\lambda }_{st}^{p_t}\right)}^{\mathrm{1}/pt}.\end{array}$$

Like in [31], the power mean of order $r$ for $n$ positive numbers ${\lambda }_{\mathrm{1}},{\lambda }_{\mathrm{2}},\dots ,{\lambda }_n$ is defined by (9) M n λ , r = 1 n ∑ k = 1 n λ k r 1 / r , r ≠ 0 , λ 1 λ 2 ⋯ λ n n , r = 0 , where $\mathit{\lambda }=({\lambda }_{\mathrm{1}},{\lambda }_{\mathrm{2}},\dots ,{\lambda }_n)$.

Lemma 8 (see [32]).

Let ${\lambda }_{\mathrm{1}},{\lambda }_{\mathrm{2}},\dots ,{\lambda }_n$ be a positive sequence, $n\in \mathbb{N}$. If $r\in \mathbb{R}$, then the function $r\mapsto M_n(\mathit{\lambda },r)$ is increasing for fixed ${\lambda }_{\mathrm{1}},{\lambda }_{\mathrm{2}},\dots ,{\lambda }_n$.

Lemma 9 (see [31]).

Let ${\lambda }_s>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n)$ be positive numbers, and let $p>\mathrm{0}$. Then$$\begin{array}{}\text{(10)}& {\displaystyle \sum }_{s=\mathrm{1}}^n{\lambda }_s^p\le n^{\mathrm{1}-\min \left\{p,\mathrm{1}\right\}}{\left(\sum _{s=\mathrm{1}}^n{\lambda }_s\right)}^p.\end{array}$$

Putting $p\ge \mathrm{1}$, we get the following inequality:$$\begin{array}{}\text{(11)}& {\displaystyle \sum }_{s=\mathrm{1}}^n{\lambda }_s^p\le {\left(\sum _{s=\mathrm{1}}^n{\lambda }_s\right)}^p.\end{array}$$Let $$\begin{array}{c}\text{(12)}& {\mathit{\lambda }}_t=\left({\lambda }_{\mathrm{1}t},{\lambda }_{\mathrm{2}t},\dots ,{\lambda }_{nt}\right),\hspace{1em}\left(t=\mathrm{1,2},\dots ,m\right),M_m\left({\mathit{\lambda }}_t^p,r\right)={\left(\frac{\mathrm{1}}{m}\sum _{s=\mathrm{1}}^m{\lambda }_{st}^{pr}\right)}^{\mathrm{1}/r},\\ M_{m,n}\left({\mathit{\lambda }}_t^p,r\right)={\left(\frac{\mathrm{1}}{n-m+\mathrm{1}}\sum _{s=m}^n{\lambda }_{st}^{pr}\right)}^{\mathrm{1}/r},\\ M_{m,n}^{\chi }\left({\mathit{\lambda }}_j,r\right)={\left(\frac{\mathrm{1}}{n-m+\mathrm{1}}\sum _{i=m}^n{\lambda }_{st}^r\right)}^{\chi /r}.\end{array}$$

The main conclusions of this paper are as follows.

Theorem 10.

Let $r>\mathrm{0}$, $p_t>\mathrm{0}$, $M_{\mathrm{1}}^{p_t}({\mathit{\lambda }}_t,r)-M_{\mathrm{2},n}^{p_t}({\mathit{\lambda }}_t,r)>\mathrm{0}$, and ${\prod }_{t=\mathrm{1}}^m{\mathit{\lambda }}_t=({\prod }_{t=1}^m{\lambda }_{\mathrm{1}t},{\prod }_{t=1}^m{\lambda }_{\mathrm{2}t},\dots ,{\prod }_{t=1}^m{\lambda }_{nt})$, and let $\tau =\min \{{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right),\mathrm{1}\}$ and $\alpha =\mathrm{1}+{\sum }_{t=\mathrm{1}}^m\mathrm{m}\mathrm{i}\mathrm{n}\{\mathrm{1}/r,\mathrm{1}/p_t\}-{\sum }_{t=1}^m\left(\mathrm{1}/p_t\right)-m/r$. Then$$\begin{array}{}\text{(13)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[M_{\mathrm{1}}\left({\mathit{\lambda }}_t^{p_t},r\right)-M_{\mathrm{2},n}^{p_t}\left({\mathit{\lambda }}_t,r\right)\right]}^{\mathrm{1}/p_t}\le n^{\mathrm{1}-\tau }{M}_{\mathrm{1}}\left({\displaystyle \prod }_{t=\mathrm{1}}^m{\mathit{\lambda }}_t,\mathrm{1}\right)-{\left(n-\mathrm{1}\right)}^{\alpha }{M}_{\mathrm{2},n}\left({\displaystyle \prod }_{t=\mathrm{1}}^m{\mathit{\lambda }}_t,\mathrm{1}\right).\end{array}$$

Proof.

Simple computations lead to $$\begin{array}{}\text{(14)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[M_{\mathrm{1}}\left({\mathit{\lambda }}_t^{p_t},r\right)-M_{\mathrm{2},n}^{p_t}\left({\mathit{\lambda }}_t,r\right)\right]}^{\mathrm{1}/p_t}={\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}.\end{array}$$ Since $p_t>\mathrm{0}$, ${\lambda }_{\mathrm{1}t}^{p_t}-(\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^r{)}^{p_t/r}>\mathrm{0}$, let$$\begin{array}{}\text{(15)}& {\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right)}^{p_t/r}={\xi }_t^{p_t},\hspace{1em}\left({\xi }_t>\mathrm{0},\hspace{0.17em}\hspace{0.17em}t=\mathrm{1,2},\dots ,m-\mathrm{1}\right),\text{(16)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-n^{\tau -\mathrm{1}}{\left(n-\mathrm{1}\right)}^{{\sum }_{t=\mathrm{1}}^m\min \left\{\mathrm{1}/r,\mathrm{1}/p_t\right\}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)-m/r}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}=n^{\tau -\mathrm{1}}{\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t.\end{array}$$Combining (15) and inequalities (10) and (8), we have$$\begin{array}{}\text{(17)}& {\left[{\xi }_m^{p_m}+{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{sm}^r\right)}^{p_m/r}\right]}^{\mathrm{1}/p_m}{\displaystyle \prod }_{t=\mathrm{1}}^{m-\mathrm{1}}{\lambda }_{\mathrm{1}t}={\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\xi }_t^{p_t}+{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}\ge {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\xi }_t^{p_t}+{\left(n-\mathrm{1}\right)}^{\min \left\{p_t/r,\mathrm{1}\right\}-p_t/r-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}\right]}^{\mathrm{1}/p_t}={\displaystyle \prod }_{t=\mathrm{1}}^m{\left\{{\xi }_t^{p_t}+\sum _{s=\mathrm{2}}^n{\left({\left(n-\mathrm{1}\right)}^{\min \left\{\mathrm{1}/r,\mathrm{1}/p_t\right\}-\mathrm{1}/r-\mathrm{1}/p_t}{\lambda }_{st}\right)}^{p_t}\right\}}^{\mathrm{1}/p_t}\ge n^{\tau -\mathrm{1}}\left\{{\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t+{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m\left({\left(n-\mathrm{1}\right)}^{\min \left\{\mathrm{1}/r,\mathrm{1}/p_t\right\}-\mathrm{1}/r-\mathrm{1}/p_t}{\lambda }_{st}\right)\right\}=n^{\tau -\mathrm{1}}\left[{\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t+{\left(n-\mathrm{1}\right)}^{{\sum }_{t=\mathrm{1}}^m\min \left\{\mathrm{1}/r,\mathrm{1}/p_t\right\}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)-m/r}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}\right]={\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}.\end{array}$$Therefore, from (15), (16), and (17), we obtain$$\begin{array}{}\text{(18)}& {\left[{\xi }_m^{p_m}+{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{sm}^r\right)}^{p_m/r}\right]}^{\mathrm{1}/p_m}{\displaystyle \prod }_{t=\mathrm{1}}^{m-\mathrm{1}}{\lambda }_{\mathrm{1}t}\ge {\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}.\end{array}$$ Hence, we obtain$$\begin{array}{}\text{(19)}& {\xi }_m\ge {\left[{\lambda }_{\mathrm{1}m}^{p_m}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{sm}^r\right)}^{p_m/r}\right]}^{\mathrm{1}/p_m}.\end{array}$$ Therefore, we have$$\begin{array}{}\text{(20)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t\ge {\left[{\lambda }_{\mathrm{1}m}^{p_m}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{sm}^r\right)}^{p_m/r}\right]}^{\mathrm{1}/p_m}{\displaystyle \prod }_{t=\mathrm{1}}^{m-\mathrm{1}}{\xi }_t={\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}.\end{array}$$By using (16), we immediately obtain$$\begin{array}{}\text{(21)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{\alpha }\frac{\mathrm{1}}{n-\mathrm{1}}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st},\end{array}$$ which leads to inequality (13). The proof of Theorem 10 is accomplished.

According to Theorem 10 we can get the following corollaries.

Corollary 11.

Let $r\ge p_t>\mathrm{0}$ and ${\lambda }_{\mathrm{1}t}^{p_t}-(\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^r{)}^{p_t/r}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let $\tau =\mathrm{m}\mathrm{i}\mathrm{n}\{{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right),\mathrm{1}\}$. Then$$\begin{array}{}\text{(22)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$

Proof.

From inequality (21), we can get$$\begin{array}{}\text{(23)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{\alpha }\frac{\mathrm{1}}{n-\mathrm{1}}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{\alpha -\mathrm{1}}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$ The proof of Corollary 11 is accomplished.

Corollary 12.

Let $\mathrm{1}\ge p_t>\mathrm{0}$ and ${\lambda }_{\mathrm{1}t}^{p_t}-(\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}{)}^{p_t}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$. Then$$\begin{array}{}\text{(24)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}\right)}^{p_t}\right]}^{\mathrm{1}/p_t}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$

If we put ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)=\mathrm{1}$ in Corollary 11, the conclusions that we can draw from this are as follows.

Corollary 13.

Let $r\ge p_t>\mathrm{0}$ and ${\lambda }_{\mathrm{1}t}^{p_t}-(\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^r{)}^{p_t/r}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)=\mathrm{1}$. Then$$\begin{array}{}\text{(25)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-\frac{\mathrm{1}}{n-\mathrm{1}}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$

Let $p_t=r\hspace{0.17em}\hspace{0.17em}(t=\mathrm{1,2},\dots ,m)$; then from Theorem 10, we get

Corollary 14.

Let $r>\mathrm{0}$, ${\lambda }_{\mathrm{1}t}^r-\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^r>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let $\tau =\min \{m/r,\mathrm{1}\}$. Then$$\begin{array}{}\text{(26)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^r-\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^r\right]}^{\mathrm{1}/r}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{-m/r}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$

Using the obtained inequality of inequality (26), we can get the following result of Wu [13].

Corollary 15.

Let $r>\mathrm{0}$ and ${\psi }_{\mathrm{1}t}^r-{\sum }_{s=\mathrm{2}}^n{\psi }_{st}^r>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let $\tau =\min \{m/r,\mathrm{1}\}$. Then$$\begin{array}{}\text{(27)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\psi }_{\mathrm{1}t}^r-\sum _{s=\mathrm{2}}^n{\psi }_{st}^r\right)}^{\mathrm{1}/r}\le n^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_{\mathrm{1}t}-{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_{st}.\end{array}$$

Theorem 16.

Let $r>\mathrm{0}$, $p_t>\mathrm{0}$, and $M_{\mathrm{1}}({\mathit{\lambda }}_t^{p_t},r)-M_{\mathrm{2},n}({\mathit{\lambda }}_t^{p_t},r)>\mathrm{0}$, let ${\mathit{\lambda }}_t=({\lambda }_{\mathrm{1}t},{\lambda }_{\mathrm{2}t},\dots ,{\lambda }_{nt})$ and ${\prod }_{t=\mathrm{1}}^m{\mathit{\lambda }}_t=({\prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t},{\prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{2}t},\dots ,{\prod }_{t=\mathrm{1}}^m{\lambda }_{nt})$, and denote $\tau =\min \{{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right),\mathrm{1}\}$. Then$$\begin{array}{}\text{(28)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[M_{\mathrm{1}}\left({\mathit{\lambda }}_t^{p_t},r\right)-M_{\mathrm{2},n}\left({\mathit{\lambda }}_t^{p_t},r\right)\right]}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-\tau }{M}_{\mathrm{1}}\left({\displaystyle \prod }_{t=\mathrm{1}}^m{\mathit{\lambda }}_t,r\right)-{\left(n-\mathrm{1}\right)}^{\left(\tau -{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\right)/r}{M}_{\mathrm{2},n}\left({\displaystyle \prod }_{t=\mathrm{1}}^m{\mathit{\lambda }}_t,r\right).\end{array}$$

Proof.

For the hypotheses $p_t>\mathrm{0}$ and $M_{\mathrm{1}}({\mathit{\lambda }}_t^{p_t},r)-M_{\mathrm{2},n}({\mathit{\lambda }}_t^{p_t},r)>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(t=\mathrm{1,2},\dots ,m)$, we have $$\begin{array}{}\text{(29)}& {\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r}>\mathrm{0}.\end{array}$$ Denote$$\begin{array}{}\text{(30)}& {\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r}={\xi }_t^{p_t},\hspace{1em}\left({\xi }_t>\mathrm{0},\hspace{0.17em}\hspace{0.17em}t=\mathrm{1,2},\dots ,m-\mathrm{1}\right),\text{(31)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\mathrm{2}}^{\tau -\mathrm{1}}{\left(n-\mathrm{1}\right)}^{\left(\mathrm{1}/r\right)\left(\tau -{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\right)}{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n\prod _{t=\mathrm{1}}^m{\lambda }_{st}^r\right)}^{\mathrm{1}/r}={\mathrm{2}}^{\tau -\mathrm{1}}{\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t.\end{array}$$

By using inequality (8), simple computations lead to$$\begin{array}{}\text{(32)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\xi }_t^{p_t}+{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_t}\ge {\mathrm{2}}^{\tau -\mathrm{1}}\left[{\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t+{\displaystyle \prod }_{t=\mathrm{1}}^m{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r{p}_t}\right]={\mathrm{2}}^{\tau -\mathrm{1}}\left[{\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t+{\left(n-\mathrm{1}\right)}^{-\left(\mathrm{1}/r\right){\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \prod }_{t=\mathrm{1}}^m{\left(\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r{p}_t}\right]\ge {\mathrm{2}}^{\tau -\mathrm{1}}\left[{\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t+{\left(n-\mathrm{1}\right)}^{\left(\mathrm{1}/r\right)\left(\tau -{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\right)}{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n\prod _{t=\mathrm{1}}^m{\lambda }_{st}^r\right)}^{\mathrm{1}/r}\right].\end{array}$$Therefore, from (30), (31), and (32), we obtain $$\begin{array}{}\text{(33)}& {\left[{\xi }_m^{p_m}+{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{sm}^{r{p}_m}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_m}{\displaystyle \prod }_{t=\mathrm{1}}^{m-\mathrm{1}}{\left[{\xi }_t^{p_t}+{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_t}={\left[{\xi }_m^{p_m}+{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{sm}^{r{p}_m}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_m}{\displaystyle \prod }_{t=\mathrm{1}}^{m-\mathrm{1}}{\lambda }_{\mathrm{1}t}\ge {\mathrm{2}}^{\tau -\mathrm{1}}\left[{\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t+{\left(n-\mathrm{1}\right)}^{\left(\mathrm{1}/r\right)\left(\tau -{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\right)}{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n\prod _{t=\mathrm{1}}^m{\lambda }_{st}^r\right)}^{\mathrm{1}/r}\right]={\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}.\end{array}$$ Hence $$\begin{array}{}\text{(34)}& {\xi }_m\ge {\left[{\lambda }_{\mathrm{1}m}^{p_m}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{sm}^{r{p}_m}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_m}.\end{array}$$ Thus$$\begin{array}{}\text{(35)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\xi }_t\ge {\left[{\lambda }_{\mathrm{1}m}^{p_m}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{sm}^{r{p}_m}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_m}{\displaystyle \prod }_{t=\mathrm{1}}^{m-\mathrm{1}}{\xi }_t={\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_m}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_t}.\end{array}$$

Combining (31) and inequality (35), we get $$\begin{array}{}\text{(36)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_m}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{\left(\mathrm{1}/r\right)\left(\tau -{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\right)}{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n\prod _{t=\mathrm{1}}^m{\lambda }_{st}^r\right)}^{\mathrm{1}/r},\end{array}$$ which is equivalent to inequality (28). Successful proof of Theorem 16 is as follows.

Corollary 17.

Let ${\lambda }_{st},p_t>\mathrm{0}$ and ${\lambda }_{\mathrm{1}t}^{p_t}-(\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}{)}^{\mathrm{1}/r}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)>\mathrm{1}$. Then$$\begin{array}{}\text{(37)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_t}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{\left(\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\right)/r}{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n\prod _{t=\mathrm{1}}^m{\lambda }_{st}^r\right)}^{\mathrm{1}/r}.\end{array}$$

In particular, putting $r=\mathrm{1}$ in Corollary 17, we obtain a new version of the Aczél-Vasić-Pečarić inequality (3).$$\begin{array}{}\text{(38)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}\right]}^{\mathrm{1}/p_t}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$

Corollary 18.

Let ${\lambda }_{st},p_t>\mathrm{0}$ and ${\lambda }_{\mathrm{1}t}^{p_t}-(\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}{)}^{\mathrm{1}/r}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\le \mathrm{1}$. Then$$\begin{array}{}\text{(39)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n\prod _{t=\mathrm{1}}^m{\lambda }_{st}^r\right)}^{\mathrm{1}/r}.\end{array}$$

Corollary 19.

Let ${\lambda }_{st},p_t>\mathrm{0}$ and ${\lambda }_{\mathrm{1}t}^{p_t}-\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\le \mathrm{1}$. Then$$\begin{array}{}\text{(40)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\lambda }_{\mathrm{1}t}^{p_t}-\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}\right)}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-\frac{\mathrm{1}}{n-\mathrm{1}}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$

If we put $r=\mathrm{1}$ in Theorem 16, then we have the following corollary.

Corollary 20.

Let ${\lambda }_{st},p_t>\mathrm{0}$ and ${\lambda }_{\mathrm{1}t}^{p_t}-\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let $\tau =\min \{{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right),\mathrm{1}\}$. Then$$\begin{array}{}\text{(41)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\lambda }_{\mathrm{1}t}^{p_t}-\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}\right)}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{\tau -{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)-\mathrm{1}}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}.\end{array}$$

Using the substitutions ${\psi }_{\mathrm{1}t}\to {\lambda }_{\mathrm{1}t}$ and $(n-\mathrm{1}{)}^{\mathrm{1}/p_t}{\psi }_{st}\to {\lambda }_{st}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{2,3},\dots ,n,t=\mathrm{1,2},\dots ,m)$ in Corollary 20, we obtain the following refinement of the Aczél-Vasić-Pečarić inequality (3).

Corollary 21.

Let ${\psi }_{st},p_t>\mathrm{0}$ and ${\psi }_{\mathrm{1}t}^{p_t}-{\sum }_{s=\mathrm{2}}^n{\psi }_{st}^{p_t}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let $\tau =\mathrm{m}\mathrm{i}\mathrm{n}\{{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right),\mathrm{1}\}$. Then$$\begin{array}{}\text{(42)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\psi }_{\mathrm{1}t}^{p_t}-\sum _{s=\mathrm{2}}^n{\psi }_{st}^{p_t}\right)}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-\tau }{\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_{\mathrm{1}t}-{\left(n-\mathrm{1}\right)}^{\tau -\mathrm{1}}{\displaystyle \sum }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_{st}.\end{array}$$

If we put ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\ge \mathrm{1}$ in Corollary 21, then we can derive inequality (3).

In particular, putting $m=\mathrm{2}$, $p_{\mathrm{1}}=p$, $p_{\mathrm{2}}=q$, ${\lambda }_{s\mathrm{1}}={\lambda }_s$, and ${\lambda }_{s\mathrm{2}}={\nu }_s\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n)$ in Corollary 21, we obtain a fresh improvement and promotion of nonequality (2).

Corollary 22.

Let ${\lambda }_s,{\nu }_s>\mathrm{0}$, $p,q>\mathrm{0}$, and $\tau =\min \{\mathrm{1}/p+\mathrm{1}/q,\mathrm{1}\}$, and let ${\lambda }_s^p-{\sum }_{s=\mathrm{2}}^n{\lambda }_s^p>\mathrm{0}$ and ${\nu }_s^q-{\sum }_{s=\mathrm{2}}^n{\nu }_s^q>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$. Then$$\begin{array}{}\text{(43)}& {\left({\lambda }_s^p-\sum _{s=\mathrm{2}}^{\mathrm{n}}{\lambda }_s^p\right)}^{\mathrm{1}/p}{\left({\nu }_s^q-\sum _{s=\mathrm{2}}^n{\nu }_s^q\right)}^{\mathrm{1}/q}\le {\mathrm{2}}^{\mathrm{1}-\tau }{\lambda }_{\mathrm{1}}{\nu }_{\mathrm{1}}-{\left(n-\mathrm{1}\right)}^{\tau -\mathrm{1}}{\displaystyle \sum }_{s=\mathrm{2}}^n{\lambda }_s{\nu }_s.\end{array}$$

Theorem 23.

Let ${\lambda }_{st},p_t>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$ and ${\lambda }_{\mathrm{1}t}^{p_t}-{\prod }_{s=\mathrm{2}}^n{\lambda }_{st}^{p_t/\left(n-\mathrm{1}\right)}>\mathrm{0}$, and let ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\le \mathrm{1}$. Then$$\begin{array}{}\text{(44)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\lambda }_{\mathrm{1}t}^{p_t}-\prod _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t/\left(n-\mathrm{1}\right)}\right)}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\displaystyle \prod }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}^{\mathrm{1}/\left(n-\mathrm{1}\right)}.\end{array}$$

Proof.

For ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\le \mathrm{1}$, by inequality (28), we obtain$$\begin{array}{}\text{(45)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\lambda }_{\mathrm{1}t}^{p_t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{r{p}_t}\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\left(\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n\prod _{t=\mathrm{1}}^m{\lambda }_{st}^r\right)}^{\mathrm{1}/r}.\end{array}$$We get this inequality (44) by adding $r\to \mathrm{0}$ on both sides of inequality (45).

For the hypotheses ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)\le \mathrm{1}$ and ${\lambda }_{\mathrm{1}t}^{p_t}-{\prod }_{s=\mathrm{2}}^n{\lambda }_{st}^{p_t/\left(n-\mathrm{1}\right)}>\mathrm{0}$ in Theorem 23, denote ${\zeta }_t=({\prod }_{s=\mathrm{2}}^n{\lambda }_{st}/{\lambda }_{\mathrm{1}t}{)}^{\mathrm{1}/\left(n-\mathrm{1}\right)}$, then we have $\mathrm{0}<{\zeta }_t^{p_t}<{\zeta }_t<\mathrm{1}$. Thus, we obtain the following corollaries.

Corollary 24.

Let $p_t>\mathrm{0}$, $\mathrm{0}<{\zeta }_t<\mathrm{1}\hspace{0.17em}\hspace{0.17em}(t=\mathrm{1,2},\dots ,m)$, and let ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)<\mathrm{1}$. Then$$\begin{array}{}\text{(46)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left(\mathrm{1}-{\zeta }_t^{p_t}\right)}^{\mathrm{1}/p_t}+{\displaystyle \prod }_{t=\mathrm{1}}^m{\zeta }_t\le {\mathrm{2}}^{\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}.\end{array}$$

Denote $\Theta (r)={\prod }_{t=\mathrm{1}}^m[M_{\mathrm{1}}({\mathit{\lambda }}_t^p,r)-M_{\mathrm{2},n}({\mathit{\lambda }}_t^p,r){]}^{\mathrm{1}/p_t}$, by Lemma 8, we realize that $\Theta (r)$ is decreasing on $(\mathrm{0},+\infty )$. Combining Theorems 16 and 23, we obtain the following corollary.

Corollary 25.

Let ${\lambda }_{st}>\mathrm{0}$ and $p_t>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(s=\mathrm{1,2},\dots ,n,t=\mathrm{1,2},\dots ,m)$, and let ${\lambda }_{\mathrm{1}t}^{p_t}-\left(\mathrm{1}/\left(n-\mathrm{1}\right)\right){\sum }_{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}>\mathrm{0}$. If ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)<\mathrm{1}$, then$$\begin{array}{}\text{(47)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\lambda }_{\mathrm{1}t}^{p_t}-\frac{\mathrm{1}}{n-\mathrm{1}}\sum _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t}\right)}^{\mathrm{1}/p_t}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\lambda }_{\mathrm{1}t}^{p_t}-\prod _{s=\mathrm{2}}^n{\lambda }_{st}^{p_t/\left(n-\mathrm{1}\right)}\right)}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{\mathrm{1}t}-{\displaystyle \prod }_{s=\mathrm{2}}^n{\displaystyle \prod }_{t=\mathrm{1}}^m{\lambda }_{st}^{\mathrm{1}/\left(n-\mathrm{1}\right)}.\end{array}$$

3. Applications

As is well known, analytic inequalities have various important applications in many branches of mathematics [33–37]. In this section, we will get some applications for this new inequality in Section 2.

Theorem 26.

Let ${\psi }_t>\mathrm{0}$, $r\ge p_t>\mathrm{0}$, ${\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)=\mathrm{1}$, $(t=\mathrm{1,2},\dots ,m)$, and ${\psi }_t^{p_t}-({\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^r(x)dx{)}^{p_t/r}>\mathrm{0}$. Then$$\begin{array}{}\text{(48)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\psi }_t^{p_t}-{\left({\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^r\left(x\right)dx\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_t-{{\displaystyle \int }}_{\mathrm{0}}^{\mathrm{1}}{\displaystyle \prod }_{t=\mathrm{1}}^m{f}_t\left(x\right)dx,\end{array}$$where $f_t(x)$ can be integrated on $[\mathrm{0,1}]$ and $f_t(x)>\mathrm{0}$.

Proof.

For all positive integers $n$, the equidistance of the partition of $[\mathrm{0,1}]$ is selected as$$\begin{array}{c}\text{(49)}& \mathrm{0}<\frac{\mathrm{1}}{n}<\cdots <\frac{n-\mathrm{1}}{n}<\mathrm{1},x_s=\frac{s}{n},\\ △x_s=\frac{\mathrm{1}}{n},\\ s=\mathrm{0,1},\dots ,n-\mathrm{1}.\end{array}$$ Noting that ${\psi }_t^{p_t}-({\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^r(x)dx{)}^{p_t/r}>\mathrm{0}\hspace{0.17em}\hspace{0.17em}(t=\mathrm{1,2},\dots ,m)$, we have $$\begin{array}{}\text{(50)}& {\psi }_t^{p_t}-{\left(\underset{n\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\sum _{s=\mathrm{0}}^{n-\mathrm{1}}{f}_t^r\left(\mathrm{0}+\frac{s}{n}\right)\frac{\mathrm{1}}{n}\right)}^{p_t/r}>\mathrm{0}\hspace{1em}\left(t=\mathrm{1,2},\dots ,m\right).\end{array}$$ Therefore, there is a positive integer $N$, like that $$\begin{array}{}\text{(51)}& {\psi }_t^{p_t}-{\left(\sum _{s=\mathrm{0}}^{n-\mathrm{1}}{f}_t^r\left(\mathrm{0}+\frac{s}{n}\right)\frac{\mathrm{1}}{n}\right)}^{p_t/r}>\mathrm{0},\end{array}$$ for all $n>N$ and $t=\mathrm{1,2},\dots ,m$.

Moreover, for any $n>N$, it follows from Corollary 13 that$$\begin{array}{}\text{(52)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\psi }_t^{p_t}-{\left(\frac{\mathrm{1}}{n}\sum _{s=\mathrm{0}}^{n-\mathrm{1}}{f}_t^r\left(\frac{s}{n}\right)\right)}^{p_t/r}\right]}^{\mathrm{1}/p_t}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_t-\frac{\mathrm{1}}{n}{\displaystyle \sum }_{s=\mathrm{0}}^{n-\mathrm{1}}{\displaystyle \prod }_{t=\mathrm{1}}^m{f}_t\left(\frac{s}{n}\right).\end{array}$$

We may find that $f_t(x)\hspace{0.17em}\hspace{0.17em}(t=\mathrm{1,2},\dots ,m)$ are positive Riemann integrable functions on $[\mathrm{0,1}]$; we know that ${\prod }_{t=\mathrm{1}}^m{f}_t(x)$ and $f_t^r(x)$ are also integrable on $[\mathrm{0,1}]$. Letting $n\to \infty$ on both sides of inequality (52), we get the desired inequality (48). The proof of Theorem 26 is achieved.

Theorem 27.

Let ${\psi }_t>\mathrm{0}$ and $r\ge m$, and let $f_t(x)\hspace{0.17em}\hspace{0.17em}(t=\mathrm{1,2},\dots ,m)$ be positive Riemann integrable functions on $[\mathrm{0,1}]$ such that ${\psi }_t^r-{\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^r(x)dx>\mathrm{0}$. Then$$\begin{array}{}\text{(53)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\psi }_t^r-{\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^r\left(x\right)dx\right]}^{\mathrm{1}/r}\le {\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_t-{{\displaystyle \int }}_{\mathrm{0}}^{\mathrm{1}}{\displaystyle \prod }_{t=\mathrm{1}}^m{f}_t\left(x\right)dx.\end{array}$$

Proof.

Applying Corollary 14 and along the lines of the proof Theorem 26, Theorem 27 is simply given.

Theorem 28.

Let $r>\mathrm{0}$, ${\psi }_t>\mathrm{0}$, $p_t>\mathrm{0}$, ${\sum }_{t=\mathrm{1}}^m\left(1/p_t\right)\le \mathrm{1}$, $(t=\mathrm{1,2},\dots ,m)$, and ${\psi }_t^{p_t}-({\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^{r{p}_t}(x)dx{)}^{\mathrm{1}/r}>\mathrm{0}$. Then$$\begin{array}{}\text{(54)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left[{\psi }_t^{p_t}-{\left({\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^{r{p}_t}\left(x\right)dx\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_t-{\left({\int }_{\mathrm{0}}^{\mathrm{1}}\prod _{t=\mathrm{1}}^m{f}_t^r\left(x\right)dx\right)}^{\mathrm{1}/r},\end{array}$$where $f_t(x)$ can be integrated on $[\mathrm{0,1}]$ and $f_t(x)>\mathrm{0}$.

Proof.

Combining the proof Theorem 26 and Corollary 18, it is easy to get Theorem 28.

Corollary 29.

Let $p_t>\mathrm{0}$, ${\sum }_{t=\mathrm{1}}^m\left(1/p_t\right)\le \mathrm{1}$, $(t=\mathrm{1,2},\dots ,m)$, and ${\psi }_t^{p_t}-{\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^{p_t}(x)dx>\mathrm{0}$. Then$$\begin{array}{}\text{(55)}& {\displaystyle \prod }_{t=\mathrm{1}}^m{\left({\psi }_t^{p_t}-{\int }_{\mathrm{0}}^{\mathrm{1}}{f}_t^{p_t}\left(x\right)dx\right)}^{\mathrm{1}/p_t}\le {\mathrm{2}}^{\mathrm{1}-{\sum }_{t=\mathrm{1}}^m\left(\mathrm{1}/p_t\right)}{\displaystyle \prod }_{t=\mathrm{1}}^m{\psi }_t-{{\displaystyle \int }}_{\mathrm{0}}^{\mathrm{1}}{\displaystyle \prod }_{t=\mathrm{1}}^m{f}_t\left(x\right)dx,\end{array}$$where $f_t(x)$ can be integrated on $[\mathrm{0,1}]$ and $f_t(x)>\mathrm{0}$.

A direct result from Theorem 28 is given by us. Putting $m=\mathrm{2}$, $p_{\mathrm{1}}=p>\mathrm{0}$, $p_{\mathrm{2}}=q>\mathrm{0}$, ${\psi }_{\mathrm{1}}={\lambda }_{\mathrm{1}}$, ${\psi }_{\mathrm{2}}={\nu }_{\mathrm{1}}$, $f_{\mathrm{1}}=f$, and $f_{\mathrm{2}}=g$ in (54), we can get the following corollaries.

Corollary 30.

Let $r>\mathrm{0}$, ${\lambda }_{\mathrm{1}},{\nu }_{\mathrm{1}}>\mathrm{0}$, $p,q>\mathrm{0}$, $\mathrm{1}/p+\mathrm{1}/q\le \mathrm{1}$, ${\lambda }_{\mathrm{1}}^p-({\int }_{\mathrm{0}}^{\mathrm{1}}{f}^{rp}(x)dx{)}^{1/r}>\mathrm{0}$, and ${\nu }_{\mathrm{1}}^q-({\int }_{\mathrm{0}}^{\mathrm{1}}{g}^{rq}(x)dx{)}^{1/r}>\mathrm{0}$. Then$$\begin{array}{}\text{(56)}& {\left[{\lambda }_{\mathrm{1}}^p-{\left({\int }_{\mathrm{0}}^{\mathrm{1}}{f}^{rp}\left(x\right)dx\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/p}{\left[{\nu }_{\mathrm{1}}^q-{\left({\int }_{\mathrm{0}}^{\mathrm{1}}{g}^{rq}\left(x\right)dx\right)}^{\mathrm{1}/r}\right]}^{\mathrm{1}/q}\le {\mathrm{2}}^{\mathrm{1}-\mathrm{1}/p-\mathrm{1}/q}{\lambda }_{\mathrm{1}}{\nu }_{\mathrm{1}}-{\left[{\int }_{\mathrm{0}}^{\mathrm{1}}{\left(f\left(x\right)g\left(x\right)\right)}^{r}dx\right]}^{\mathrm{1}/r},\end{array}$$where $f(x),g(x)$ can be integrated on $[\mathrm{0,1}]$ and $f(x)>\mathrm{0},g(x)>\mathrm{0}$.

If we set $r=\mathrm{1}$ in Corollary 30, then the following inequality holds.

Corollary 31.

Let ${\lambda }_{\mathrm{1}},{\nu }_{\mathrm{1}}>\mathrm{0}$, $p,q>\mathrm{0}$, $\mathrm{1}/p+\mathrm{1}/q\le \mathrm{1}$, ${\lambda }_{\mathrm{1}}^p-{\int }_{\mathrm{0}}^{\mathrm{1}}{f}^p(x)dx>\mathrm{0}$, and ${\nu }_{\mathrm{1}}^q-{\int }_{\mathrm{0}}^{\mathrm{1}}{g}^q(x)dx>\mathrm{0}$. Then$$\begin{array}{}\text{(57)}& {\left({\lambda }_{\mathrm{1}}^p-{\int }_{\mathrm{0}}^{\mathrm{1}}{f}^p\left(x\right)dx\right)}^{\mathrm{1}/p}{\left({\nu }_{\mathrm{1}}^q-{\int }_{\mathrm{0}}^{\mathrm{1}}{g}^q\left(x\right)dx\right)}^{\mathrm{1}/q}\le {\mathrm{2}}^{\mathrm{1}-\mathrm{1}/p-\mathrm{1}/q}{\lambda }_{\mathrm{1}}{\nu }_{\mathrm{1}}-{{\displaystyle \int }}_{\mathrm{0}}^{\mathrm{1}}f\left(x\right)g\left(x\right)dx,\end{array}$$where $f(x),g(x)$ can be integrated on $[\mathrm{0,1}]$ and $f(x)>\mathrm{0},g(x)>\mathrm{0}$.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

Acknowledgments

This work was supported by the Fundamental Research Funds for the Central Universities (no. 2015ZD29) and the Higher School Science Research Funds of Hebei Province of China (no. Z2015137).

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