A Reverse Hardy-Hilbert’s Inequality Involving One Partial Sum as the Terms of Double Series

Abstract

In this paper, by constructing proper weight coefficients and utilizing the Euler-Maclaurin summation formula and the Abel partial summation formula, we establish reverse Hardy-Hilbert’s inequality involving one partial sum as the terms of double series. On the basis of the obtained inequality, the equivalent conditions of the best possible constant factor associated with several parameters are discussed. Finally, we illustrate that more reverse inequalities of Hardy-Hilbert type can be generated from the special cases of the present results.

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1. Introduction

The celebrated Hardy-Hilbert’s inequality asserts that if $p>1,$$1/p+1/q=1$, $a_m\ge 0$,$b_n\ge 0$$0<{\sum }_{m=1}^{\infty }{a}_m^p<\infty$, and $0<{\sum }_{n=1}^{\infty }{b}_n^q<\infty$, then $$\begin{array}{}\text{(1)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{b}_n}{m+n}<\frac{\pi }{\sin \pi /p}{\sum _{m=1}^{\infty }{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{b}_n^q}^{1/q},\end{array}$$where the constant factor $\pi /\sin \pi /p$ is best possible (see [1]).

A generalization of inequality (Equation (1)) was posted by Krnić and Pečarić in [2], as follows: $$\begin{array}{}\text{(2)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{b}_n}{{m+n}^{\lambda }}<B{\lambda }_1,{\lambda }_2{\sum _{m=1}^{\infty }{m}^{p1-{\lambda }_1-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-{\lambda }_2-1}{b}_n^q}^{1/q},\end{array}$$where ${\lambda }_i\in 0,2i=1,2,{\lambda }_1+{\lambda }_2=\lambda \in 0,4$, and also the constant factor $B{\lambda }_1,{\lambda }_2$ given by the beta function is the best possible.

In 2019, Adiyasuren et al. [3] put forwarded an analogous version of Hardy-Hilbert’s inequality containing the kernel $1/{m+n}^{\lambda }$ and partial sums. Recently, Liu [4] and Yang [5] provided some extensions of Hardy-Hilbert type inequalities.

In 2020, by introducing more parameters, a generalization of inequality (Equation (2)) was established by Yang et al. [6], which is a further generalization of inequality (Equation (1)), as follows: $$\begin{array}{}\text{(3)}& \sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_m{b}_n}{{m^{\alpha }+n^{\beta }}^{\lambda }}<{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}\times {\sum _{m=1}^{\infty }{m}^{p1-\alpha \lambda -{\lambda }_2/p+{\lambda }_1/q-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-\beta {\lambda -\lambda }_1/q+{\lambda }_2/p-1}{b}_n^q}^{1/q},\end{array}$$where $1/p+1/q=1,p>1,\alpha ,\beta \in 0,1,\lambda \in 0,6,{\lambda }_1\in 0,2/a\cap 0,\lambda ,{\lambda }_2\in 0,2/\beta \cap 0,\lambda ,k_{\lambda }{\lambda }_i≔B{\lambda }_i,\lambda -{\lambda }_{i}i=1,2$.

In 2021, Liao et al. [7] considered a variation of inequality (Equation (3)) and proposed the following inequality containing partial sums as the terms of series: $$\begin{array}{}\text{(4)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{b}_n}{{m^{\alpha }+n^{\beta }}^{\lambda }}<\lambda {\frac{1}{\beta }{k}_{\lambda +1}{\lambda }_2+1}^{1/p}{\frac{1}{\alpha }{k}_{\lambda +1}{\lambda }_1}^{1/q}\times {\sum _{m=1}^{\infty }{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-\beta 1+{\stackrel{\wedge }{\lambda }}_2-1}{B}_n^q}^{1/q},\end{array}$$where $1/p+1/q=1,p>1,\alpha ,\beta \in 0,1,\lambda \in 0,5,{\lambda }_1\in 0,2/a\cap 0,\lambda +1,{\lambda }_2\in 0,2/\beta -1\cap 0,\lambda +1,{\widehat{\lambda }}_1≔\lambda -{\lambda }_2/p+{\lambda }_1/q,{\widehat{\lambda }}_2≔\lambda -{\lambda }_1/q+{\lambda }_2/p,k_{\lambda +1}{\lambda }_i≔B{\lambda }_i,\lambda +1-{\lambda }_{i}i=1,2,B_n≔{\sum }_{k=1}^n{b}_k$.

Inspired by the work [4–7] mentioned above, in this paper, we establish reverse Hardy-Hilbert’s inequality which contains one partial sum as the terms of double series. Our method is mainly based on real analysis techniques and the applications of the Euler-Maclaurin summation formula and the Abel partial summation formula. Moreover, the equivalent conditions of the best possible constant factor associated with several parameters are discussed. As applications, we deal with some equivalent forms of the obtained inequality and illustrate that more reverse inequalities of Hardy-Hilbert type can be derived from the special cases of the current inequality.

2. Some Lemmas

For convenience, let us first state the conditions (C1) below, which will be used repeatedly in what follows.

(C1) $0<p<1,q<0,1/p+1/q=1,\alpha ,\beta \in 0,1,\lambda \in 0,6,{\lambda }_1\in 0,2/\alpha \cap 0,\lambda ,{\lambda }_2\in 0,2/\beta \cap 0,\lambda ,{\widehat{\lambda }}_1≔\lambda -{\lambda }_2/p+{\lambda }_1/q,{\widehat{\lambda }}_2≔\lambda -{\lambda }_1/q+{\lambda }_2/p,k_{\lambda }{\lambda }_i≔B{\lambda }_i,\lambda -{\lambda }_{i}i=1,2$. Furthermore, for $a_m\ge 0$ and $b_n\ge 0$, we define the partial sums $B_n≔{\sum }_{k=1}^n{b}_{k}n\in N=1,2,\cdots$, which satisfies $\underset{n⟶\infty }{\lim }{B}_n/e^{{tn}^{\beta }}=0$ denoted as $B_n=o{e}^{{tn}^{\beta }}t>0;n⟶\infty$ and $$\begin{array}{c}\text{(5)}& 0<\sum _{m=1}^{\infty }{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p<\infty ,0<\sum _{n=1}^{\infty }{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q<\infty .\end{array}$$

Lemma 1 (see [8]).

(i) Let ${-1}^i{d}^i/{dt}^i\psi t>0,t\in m,\infty m\in N$ and ${\psi }^i\infty =0$ and $i=0,1,2,3$, and let $P_{i}t,B_{i}i\in N$ denote, respectively, the Bernoulli functions and Bernoulli numbers of $i$-order. Then,

$$\begin{array}{}\text{(6)}& {\int }_m^{\infty }{P}_{2q-1}t\psi tdt=-{\epsilon }_q\frac{B_{2q}}{2q}\psi m,\hspace{1em}0<{\epsilon }_q<1,q=1,2,\cdots .\end{array}$$

In particular, for $q=1$ and $B_2=1/6$, we have the inequality: $$\begin{array}{}\text{(7)}& -\frac{1}{12}\psi m<{\int }_m^{\infty }{P}_{1}t\psi tdt<0.\end{array}$$

For $q=2$ and $B_4=-1/30$, we have the inequality: $$\begin{array}{}\text{(8)}& 0<{\int }_m^{\infty }{P}_{3}t\psi tdt<\frac{1}{120}\psi m.\end{array}$$

(ii) If $ft>0\in C^{3}m,\infty ,f^i\infty =0i=0,1,2,3$, then we have the following Euler-Maclaurin summation formula:

$$\begin{array}{}\text{(9)}& \sum _{k=m}^{\infty }fk={\int }_m^{\infty }ftdt+\frac{1}{2}fm+{\int }_m^{\infty }{P}_{1}t{f}^{\prime }tdt,\text{(10)}& {\int }_m^{\infty }{P}_{1}t{f}^{\prime }tdt=-\frac{1}{12}{f}^{\prime }m+\frac{1}{6}{\int }_m^{\infty }{P}_{3}t{f}^{‴}tdt.\end{array}$$

Lemma 2.

For $s\in 0,6,s_2\in 0,2/\beta \cap 0,s,k_s{s}_2=B{s}_2,s-s_2$, we define the following weight coefficient: $$\begin{array}{}\text{(11)}& {\varpi }_{\alpha }{s}_2,m≔m^{\alpha s-s_2}\sum _{n=1}^{\infty }\frac{\beta n^{\beta s_2-1}}{{m^{\alpha }+n^{\beta }}^s},\hspace{1em}m\in N.\end{array}$$

Then, the following inequalities hold: $$\begin{array}{}\text{(12)}& 0<k_s{s}_{2}1-O\frac{1}{m^{\alpha s_2}}<{\varpi }_{\alpha }{s}_2,m<k_s{s}_2,\hspace{1em}m\in N,\end{array}$$where $O1/m^{\alpha s_2}≔1/k_s{s}_2{\int }_0^{1/m^{\alpha }}{u}^{s_2-1}/{1+u}^{s}du>0.$

Proof.

For fixed $m\in N$, we define a function $\psi m,t$ as follows: $$\begin{array}{}\text{(13)}& \psi m,t≔\frac{\beta t^{\beta s_2-1}}{{m^{\alpha }+t^{\beta }}^s},\hspace{1em}t>0.\end{array}$$

Using equality (Equation (9)), we have the following: $$\begin{array}{}\text{(14)}& \sum _{n=1}^{\infty }\psi m,n={\int }_1^{\infty }\psi m,tdt+\frac{1}{2}\psi m,1+{\int }_1^{\infty }{P}_{1}t{\psi }^{\prime }m,tdt={\int }_0^{\infty }\psi m,tdt-\hslash m,\hslash m≔{\int }_0^1\psi m,tdt-\frac{1}{2}\psi m,1-{\int }_1^{\infty }{P}_{1}t{\psi }^{\prime }m,tdt.\end{array}$$

It is easy to observe that $-1/2\psi m,1=-\beta /2{m^{\alpha }+1}^s$. In addition, integration by parts, it follows that $$\begin{array}{}\text{(15)}& {{\int }_0^1\psi m,tdt=\beta {\int }_0^1\frac{t^{\beta s_2-1}}{{m^{\alpha }+t^{\beta }}^s}dt\stackrel{u=t^{\beta }}{=}{\int }_0^1\frac{u^{s_2-1}}{{m^{\alpha }+u}^s}du=\frac{1}{s_2}{\int }_0^1\frac{{du}^{s_2}}{{m^{\alpha }+u}^s}=\frac{1}{s_2}\frac{u^{s_2}}{{m^{\alpha }+u}^s}}_0^1+\frac{s}{s_2}{\int }_0^1\frac{u^{s_2}}{{m^{\alpha }+u}^{s+1}}du=\frac{1}{s_2}\frac{1}{{m^{\alpha }+1}^s}+\frac{s}{s_2{s}_2+1}{\int }_0^1\frac{{du}^{s_2+1}}{{m^{\alpha }+u}^{s+1}}>\frac{1}{s_2}\frac{1}{{m^{\alpha }+1}^s}+\frac{s}{s_2{s}_2+1}{\frac{u^{s_2+1}}{{m^{\alpha }+u}^{s+1}}}_0^1+\frac{ss+1}{s_2{s}_2+1{m^{\alpha }+1}^{s+2}}{\int }_0^1{u}^{s_2+1}du=\frac{1}{s_2}\frac{1}{{m^{\alpha }+1}^s}+\frac{\lambda }{s_2{s}_2+1}\frac{1}{{m^{\alpha }+1}^{s+1}}+\frac{ss+1}{s_2{s}_2+1s_2+2}\frac{1}{{m^{\alpha }+1}^{s+2}},-{\psi }^{\prime }m,t=-\frac{\beta \beta s_2-1t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^s}+\frac{{\beta }^2{st}^{\beta +\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}=-\frac{\beta \beta s_2-1t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^s}+\frac{{\beta }^{2}s{m}^{\alpha }+t^{\beta }-m^{\alpha }{t}^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}=\frac{\beta \beta s-\beta s_2+1t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^s}-\frac{{\beta }^2{sm}^{\alpha }{t}^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}.\end{array}$$Further, for $0<s_2\le 2/\beta ,0<\beta \le 1,s_2<s\le 6$, it is not difficult to verify that $$\begin{array}{}\text{(16)}& {-1}^i\frac{d^i}{{dt}^i}\frac{t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^s}>0,{-1}^i\frac{d^i}{{dt}^i}\frac{t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}>0\hspace{1em}i=0,1,2,3.\end{array}$$

By Equations (7)–(10), we obtain the following: $$\begin{array}{}\text{(17)}& \beta \beta s-\beta s_2+1{\int }_1^{\infty }{P}_{1}t\frac{t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^s}dt>-\frac{\beta \beta s-\beta s_2+1}{12{m^{\alpha }+1}^s},-{\beta }^2{m}^{\alpha }s{\int }_1^{\infty }{P}_{1}t\frac{t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}dt=\frac{{\beta }^2{m}^{\alpha }s}{12{m^{\alpha }+1}^{s+1}}-\frac{{\beta }^2{m}^{\alpha }s}{6}{\int }_1^{\infty }{P}_{3}t{\frac{t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}}^{\prime \prime }dt>\frac{{\beta }^2{m}^{\alpha }s}{12{m^{\alpha }+1}^{s+1}}-\frac{{\beta }^2{m}^{\alpha }s}{720}{{\frac{t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}}^{\prime \prime }}_{t=1}>\frac{{\beta }^2{m}^{\alpha }+1-1s}{12{m^{\alpha }+1}^{s+1}}-\frac{{\beta }^2{m}^{\alpha }+1s}{720}\frac{s+1s+2{\beta }^2}{{m^{\alpha }+1}^{s+3}}+\frac{\beta s+15-\beta -2\beta s_2}{{m^{\alpha }+1}^{s+2}}+\frac{2-\beta s_{2}3-\beta s_2}{{m^{\alpha }+1}^{s+1}}=\frac{{\beta }^{2}s}{12{m^{\alpha }+1}^s}-\frac{{\beta }^{2}s}{12{m^{\alpha }+1}^{s+1}}-\frac{{\beta }^{2}s}{720}\frac{s+1s+2{\beta }^2}{{m^{\alpha }+1}^{s+2}}+\frac{\beta s+15-\beta -2\beta s_2}{{m^{\alpha }+1}^{s+1}}+\frac{2-\beta s_{2}3-\beta s_2}{{m^{\alpha }+1}^s},\end{array}$$and then we get the following: $$\begin{array}{}\text{(18)}& \hslash m>\frac{1}{{m^{\alpha }+1}^s}{\hslash }_1+\frac{\lambda }{{m^{\alpha }+1}^{s+1}}{\hslash }_2+\frac{ss+1}{{m^{\alpha }+1}^{s+2}}{\hslash }_3,\end{array}$$where $$\begin{array}{c}\text{(19)}& {\hslash }_1≔\frac{1}{s_2}-\frac{\beta }{2}-\frac{\beta -{\beta }^2{s}_2}{12}-\frac{{\beta }^{2}s2-\beta s_{2}3-\beta s_2}{720},{\hslash }_2≔\frac{1}{s_2{s}_2+1}-\frac{{\beta }^2}{12}-\frac{{\beta }^{3}s+15-\beta -2\beta s_2}{720},\\ {\hslash }_3≔\frac{1}{s_2{s}_2+1s_2+2}-\frac{{\beta }^{4}s+2}{720}.\end{array}$$

It is easy to see that $$\begin{array}{}\text{(20)}& {\hslash }_1\ge \frac{1}{s_2}-\frac{\beta }{2}-\frac{\beta -{\beta }^2{s}_2}{12}-\frac{s{\beta }^{2}2-\beta s_{2}3-\beta s_2}{720}=\frac{\psi s_2}{720s_2},\end{array}$$where $\psi \sigma \sigma \in 0,2/\beta$ is defined by the following: $$\begin{array}{}\text{(21)}& \psi \sigma ≔720-420\beta +6s{\beta }^2\sigma +60{\beta }^2+5s{\beta }^3{\sigma }^2-s{\beta }^4{\sigma }^3.\end{array}$$

Further, we obtain that for $\beta \in 0,1,s\in 0,6$, $$\begin{array}{}\text{(22)}& {\psi }^{\prime }\sigma =-420\beta +6s{\beta }^2+260{\beta }^2+5s{\beta }^3\sigma -3{\beta }^4{\sigma }^2\le -420\beta -6s{\beta }^2+260{\beta }^2+5s{\beta }^3\frac{2}{\beta }=14s\beta -180\beta <0.\end{array}$$

Thus, it follows that $$\begin{array}{}\text{(23)}& {\hslash }_1\ge \frac{\psi s_2}{720s_2}\ge \frac{\psi 2/\beta }{720s_2}=\frac{1}{6s_2}>0.\end{array}$$

Also, we find that for $s_2\in 0,2/\beta$, $$\begin{array}{c}\text{(24)}& {\hslash }_2>\frac{{\beta }^2}{6}-\frac{{\beta }^2}{12}-\frac{5s+1{\beta }^2}{720}=\frac{1}{12}-\frac{s+1}{140}{\beta }^2>0,{\hslash }_3\ge \frac{1}{24}-\frac{s+2}{720}{\beta }^3>0\hspace{1em}0<s\le 6.\end{array}$$

Hence, we have $\hslash m>0$. Setting $t=m^{\alpha /\beta }{u}^{1/\beta }$, it follows that $$\begin{array}{}\text{(25)}& {\varpi }_{\alpha }{s}_2,m=m^{\alpha s-s_2}\sum _{n=1}^{\infty }\psi m,n<m^{\alpha s-s_2}{\int }_0^{\infty }\psi m,tdt=\beta m^{\alpha s-s_2}{\int }_0^{\infty }\frac{t^{\beta s_2-1}}{{m^{\alpha }+t^{\beta }}^s}dt={\int }_0^{\infty }\frac{u^{s_2-1}}{{1+u}^s}du=B{s}_2,s-s_2=k_s{s}_2.\end{array}$$

On the other hand, by Equation (9), we have the following: $$\begin{array}{}\text{(26)}& \sum _{n=1}^{\infty }\psi m,n={\int }_1^{\infty }\psi m,tdt+\frac{1}{2}\psi m,1+{\int }_1^{\infty }{P}_{1}t{\psi }^{\prime }m,tdt={\int }_1^{\infty }\psi m,tdt+Hm,Hm≔\frac{1}{2}\psi m,1+{\int }_1^{\infty }{P}_{1}t{\psi }^{\prime }m,tdt.\end{array}$$

Note that $1/2\psi m,1=\beta /2{m^{\alpha }+1}^s$, and $$\begin{array}{}\text{(27)}& {\psi }^{\prime }m,t=-\frac{\beta \beta s-\beta s_2+1t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^s}+\frac{{\beta }^2{sm}^{\alpha }{t}^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}.\end{array}$$

For $s_2\in 0,2/\beta \cap 0,s,0<s\le 6$, by Equation (7), we find $-\beta \beta s-\beta s_2+1{\int }_1^{\infty }{P}_{1}t{t}^{\beta s_2-2}/{m^{\alpha }+t^{\beta }}^{s}dt>0,$ and $$\begin{array}{}\text{(28)}& {\beta }^2{m}^{\alpha }s{\int }_1^{\infty }{P}_{1}t\frac{t^{\beta s_2-2}}{{m^{\alpha }+t^{\beta }}^{s+1}}dt>-\frac{{\beta }^2{m}^{\alpha }s}{12{m^{\alpha }+1}^{s+1}}>-\frac{{\beta }^{2}s}{12{m^{\alpha }+1}^s}.\end{array}$$

Hence, we have the following: $$\begin{array}{}\text{(29)}& Hm>\frac{\beta }{2{m^{\alpha }+1}^s}-\frac{{\beta }^{2}s}{12{m^{\alpha }+1}^s}\ge \frac{\beta }{2{m^{\alpha }+1}^s}-\frac{6\beta }{12{m^{\alpha }+1}^s}=0,\end{array}$$and then we obtain the following: $$\begin{array}{}\text{(30)}& {\varpi }_{\alpha }{\lambda }_2,m=m^{\alpha s-s_2}\sum _{n=1}^{\infty }\psi m,n>m^{\alpha s-s_2}{\int }_1^{\infty }\psi m,tdt=m^{\alpha s-s_2}{\int }_0^{\infty }\psi m,tdt-m^{\alpha s-s_2}{\int }_0^1\psi m,tdt=k_s{s}_{2}1-O\frac{1}{m^{\alpha s_2}}>0,\end{array}$$where $O1/m^{\alpha s_2}=1/k_s{s}_2{\int }_0^{1/m^{\alpha }}{u}^{s_2-1}du/{1+u}^s,$ which satisfies the following: $$\begin{array}{}\text{(31)}& 0<{\int }_0^{1/m^{\alpha }}\frac{u^{s_2-1}du}{{1+u}^s}<{\int }_0^{1/m^{\alpha }}{u}^{s_2-1}du=\frac{1}{s_2{m}^{\alpha s_2}}.\end{array}$$

Therefore, the two-sided inequalities in Equation (12) follow. This completes the proof of Lemma 2.

Moreover, by the same way as above, for $s\in 0,6,s_1\in 0,2/\alpha \cap 0,s,k_s{s}_1=B{s}_1,s-s_1$, one can obtain the following inequalities for another weight coefficient: $$\begin{array}{}\text{(32)}& k_s{s}_{1}1-O\frac{1}{n^{\beta s_1}}<{\omega }_{\beta }{s}_1,n≔n^{\beta s-s_1}\sum _{m=1}^{\infty }\frac{\alpha m^{\alpha s_1-1}}{{m^{\alpha }+n^{\beta }}^s}<k_s{s}_{1}n\in N.\end{array}$$

Lemma 3.

Under the assumption (C1), we have the following reverse Hardy-Hilbert’s inequality: $$\begin{array}{}\text{(33)}& I≔\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_m{b}_n}{{m^{\alpha }+n^{\beta }}^{\lambda }}>{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}\times {\sum _{m=1}^{\infty }1-O\frac{1}{m^{{\alpha \lambda }_2}}{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q}.\end{array}$$

Proof.

By using the reverse Hӧlder’s inequality [9], we obtain the following: $$\begin{array}{}\text{(34)}& I=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{1}{{m^{\alpha }+n^{\beta }}^{\lambda }}\frac{m^{\alpha 1-{\lambda }_1/q}{\beta n^{\beta -1}}^{1/p}}{n^{\beta 1-{\lambda }_2/p}{\alpha m^{\alpha -1}}^{1/q}}{a}_m\frac{n^{\beta 1-{\lambda }_2/p}{\alpha m^{\alpha -1}}^{1/q}}{m^{\alpha 1-{\lambda }_1/q}{\beta n^{\beta -1}}^{1/p}}{b}_n\ge {\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{\beta }{{m^{\alpha }+n^{\beta }}^{\lambda }}\frac{m^{\alpha 1-{\lambda }_{1}p-1}{n}^{\beta -1}{a}_m^p}{n^{\beta 1-{\lambda }_2}{\alpha m^{\alpha -1}}^{p-1}}}^{1/p}{\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{\alpha }{{m^{\alpha }+n^{\beta }}^{\lambda }}\frac{n^{\beta 1-{\lambda }_{2}q-1}{m}^{\alpha -1}{b}_n^q}{m^{\alpha 1-{\lambda }_1}{\beta n^{\beta -1}}^{q-1}}}^{1/q}={\frac{1}{\beta }\sum _{m=1}^{\infty }{\varpi }_{\alpha }{\lambda }_2,m{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p}^{1/p}{\frac{1}{\alpha }\sum _{n=1}^{\infty }{\omega }_{\beta }{\lambda }_1,n{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q}.\end{array}$$

Hence, by virtue of Equations (12) and (32), for $s=\lambda ,s_i={\lambda }_{i}i=1,2$, we get inequality (Equation (33)). The proof of Lemma 3 is complete.

Lemma 4.

Let $t>0$, then we have the inequality: $$\begin{array}{}\text{(35)}& \sum _{n=1}^{\infty }{e}^{-{tn}^{\beta }}{b}_n\le t\sum _{n=1}^{\infty }{e}^{-{tn}^{\beta }}{B}_n.\end{array}$$

Proof.

In view of $B_n{e}^{-{tn}^{\beta }}=o1n⟶\infty$, applying the Abel partial summation formula, we find the following: $$\begin{array}{}\text{(36)}& \sum _{n=1}^{\infty }{e}^{-{tn}^{\beta }}{b}_n=\underset{n⟶\infty }{\lim }{B}_n{e}^{-{tn}^{\beta }}+\sum _{n=1}^{\infty }{B}_n{e}^{-{tn}^{\beta }}-e^{-t{n+1}^{\beta }}=\sum _{n=1}^{\infty }{B}_n{e}^{-{tn}^{\beta }}-e^{-t{n+1}^{\beta }}.\end{array}$$

Since $1-e^{-t}<tt>0$, and for $\beta \in 0,1$, $$\begin{array}{}\text{(37)}& e^{-t{n+1}^{\beta }}\ge e^{-t{n}^{\beta }+1}\iff e^{t{n+1}^{\beta }-n^{\beta }-1}\le 1\iff {n+1}^{\beta }-n^{\beta }-1=\alpha {n+{\theta }_n}^{\beta -1}-1\le 0{\theta }_n\in 0,1,\end{array}$$we acquire that $$\begin{array}{}\text{(38)}& \sum _{n=1}^{\infty }{e}^{-{tn}^{\beta }}{b}_n\le \sum _{n=1}^{\infty }{B}_n{e}^{-{tn}^{\beta }}-e^{-t{n}^{\beta }+1}=1-e^{-t}\sum _{n=1}^{\infty }{B}_n{e}^{-{tn}^{\beta }}\le t\sum _{n=1}^{\infty }{B}_n{e}^{-{tn}^{\beta }},\end{array}$$which implies inequality (Equation (35)). Lemma 4 is proved.

3. Main Results

Theorem 5.

Under the assumption (C1), we have the following reverse Hardy-Hilbert’s inequality with one partial sum as the terms of double series: $$\begin{array}{}\text{(39)}& I_1≔\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}>\frac{1}{\lambda }{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}\times {\sum _{m=1}^{\infty }1-O\frac{1}{m^{{\alpha \lambda }_2}}{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q}.\end{array}$$

In particular, for ${\lambda }_1+{\lambda }_2=\lambda$, we have the following: $$\begin{array}{c}\text{(40)}& 0<\sum _{m=1}^{\infty }{m}^{p1-{\alpha \lambda }_1-1}{a}_m^p<\infty ,0<\sum _{n=1}^{\infty }{n}^{q1-{\beta \lambda }_2-1}{b}_n^q<\infty ,\end{array}$$and the following inequality: $$\begin{array}{}\text{(41)}& I_1=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}>\frac{B{\lambda }_1,{\lambda }_2}{{\beta }^{1/p}{\alpha }^{1/q}\lambda }\times {\sum _{m=1}^{\infty }1-O\frac{1}{m^{{\alpha \lambda }_2}}{m}^{p1-{\alpha \lambda }_1-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-{\beta \lambda }_2-1}{b}_n^q}^{1/q}.\end{array}$$

Proof.

In view of the fact that $$\begin{array}{}\text{(42)}& \frac{1}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}=\frac{1}{\Gamma \lambda +1}{\int }_0^{\infty }{t}^{\lambda +1-1}{e}^{-m^{\alpha }+n^{\beta }t}dt,\end{array}$$by using inequality (Equation (35)), it follows that $$\begin{array}{}\text{(43)}& I_1=\frac{1}{\Gamma \lambda +1}\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{a}_m{B}_n{\int }_0^{\infty }{t}^{\lambda +1-1}{e}^{-m^{\alpha }+n^{\beta }t}dt=\frac{1}{\Gamma \lambda +1}{\int }_0^{\infty }{t}^{\lambda -1}\sum _{m=1}^{\infty }{e}^{-m^{\alpha }t}{a}_{m}t\sum _{n=1}^{\infty }{e}^{-n^{\beta }t}{B}_{n}dt\ge \frac{1}{\Gamma \lambda +1}{\int }_0^{\infty }{t}^{\lambda -1}\sum _{m=1}^{\infty }{e}^{-m^{\alpha }t}{a}_m\sum _{n=1}^{\infty }{e}^{-n^{\beta }t}{b}_{n}dt=\frac{1}{\Gamma \lambda +1}\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{a}_m{b}_n{\int }_0^{\infty }{t}^{\lambda -1}{e}^{-m^{\alpha }+n^{\beta }t}dt=\frac{\Gamma \lambda }{\Gamma \lambda +1}\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{b}_n}{{m^{\alpha }+n^{\beta }}^{\lambda }}=\frac{1}{\lambda }I.\end{array}$$

Then by Equation (33), we deduce inequality (Equation (39)). The proof of Theorem 5 is complete.

Remark 6.

For $s=\lambda +1\in 1,6,\lambda \in 0,5,\beta =1,s_1={\widehat{\lambda }}_1\in 0,2/a\cap 0,\lambda +1$, in Equation (32), we have the following: $$\begin{array}{}\text{(44)}& k_{\lambda +1}{\tilde{\lambda }}_{1}1-O\frac{1}{n^{\beta {\tilde{\lambda }}_1}}<{\omega }_1{\tilde{\lambda }}_1,n=n^{\lambda -{\tilde{\lambda }}_1+1}\sum _{m=1}^{\infty }\frac{\alpha m^{\alpha {\tilde{\lambda }}_1-1}}{{m^{\alpha }+n}^{\lambda +1}}<k_{\lambda +1}{\tilde{\lambda }}_1\hspace{1em}n\in N.\end{array}$$

Theorem 7.

Under the assumption (C1), if ${\lambda }_1+{\lambda }_2=\lambda \in 0,5$, then for $\beta =1$, ${\lambda }_1\in 0,2/\alpha \cap 0,\lambda ,{\lambda }_2\in 0,1\cap 0,\lambda$, the constant factor$1/\lambda {1/\beta k_{\lambda }{\lambda }_2}^{1/p}{1/\alpha k_{\lambda }{\lambda }_1}^{1/q}$ in Equation (39) is the best possible.

Proof.

We may divide two cases of ${\lambda }_2\in 0,1\cap 0,\lambda$ and${\lambda }_2=1<\lambda$ to prove that the constant factor $1/{\lambda \alpha }^{1/q}B{\lambda }_1,{\lambda }_2$ in Equation (41) (for $\beta =1$) is the best possible.

Case (i) ${\lambda }_2\in 0,1\cap 0,\lambda$. For any $0<\epsilon <\min p{\lambda }_1,q1-{\lambda }_2$, we set the following: $$\begin{array}{}\text{(45)}& {\tilde{a}}_m≔m^{\alpha {\lambda }_1-\epsilon /p-1},{\tilde{b}}_n≔n^{{\lambda }_2-\epsilon /q-1}\hspace{1em}m,n\in N.\end{array}$$

Since $0<{\lambda }_2-\epsilon /q<1$, by the decreasingness property of series, we have the following: $$\begin{array}{}\text{(46)}& {\tilde{B}}_n≔\sum _{k=1}^n{\tilde{b}}_k=\sum _{k=1}^n{k}^{{\lambda }_2-\epsilon /q-1}<{\int }_0^n{t}^{{\lambda }_2-\epsilon /q-1}dt=\frac{1}{{\lambda }_2-\epsilon /q}{n}^{{\lambda }_2-\epsilon /q}\hspace{1em}n\in N.\end{array}$$

If there exists a constant $M$, $M\ge 1/{\lambda \alpha }^{1/q}B{\lambda }_1,{\lambda }_2$, such that Equation (41) for $\beta =1$ is valid when $1/{\lambda \alpha }^{1/q}B{\lambda }_1,{\lambda }_2$ is replaced by $M$; then, in particular, by substitution of $a_m={\tilde{a}}_m,b_n={\tilde{b}}_n$, and $B_n={\tilde{B}}_n$ in Equation (41) for $\beta =1$, we have the following: $$\begin{array}{}\text{(47)}& {\tilde{I}}_1≔\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{1}{{m^{\alpha }+n}^{\lambda +1}}{\tilde{a}}_m{\tilde{B}}_n>M{\sum _{m=1}^{\infty }1-O\frac{1}{m^{{\alpha \lambda }_2}}{m}^{p1-{\alpha \lambda }_1-1}{\tilde{a}}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-{\lambda }_2-1}{\tilde{b}}_n^q}^{1/q}.\end{array}$$

By using inequality (Equation (47)) and the decreasingness property of series, we obtain the following: $$\begin{array}{c}\text{(48)}& {\tilde{I}}_1>M{\sum _{m=1}^{\infty }1-O\frac{1}{m^{\alpha {\lambda }_2}}{m}^{p1-\alpha {\lambda }_1-1}{m}^{p\alpha {\lambda }_1-\alpha \epsilon -p}}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-{\lambda }_2-1}{n}^{q{\lambda }_2-\epsilon -q}}^{1/q}=M{\sum _{m=1}^{\infty }{m}^{-\alpha \epsilon -1}-\sum _{m=1}^{\infty }O\frac{1}{m^{\alpha {\lambda }_2+\epsilon +1}}}^{\frac{1}{p}}{1+\sum _{n=2}^{\infty }{n}^{-\epsilon -1}}^{\frac{1}{q}}\\ >M{{\int }_1^{\infty }{x}^{-\alpha \epsilon -1}dx-O1}^{\frac{1}{p}}{1+{\int }_1^{\infty }{y}^{-\epsilon -1}dy}^{\frac{1}{q}}\\ =\frac{M}{\epsilon }{\frac{1}{\alpha }-\epsilon O1}^{\frac{1}{p}}{\epsilon +1}^{\frac{1}{q}}.\end{array}$$

By utilizing inequality (Equation (44)) and setting ${\widehat{\lambda }}_1={\lambda }_1-\epsilon /p\in 0,2/a\cap 0,\lambda ,{\widehat{\lambda }}_2=\lambda -{\widehat{\lambda }}_1$, we find the following: $$\begin{array}{}\text{(49)}& {\tilde{I}}_2<\frac{1}{\alpha {\lambda }_2-\epsilon /q}\sum _{n=1}^{\infty }{n}^{{\lambda }_2+\epsilon /p+1}\sum _{m=1}^{\infty }\frac{\alpha }{{m+n}^{\lambda +1}}{m}^{\alpha {\lambda }_1-\epsilon /p-1}{n}^{-\epsilon -1}=\frac{1}{\alpha {\lambda }_2-\epsilon /q}\sum _{n=1}^{\infty }{\omega }_1{\tilde{\lambda }}_1,n{n}^{-\epsilon -1}<\frac{1}{\alpha {\lambda }_2-\epsilon /q}{k}_{\lambda +1}{\tilde{\lambda }}_{1}1+\sum _{n=2}^{\infty }{n}^{-\epsilon -1}<\frac{\Gamma {\tilde{\lambda }}_1\Gamma {\tilde{\lambda }}_2+1}{\alpha {\lambda }_2-\epsilon /q\Gamma \lambda +1}1+{\int }_1^{\infty }{y}^{-\epsilon -1}dy=\frac{{\tilde{\lambda }}_2\Gamma {\tilde{\lambda }}_1\Gamma {\tilde{\lambda }}_2}{\epsilon \alpha {\lambda }_2-\epsilon /q\lambda \Gamma \lambda }\epsilon +1.\end{array}$$

Then, we have $$\begin{array}{}\text{(50)}& \frac{{\tilde{\lambda }}_2\Gamma {\tilde{\lambda }}_1\Gamma {\tilde{\lambda }}_2}{\alpha {\lambda }_2-\epsilon /q\lambda \Gamma \lambda }\epsilon +1>\epsilon \tilde{I}>M{1/\alpha -\epsilon O1}^{1/p}{\epsilon +1}^{1/q}.\end{array}$$

Letting $\epsilon ⟶0^{+}$ along with the continuity of the gamma function, we deduce that $1/{\lambda \alpha }^{1/q}B{\lambda }_1,{\lambda }_2\ge M$. Hence, $M=1/{\lambda \alpha }^{1/q}B{\lambda }_1,{\lambda }_2$ is the best possible constant factor of Equation (41) (for $\beta =1$).

Case (ii) ${\lambda }_2=1<\lambda$. For $0<\epsilon <1$, replacing $\lambda$ by $\lambda -\epsilon$ in Equation (41) and taking ${\lambda }_1=\lambda -1,{\lambda }_2=1-\epsilon$, and $\beta =1$, we obtain the following: $$\begin{array}{}\text{(51)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n}^{\lambda -\epsilon +1}}>\frac{B\lambda -1,1-\epsilon }{{\alpha }^{1/q}\lambda -\epsilon }{K}_{p,\epsilon }{\sum _{n=1}^{\infty }{n}^{q\epsilon -1}{b}_n^q}^{1/q},\end{array}$$where $K_{p,\epsilon }≔{{\sum }_{m=1}^{\infty }1-O1/m^{\alpha 1-\epsilon }{m}^{q1-\alpha \lambda -1-1}{a}_m^p}^{1/p}$.

If ${\lambda }_1=1$, then we have the following: $$\begin{array}{c}\text{(52)}& n^{q\epsilon -1}\le n^{-1}=n^{q1-{\lambda }_2-1},\sum _{n=1}^{\infty }{n}^{q\epsilon -1}{b}_n^q\le \sum _{n=1}^{\infty }{n}^{q1-{\lambda }_2-1}{b}_n^q<\infty .\end{array}$$It follows that $\underset{\epsilon ⟶0^{+}}{\lim }{\sum }_{n=1}^{\infty }{n}^{q\epsilon -1}{b}_n^q={\sum }_{n=1}^{\infty }{n}^{q1-1-1}{b}_n^q,$, and similarly, $\underset{\epsilon ⟶0^{+}}{\lim }{K}_{p,\epsilon }=K_{p,0}$.

If there exists a constant factor $M$, $M\ge 1/{\alpha }^{1/q}\lambda B\lambda -1,1=1/{\alpha }^{1/q}\lambda \lambda -1$, such that (Equation (41)) for $\beta =1,{\lambda }_2=1,{\lambda }_1=\lambda -1$ is valid when we replace $1/{\alpha }^{1/q}\lambda \lambda -1$ by $M$, namely, $$\begin{array}{}\text{(53)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n}^{\lambda +1}}>M\cdot K_{p,0}{\sum _{n=1}^{\infty }{n}^{-1}{b}_n^q}^{1/q},\end{array}$$

then, by applying Fatou’s lemma [10] and Equation (53), we deduce that $$\begin{array}{}\text{(54)}& \underset{\epsilon ⟶0^{+}}{\lim }\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n/{m^{\alpha }+n}^{\lambda -\epsilon +1}}{K_{p,\epsilon }}{\sum _{n=1}^{\infty }{n}^{q\epsilon -1}{b}_n^q}^{-1/q}\ge \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n/{m^{\alpha }+n}^{\lambda +1}}{K_{p,0}}{\sum _{n=1}^{\infty }{n}^{-1}{b}_n^q}^{-1/q}>M.\end{array}$$

By the property of limitation, there exists a constant ${\delta }_0\in 0,1,$ such that for any $\delta \in 0,{\delta }_0$, $$\begin{array}{}\text{(55)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n/{m^{\alpha }+n}^{\lambda -\delta +1}}{K_{q,\delta }}{\sum _{n=1}^{\infty }{n}^{q1-1-\delta -1}{b}_n^q}^{-1/q}>M,\end{array}$$namely, $$\begin{array}{}\text{(56)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n}^{\lambda -\delta +1}}>M\cdot K_{p,\delta }{\sum _{n=1}^{\infty }{n}^{q1-1-\delta -1}{b}_n^q}^{1/q}.\end{array}$$

By Case (i), if the constant factor $1/{\alpha }^{1/q}\lambda -\delta B\lambda -1,1-\delta$ (Equation (51)) $\epsilon =\delta$ is the best possible, we have $1/{\alpha }^{1/q}\lambda -\delta B\lambda -1,1-\delta \ge M$. Then, for $\delta ⟶0^{+}$, we have the following: $$\begin{array}{}\text{(57)}& \frac{1}{{\alpha }^{1/q}\lambda \lambda -1}=\frac{1}{{\alpha }^{1/q}\lambda }B\lambda -1,1\ge M,\end{array}$$which implies that $M=1/{\alpha }^{1/q}\lambda \lambda -1$ is the best possible factor of Equation (41) (for $\beta ={\lambda }_1=1,{\lambda }_2=\lambda -1$). This completes the proof of Theorem 7.

Theorem 8.

Under the assumption (C1), if the constant factor $1/\lambda {1/\beta k_{\lambda }{\lambda }_2}^{1/p}{1/\alpha k_{\lambda }{\lambda }_1}^{1/q}$ in Equation (39) is the best possible, then for $$\begin{array}{}\text{(58)}& \lambda -{\lambda }_1-{\lambda }_2\in -{\lambda }_{1}p,\frac{2}{\alpha }-{\lambda }_{1}p\cap \lambda -{\lambda }_1-\frac{2}{\beta }p,\lambda -{\lambda }_{1}p\supset 0,\end{array}$$we have ${\lambda }_1+{\lambda }_2=\lambda .$.

Proof.

For ${\widehat{\lambda }}_1=\lambda -{\lambda }_2/p+{\lambda }_1/q,{\widehat{\lambda }}_2=\lambda -{\lambda }_1/q+{\lambda }_2/p$, we have the following: $$\begin{array}{}\text{(59)}& {\widehat{\lambda }}_1+{\widehat{\lambda }}_2=\frac{\lambda -{\lambda }_2}{p}+\frac{{\lambda }_1}{q}+\frac{\lambda -{\lambda }_1}{q}+\frac{{\lambda }_2}{p}=\lambda .\end{array}$$

If $\lambda -{\lambda }_1-{\lambda }_2\in -{\lambda }_{1}p,2/\alpha -{\lambda }_{1}p\supset 0$, then $0<{\widehat{\lambda }}_1=\lambda -{\lambda }_2/p+{\lambda }_1/q\le 2/\alpha$; if $$\begin{array}{}\text{(60)}& \lambda -{\lambda }_1-{\lambda }_2\in \lambda -{\lambda }_1-\frac{2}{\beta }p,\lambda -{\lambda }_{1}p\supset 0,\end{array}$$then $0<{\widehat{\lambda }}_2\le 2/\beta$.

It is easy to observe that ${\widehat{\lambda }}_i<\lambda i=1,2$, and then, by Equation (58), it follows that ${\widehat{\lambda }}_1\in 0,2/\alpha \cap 0,\lambda$, ${\widehat{\lambda }}_2\in 0,2/\beta \cap 0,\lambda$, and $1/\lambda B{\widehat{\lambda }}_1,{\widehat{\lambda }}_2\in {\text{R}}_{+}=0,\infty$. By Equation (41), we have the following: $$\begin{array}{}\text{(61)}& I_1=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}>\frac{B{\widehat{\lambda }}_1,{\widehat{\lambda }}_2}{{\beta }^{1/p}{\alpha }^{1/q}\lambda }\times {\sum _{m=1}^{\infty }1-O\frac{1}{m^{\alpha {\stackrel{\wedge }{\lambda }}_2}}{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q}.\end{array}$$

If the constant factor $1/\lambda {1/\beta k_{\lambda }{\lambda }_2}^{1/p}{1/\alpha k_{\lambda }{\lambda }_1}^{1/q}$ in Equation (39) is the best possible, then by Equation (61), we have the inequality: $$\begin{array}{}\text{(62)}& \frac{1}{\lambda }{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}\ge \frac{B{\widehat{\lambda }}_1,{\widehat{\lambda }}_2}{{\beta }^{1/p}{\alpha }^{1/q}\lambda },\end{array}$$that is $$\begin{array}{}\text{(63)}& {k_{\lambda }{\lambda }_2}^{1/p}{k_{\lambda }{\lambda }_1}^{1/q}\ge B{\widehat{\lambda }}_1,{\widehat{\lambda }}_2.\end{array}$$

By the reverse Hӧlder’s inequality [9], we obtain the following: $$\begin{array}{}\text{(64)}& B{\widehat{\lambda }}_1,{\widehat{\lambda }}_2=k_{\lambda }{\widehat{\lambda }}_1=k_{\lambda }\frac{\lambda -{\lambda }_2}{p}+\frac{{\lambda }_1}{q}={\int }_0^{\infty }\frac{1}{{1+u}^{\lambda }}{u}^{\lambda -{\lambda }_2/p+{\lambda }_1/q-1}du={\int }_0^{\infty }\frac{1}{{1+u}^{\lambda }}{u}^{\lambda -{\lambda }_2-1/p}{u}^{{\lambda }_1-1/q}du\ge {{\int }_0^{\infty }\frac{1}{{1+u}^{\lambda }}{u}^{\lambda -{\lambda }_2-1}du}^{1/p}{{\int }_0^{\infty }\frac{1}{{1+u}^{\lambda }}{u}^{{\lambda }_1-1}du}^{1/q}={{\int }_0^{\infty }\frac{1}{{1+v}^{\lambda }}{v}^{{\lambda }_2-1}dv}^{1/p}{{\int }_0^{\infty }\frac{1}{{1+u}^{\lambda }}{u}^{{\lambda }_1-1}du}^{1/q}={k_{\lambda }{\lambda }_2}^{1/p}{k_{\lambda }{\lambda }_1}^{1/q}.\end{array}$$

In light of Equation (63), we deduce that Equation (64) keeps the form of equality. Thus, there exist constants $A$ and $B$, such that they are not both zero and they satisfy ${Au}^{\lambda -{\lambda }_2-1}={Bu}^{{\lambda }_1-1}a.e.$ in $R_{+}$ (see [9]). Assuming that $A\ne 0$, we have $u^{\lambda -{\lambda }_2-{\lambda }_1}=B/A$a.e. in $R_{+}$, and hence, $\lambda -{\lambda }_2-{\lambda }_1=0$, namely, ${\lambda }_1+{\lambda }_2=\lambda$. Theorem 8 is proved.

4. Equivalent Forms and Some Particular Inequalities

Theorem 9.

Under the assumption (C1), we have the following inequality which is equivalent to Equation (39): $$\begin{array}{}\text{(65)}& J_1≔{\sum _{m=1}^{\infty }\frac{m^{q\alpha {\stackrel{\wedge }{\lambda }}_1-1}}{{1-O1/m^{{\alpha \lambda }_2}}^{q-1}}{\sum _{n=1}^{\infty }\frac{B_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}}^q}^{1/q}>\frac{1}{\lambda }{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}{\sum _{n=1}^{\infty }{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q}.\end{array}$$

In particular, for ${\lambda }_1+{\lambda }_2=\lambda$, we have the following: $$\begin{array}{c}\text{(66)}& 0<\sum _{m=1}^{\infty }{m}^{p1-{\alpha \lambda }_1-1}{a}_m^p<\infty ,0<\sum _{n=1}^{\infty }{n}^{q1-{\beta \lambda }_2-1}{b}_n^q<\infty ,\end{array}$$and the following inequality which is equivalent to Equation (41): $$\begin{array}{}\text{(67)}& {\sum _{m=1}^{\infty }\frac{m^{q{\alpha \lambda }_1-1}}{{1-O1/m^{{\alpha \lambda }_2}}^{q-1}}{\sum _{n=1}^{\infty }\frac{B_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}}^q}^{1/q}>\frac{1}{{\alpha }^{1/q}\lambda }B{\lambda }_1,{\lambda }_2{\sum _{n=1}^{\infty }{n}^{q1-{\beta \lambda }_2-1}{b}_n^q}^{1/q}.\end{array}$$

Proof.

Suppose that Equation (65) is valid. By the reverse Hӧlder’s inequality [9], we have the following: $$\begin{array}{}\text{(68)}& I_1=\sum _{m=1}^{\infty }{1-O\frac{1}{m^{{\alpha \lambda }_2}}}^{1/p}{m}^{1/q-\alpha {\stackrel{\wedge }{\lambda }}_1}{a}_m\frac{m^{-1/q+\alpha {\stackrel{\wedge }{\lambda }}_1}}{{1-O1/m^{{\alpha \lambda }_2}}^{1/p}}\sum _{n=1}^{\infty }\frac{B_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}\ge {\sum _{m=1}^{\infty }{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p}^{1/p}{J}_1.\end{array}$$

Then by Equation (65), we obtain Equation (39). On the other hand, assuming that Equation (39) is valid, we set the following: $$\begin{array}{}\text{(69)}& a_m≔\frac{m^{q\alpha {\stackrel{\wedge }{\lambda }}_1-1}}{{1-O1/m^{{\alpha \lambda }_2}}^{q-1}}{\sum _{n=1}^{\infty }\frac{B_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}}^{q-1},\hspace{1em}m\in N.\end{array}$$

Then, it follows that $J_1={{\sum }_{m=1}^{\infty }1-O1/m^{{\alpha \lambda }_2}{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p}^{1/q}$.

If $J_1=\infty$, then Equation (65) is naturally valid; if $J_1=0$, then it is impossible that makes Equation (65) valid, namely, $J_1>0$. Suppose that $0<J_1<\infty$. By Equation (39), we have the following: $$\begin{array}{c}\text{(70)}& \sum _{m=1}^{\infty }1-O\frac{1}{m^{{\alpha \lambda }_2}}{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p={J_1}^q=I_1>\frac{1}{\lambda }{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}{J_1}^{q-1}{\sum _{n=1}^{\infty }{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q},J_1>\frac{1}{\lambda }{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}{\sum _{n=1}^{\infty }{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q},\end{array}$$namely, Equation (65) follows, which is equivalent to Equation (39). Theorem 9 is proved.

Remark 10.

By the same way as above, under the assumption (C1), if $p<0,0<q<1$, we can obtain the following equivalent inequalities: $$\begin{array}{}\text{(71)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}>\frac{1}{\lambda }{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}\times {\sum _{m=1}^{\infty }{m}^{p1-\alpha {\stackrel{\wedge }{\lambda }}_1-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }1-O\frac{1}{n^{{\beta \lambda }_1}}{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q},\text{(72)}& {\sum _{m=1}^{\infty }{m}^{q\alpha {\stackrel{\wedge }{\lambda }}_1-1}{\sum _{n=1}^{\infty }\frac{B_n}{{m^{\alpha }+n^{\beta }}^{\lambda +1}}}^q}^{1/q}>\frac{1}{\lambda }{\frac{1}{\beta }{k}_{\lambda }{\lambda }_2}^{1/p}{\frac{1}{\alpha }{k}_{\lambda }{\lambda }_1}^{1/q}{\sum _{n=1}^{\infty }1-O\frac{1}{n^{{\beta \lambda }_1}}{n}^{q1-\beta {\stackrel{\wedge }{\lambda }}_2-1}{b}_n^q}^{1/q}.\end{array}$$

Theorem 11.

Under the assumption (C1), if ${\lambda }_1+{\lambda }_2=\lambda \in 0,5$, then for $\beta =1$, ${\lambda }_1\in 0,2/\alpha \cap 0,\lambda$, and ${\lambda }_2\in 0,1\cap 0,\lambda$, the constant factor $1/\lambda {1/\beta k_{\lambda }{\lambda }_2}^{1/p}{1/\alpha k_{\lambda }{\lambda }_1}^{1/q}$ in Equation (65) is the best possible. On the other hand, if the same constant factor in Equation (65) is the best possible, then for $$\begin{array}{}\text{(73)}& \lambda -{\lambda }_1-{\lambda }_2\in -{\lambda }_{1}p,\frac{2}{\alpha }-{\lambda }_{1}p\cap \lambda -{\lambda }_1-\frac{2}{\beta }p,\lambda -{\lambda }_{1}p\supset 0,\end{array}$$we have ${\lambda }_1+{\lambda }_2=\lambda .$.

Proof.

If ${\lambda }_1+{\lambda }_2=\lambda$, then by Theorem 7, the constant factor $1/{\alpha }^{1/q}\lambda B{\lambda }_1,{\lambda }_2$ in Equation (41) (for $\beta =1$) is the best possible. Then by Equation (68) (for ${\lambda }_1+{\lambda }_2=\lambda ,\beta =1$), the constant factor in Equation (65) is still the best possible.

On the other hand, if the same constant factor, $1/\lambda {1/\beta k_{\lambda }{\lambda }_2}^{1/p}{1/\alpha k_{\lambda }{\lambda }_1}^{1/q}$, in Equation (65) is the best possible, then by the equivalency of Equations (65) and (39), in view of $I=J^q$ (in the proof of Theorem 9), it follows that the same constant factor in Equation (39) is the best possible. By using Theorem 7, we obtain ${\lambda }_1+{\lambda }_2=\lambda$. The proof of Theorem 11 is complete.

Below, we illustrate that some reverse Hardy-Hilbert type inequalities can be derived from the special cases of our main results stated in Theorems 5 and 9.

Remark 12.

(i) Choosing $\beta =1,\lambda =1,{\lambda }_1=1/r,{\lambda }_2=1/sr>1,1/r+1/s=1$ in Equations (41) and (67), respectively, we acquire the following inequalities with the best possible constant factor $\pi /{\alpha }^{1/q}\sin \pi /r$: $$\begin{array}{c}\text{(74)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n}^2}>\frac{\pi }{{\alpha }^{1/q}\sin \pi /r}{\sum _{m=1}^{\infty }1-O\frac{1}{m^{\alpha /s}}{m}^{p1-\alpha /r-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q/r-1}{b}_n^q}^{1/q},{\sum _{m=1}^{\infty }\frac{m^{q\alpha /r-1}}{{1-O1/m^{\alpha /s}}^{q-1}}{\sum _{n=1}^{\infty }\frac{B_n}{{m^{\alpha }+n}^2}}^q}^{1/q}>\frac{\pi }{{\alpha }^{1/q}\sin \pi /r}{\sum _{n=1}^{\infty }{n}^{q/r-1}{b}_n^q}^{1/q}.\end{array}$$

In particular, for $r=s=2$, we deduce from Equation (68) and Equation (71) that $$\begin{array}{c}\text{(75)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n}^2}>\frac{\pi }{{\alpha }^{1/q}}{\sum _{m=1}^{\infty }1-O\frac{1}{m^{\alpha /2}}{m}^{p1-\alpha /2-1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{q/2-1}{b}_n^q}^{1/q},{\sum _{m=1}^{\infty }\frac{m^{q\alpha /2-1}}{{1-O1/m^{\alpha /2}}^{q-1}}{\sum _{n=1}^{\infty }\frac{B_n}{{m^{\alpha }+n}^2}}^q}^{1/q}>\frac{\pi }{{\alpha }^{1/q}}{\sum _{n=1}^{\infty }{n}^{q/2-1}{b}_n^q}^{1/q}.\end{array}$$

(ii) Putting $\beta =1,\lambda =3,{\lambda }_1=2,{\lambda }_2=1$ in Equations (41) and (67), respectively, we obtain the following inequalities with the best possible constant factor $1/6{\alpha }^{1/q}$$$\begin{array}{c}\text{(76)}& \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_m{B}_n}{{m^{\alpha }+n}^4}>\frac{1}{6{\alpha }^{1/q}}{\sum _{m=1}^{\infty }1-O\frac{1}{m^{\alpha }}{m}^{p1-2\alpha -1}{a}_m^p}^{1/p}{\sum _{n=1}^{\infty }{n}^{-1}{b}_n^q}^{1/q},{\sum _{m=1}^{\infty }\frac{m^{2q\alpha -1}}{{1-O1/m^{\alpha }}^{q-1}}{\sum _{n=1}^{\infty }\frac{B_n}{{m^{\alpha }+n}^4}}^q}^{1/q}>\frac{1}{6{\alpha }^{1/q}}{\sum _{n=1}^{\infty }{n}^{-1}{b}_n^q}^{1/q}.\end{array}$$

Authors’ Contributions

BY carried out the mathematical studies and drafted the manuscript. SW and XH participated in the design of the study and performed the numerical analysis. All authors contributed equally in the preparation of this paper. All authors read and approved the final manuscript.

Acknowledgments

This work is supported by the Natural Science Foundation of Fujian Province of China (No. 2020J01365).

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