A New Reverse Extended Hardy–Hilbert’s Inequality with Two Partial Sums and Parameters

Abstract

By using the methods of real analysis and the mid-value theorem, we introduce some lemmas and obtain a new reverse extended Hardy–Hilbert’s inequality with two partial sums and multi-parameters. We also give a few equivalent conditions of the best possible constant factor related to several parameters in the new inequality. Some particular inequalities are deduced.

1. Introduction

If $p>1, \frac{1}{p}+\frac{1}{q}=1, a_m, b_n\ge 0, 0<{\displaystyle {\sum }_{m=1}^{\infty }{a}_m^p}<\infty$ and $0<{\displaystyle {\sum }_{n=1}^{\infty }{b}_n^q}<\infty$, then we have the following well-known Hardy–Hilbert’s inequality (cf. [1], Theorem 315):

$${\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{a_m{b}_n}{m+n}}}<\frac{\pi }{\sin (\pi /p)}{\left({\displaystyle \sum _{m=1}^{\infty }{a}_m^p}\right)}^{\frac{1}{p}}{\left({\displaystyle \sum _{n=1}^{\infty }{b}_n^q}\right)}^{\frac{1}{q}},$$

(1)

where the constant factor $\frac{\pi }{\sin (\pi /p)}$ is best possible.

In 2006, Krnic et al. [2] obtained the following inequality, which is a generalization of (1): for ${\lambda }_i\in (0,2](i=1,2), {\lambda }_1+{\lambda }_2=\lambda \in (0,4],$

$${\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{a_m{b}_n}{{(m+n)}^{\lambda }}}}<B({\lambda }_1,{\lambda }_2){\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1-{\lambda }_1)-1}{a}_m^p}\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-{\lambda }_2)-1}{b}_n^q}\right]}^{\frac{1}{q}},$$

(2)

with the best possible constant factor $B({\lambda }_1,{\lambda }_2)$, in which

$$B(u,v)={\displaystyle {\int }_0^{\infty }\frac{t^{u-1}}{{(1+t)}^{u+v}}} dt(u,v>0)$$

(3)

is the beta function. In particular, for $p=q=2,$ ${\lambda }_1={\lambda }_2=\frac{\lambda }{2},$ (2) deduces to Yang’s inequality in [3]. In 2019, following the way of (2), by using Abel’s summation by parts formula, Adiyasuren et al. [4] provided the following extension of (2) involving two partial sums and some parameters, for

$${\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{a_m{b}_n}{{(m+n)}^{\lambda }}}}<{\lambda }_1{\lambda }_{2}B({\lambda }_1,{\lambda }_2){\left({\displaystyle \sum _{m=1}^{\infty }{m}^{-p{\lambda }_1-1}{A}_m^p}\right)}^{\frac{1}{p}}{\left({\displaystyle \sum _{n=1}^{\infty }{n}^{-q{\lambda }_2-1}{B}_n^q}\right)}^{\frac{1}{q}},$$

(4)

where ${\lambda }_1{\lambda }_{2}B({\lambda }_1,{\lambda }_2)$ is the best possible constant factor, and $A_m:={\displaystyle {\sum }_{i=1}^m{a}_i}$ and $B_n:={\displaystyle {\sum }_{k=1}^n{b}_k}$ $(m,n\in \mathrm{N}=\{1,2,\cdots \left\}\right)$, satisfy the following inequalities:

$$0<{\displaystyle \sum _{m=1}^{\infty }{m}^{-p{\lambda }_1-1}{A}_m^p}<\infty and 0<{\displaystyle \sum _{n=1}^{\infty }{n}^{-q{\lambda }_2-1}{B}_n^q}<\infty .$$

Both (1) and (2) with their integral analogues played an important role in real analysis, in which some generalizations of (1) are given and a relation between (1) and the other Hilbert-type inequality is obtained (cf. [5,6,7,8,9,10,11,12,13,14,15,16]). In 2021, Gu et. al. [17] provided a generalization of (4) with $\frac{1}{{(m^{\alpha }+n^{\beta })}^{\lambda }}(\alpha ,\beta \in (0,1])$ as the kernel of inequality. In 2016, by using the weight coefficients and the techniques of real analysis, Hong et al. [18] considered a few equivalent statements of the generalization of (1) with the best possible constant factor related to multi-parameters. The other further results were obtained by [19,20,21,22,23,24,25,26,27,28,29].

In this article, based on the idea of [17,18], by using the techniques of real analysis and the mid-value theorem, we introduce some preserving lemmas and then give a reverse of (2) with two partial sums and multi-parameters, which is a new reverse version of the inequality in [16]. We also consider a few equivalent conditions of the best possible constant factor in the reverse inequality related to multi-parameters. Furthermore, several inequalities are deduced by setting some particular parameters.

2. Some Lemmas

In what follows, we assume that $0<p<1 (q<0), \frac{1}{p}+\frac{1}{q}=1,$ $\lambda \in (0,6],$ $\alpha ,\beta \in (0,1],$ ${\lambda }_1\in (0,\frac{2}{\alpha }]\cap (0,\lambda ), {\lambda }_2\in (0,\frac{2}{\beta }]\cap (0,\lambda ),$ ${\widehat{\lambda }}_1:=\frac{\lambda -{\lambda }_2}{p}+\frac{{\lambda }_1}{q},$ ${\widehat{\lambda }}_2:=\frac{\lambda -{\lambda }_1}{q}+\frac{{\lambda }_2}{p}$. We also assume that $a_m,b_n\ge 0,$ $A_m:={\displaystyle {\sum }_{j=1}^m{a}_j},$ $B_n:={\displaystyle {\sum }_{k=1}^n{b}_k}$ $(m,n\in \mathrm{N}),$ satisfy $A_m=o(e^{t{m}^{\alpha }})$, $B_n=o(e^{t{n}^{\beta }})$ $(t>0;m,n\to \infty ),$ and the following inequalities:

$$0<{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1-\alpha {\widehat{\lambda }}_1)-1}}{a}_m^p<\infty ,0<{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-\beta {\widehat{\lambda }}_2)-1}}{b}_n^q<\infty .$$

(5)

Lemma 1.

For $t>0$, the following inequalities on the partial sums are valid:

$${\displaystyle \sum _{m=1}^{\infty }{e}^{-t{m}^{\alpha }}}{m}^{\alpha -1}{A}_m\ge \frac{1}{t\alpha }{\displaystyle \sum _{m=1}^{\infty }{e}^{-t{m}^{\alpha }}}{a}_m,$$

(6)

$${\displaystyle \sum _{n=1}^{\infty }{e}^{-t{n}^{\beta }}}{n}^{\beta -1}{B}_n\ge \frac{1}{t\beta }{\displaystyle \sum _{n=1}^{\infty }{e}^{-t{n}^{\beta }}}{b}_n.$$

(7)

Proof. 

Since $A_m{e}^{-t{m}^{\alpha }}=o(1)(m\to \infty )$, by Abel’s summation by parts formula, it follows that

$$\begin{array}{ll}{\displaystyle \sum _{m=1}^{\infty }{e}^{-t{m}^{\alpha }}}{a}_m& =\underset{m\to \infty }{\lim }{A}_m{e}^{-t{m}^{\alpha }}+{\displaystyle \sum _{m=1}^{\infty }{A}_m}\left[e^{-t{m}^{\alpha }}-e^{-t{(m+1)}^{\alpha }}\right]\\ & ={\displaystyle \sum _{m=1}^{\infty }{A}_m}\left[e^{-t{m}^{\alpha }}-e^{-t{(m+1)}^{\alpha }}\right].\end{array}$$

We set function $g(x)=e^{-t{x}^{\alpha }},x\in [m,m+1].$ Then, we find $g^{\prime }(x)=-t\alpha x^{\alpha -1}{e}^{-t{x}^{\alpha }},$ and for $\alpha \in (0,1], h(x):=x^{\alpha -1}{e}^{-t{x}^{\alpha }}$ is decreasing in $[m,m+1].$ By the mid-value theorem, we have

$$\begin{array}{ll}{\displaystyle \sum _{m=1}^{\infty }{e}^{-t{m}^{\alpha }}}{a}_m& =-{\displaystyle \sum _{m=1}^{\infty }{A}_m}\left(g(m+1)-g(m)\right)\\ & =-{\displaystyle \sum _{m=1}^{\infty }{A}_m}g\prime (m+\theta )=t\alpha {\displaystyle \sum _{m=1}^{\infty }(}m+\theta {)}^{\alpha -1}{e}^{-t{(m+\theta )}^{\alpha }}{A}_m\\ & \le t\alpha {\displaystyle \sum _{m=1}^{\infty }{m}^{\alpha -1}}{e}^{-t{m}^{\alpha }}{A}_m(\theta \in (0,1)).\end{array}$$

Hence, we have (6). In the same way, inequality (7) follows.

The lemma is proved. □

In the following lemma, for estimating the weight coefficient in Lemma 3, we introduce some results related to the Bernoulli functions and the related formulas.

Lemma 2.

(Ref. [5]’s section 2.2.3, [30]) (i) If ${(-1)}^i\frac{d^i}{d{t}^i}g(t)>0,$ $t\in [m,\infty )(m\in \mathrm{N})$ with $g^{(i)}(\infty )=0$ $(i=0,1,2,3)$, $P_i(t), B_i (i\in \mathrm{N})$ are Bernoulli functions and Bernoulli numbers of i-order, then

$${\displaystyle {\int }_m^{\infty }{P}_{2k-1}}(t)g(t)dt=-{\epsilon }_k\frac{B_{2k}}{2k}g(m)(0<{\epsilon }_k<1;k=1,2,\cdots ).$$

(8)

In particular, for $k=1,$ since $B_2=\frac{1}{6}$, we find

$$-\frac{1}{12}g(m)<{\displaystyle {\int }_m^{\infty }{P}_1}(t)g(t)dt<0;$$

(9)

For $k=2,$ based on $B_4=-\frac{1}{30}$, it follows that

$$0<{\displaystyle {\int }_m^{\infty }{P}_3}(t)g(t)dt<\frac{1}{120}g(m).$$

(10)

(ii) (Ref. [5]’s section 2.2.3, [30]) Suppose that$f(t)(>0)\in C^3[m,\infty ), f^{(i)}(\infty )=0$ $(i=0,1,2,3)$. We have the following Euler–Maclaurin summation formula:

$${\displaystyle \sum _{k=m}^{\infty }f(k)={\displaystyle {\int }_m^{\infty }f(t)dt+\frac{1}{2}}}f(m)+{\displaystyle {\int }_m^{\infty }{P}_1}(t)f^{\prime }(t)dt,$$

(11)

$${\displaystyle {\int }_m^{\infty }{P}_1}(t)f^{\prime }(t)dt=-\frac{1}{12}{f}^{\prime }(m)+\frac{1}{6}{\displaystyle {\int }_m^{\infty }{P}_3}(t)f^{‴}(t)dt.$$

(12)

Lemma 3.

For $s\in (0,6],$ $s_2\in (0,\frac{2}{\beta }]\cap (0,s),$ $k_s(s_2):=B(s_2,s-s_2)$, the weight coefficient is defined as follows:

$${\varpi }_s(s_2,m):=m^{\alpha (s-s_2)}{\displaystyle \sum _{n=1}^{\infty }\frac{\beta n^{\beta s_2-1}}{{(m^{\alpha }+n^{\beta })}^s}}(m\in \mathrm{N}).$$

(13)

Then, the following inequalities are valid: 

$$0<k_s(s_2)(1-O(\frac{1}{m^{\alpha s_2}}))<{\varpi }_s(s_2,m)<k_s(s_2)(m\in \mathrm{N}).$$

(14)

where we indicate that $O(\frac{1}{m^{\alpha s_2}}):=\frac{1}{k_s(s_2)}{\displaystyle {\int }_0^{\frac{1}{m^{\alpha }}}\frac{u^{s_2-1}}{{(1+u)}^s}du>0}.$

Proof. 

For any $m\in \mathrm{N}$, the function $g(m,t)$ is defined by: $g(m,t):=\frac{\beta t^{\beta s_2-1}}{{(m^{\alpha }+t^{\beta })}^s}(t>0).$ In view of (11), we have

$$\begin{array}{ll}{\displaystyle \sum _{n=1}^{\infty }}g(m,n)& ={\displaystyle {\int }_1^{\infty }g(m,t)dt+\frac{1}{2}g(m,1)}+{\displaystyle {\int }_1^{\infty }{P}_1(t)g^{\prime }(m,t)dt}\\ & ={\displaystyle {\int }_0^{\infty }g(m,t)dt-h(m),}\end{array}$$

where we set $h(m):={\displaystyle {\int }_0^{1}g(m,t)dt-}\frac{1}{2}g(m,1)-{\displaystyle {\int }_1^{\infty }{P}_1(t)g^{\prime }(m,t)dt}$.

We obtain $-\frac{1}{2}g(m,1)=\frac{-\beta }{2{(m^{\alpha }+1)}^s}$. By integration by parts, it follows that

$$\begin{array}{l}{\displaystyle {\int }_0^{1}g(m,t)dt}=\beta {\displaystyle {\int }_0^1\frac{t^{\beta s_2-1}}{{(m^{\alpha }+t^{\beta })}^s}dt\stackrel{u=t^{\beta }}{=}}{\displaystyle {\int }_0^1\frac{u^{s_2-1}}{{(m^{\alpha }+u)}^s}}du\\ =\frac{1}{s_2}{\displaystyle {\int }_0^1\frac{d{u}^{s_2}}{{(m^{\alpha }+u)}^s}}=\frac{1}{s_2}\frac{u^{s_2}}{{(m^{\alpha }+u)}^s}{|}_0^1+\frac{s}{s_2}{\displaystyle {\int }_0^1\frac{u^{s_2}}{{(m^{\alpha }+u)}^{s+1}}}du\\ =\frac{1}{s_2}\frac{1}{{(m^{\alpha }+1)}^s}+\frac{s}{s_2(s_2+1)}{\displaystyle {\int }_0^1\frac{d{u}^{s_2+1}}{{(m^{\alpha }+u)}^{s+1}}}\\ >\frac{1}{s_2}\frac{1}{{(m^{\alpha }+1)}^s}+\frac{s}{s_2(s_2+1)}{\left[\frac{u^{s_2+1}}{{(m^{\alpha }+u)}^{s+1}}\right]}_0^1+\frac{s(s+1)}{s_2(s_2+1){(m^{\alpha }+1)}^{s+2}}{\displaystyle {\int }_0^1{u}^{s_2+1}}du\\ =\frac{1}{s_2}\frac{1}{{(m^{\alpha }+1)}^s}+\frac{\lambda }{s_2(s_2+1)}\frac{1}{{(m^{\alpha }+1)}^{s+1}}+\frac{s(s+1)}{s_2(s_2+1)(s_2+2)}\frac{1}{{(m^{\alpha }+1)}^{s+2}},\end{array}$$

$$\begin{array}{l}\hspace{1em}-g^{\prime }(m,t)=-\frac{\beta (\beta s_2-1)t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^s}+\frac{{\beta }^{2}s{t}^{\beta +\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}}\\ =-\frac{\beta (\beta s_2-1)t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^s}+\frac{{\beta }^{2}s(m^{\alpha }+t^{\beta }-m^{\alpha })t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}}=\frac{\beta (\beta s-\beta s_2+1)t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^s}-\frac{{\beta }^{2}s{m}^{\alpha }{t}^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}},\end{array}$$

and for $0<s_2\le \frac{2}{\beta }, 0<\beta \le 1, s_2<s\le 6$, it follows that

$${(-1)}^i\frac{d^i}{d{t}^i}[\frac{t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^s}]>0,{(-1)}^i\frac{d^i}{d{t}^i}[\frac{t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}}]>0 (i=0,1,2,3).$$

By (9), (10), (11) and (12), we obtain

$$\beta (\beta s-\beta s_2+1){\displaystyle {\int }_1^{\infty }{P}_1(t)\frac{t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^s}}dt>-\frac{\beta (\beta s-\beta s_2+1)}{12{(m^{\alpha }+1)}^s},$$

$$\begin{array}{l}\hspace{1em}-{\beta }^2{m}^{\alpha }s{\displaystyle {\int }_1^{\infty }{P}_1(t)\frac{t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}}}dt\\ =\frac{{\beta }^2{m}^{\alpha }s}{12{(m^{\alpha }+1)}^{s+1}}-\frac{{\beta }^2{m}^{\alpha }s}{6}{\displaystyle {\int }_1^{\infty }{P}_3}(t)[\frac{t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}}{]}^{″}dt\\ >\frac{{\beta }^2{m}^{\alpha }s}{12{(m^{\alpha }+1)}^{s+1}}-\frac{{\beta }^2{m}^{\alpha }s}{720}{\left[\frac{t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}}\right]}_{t=1}^{″}\\ >\frac{{\beta }^2(m^{\alpha }+1-1)s}{12{(m^{\alpha }+1)}^{s+1}}-\frac{{\beta }^2(m^{\alpha }+1)s}{720}[\frac{(s+1)(s+2){\beta }^2}{{(m^{\alpha }+1)}^{s+3}}+\frac{\beta (s+1)(5-\beta -2\beta s_2)}{{(m^{\alpha }+1)}^{s+2}}+\frac{(2-\beta s_2)(3-\beta s_2)}{{(m^{\alpha }+1)}^{s+1}}]\\ =\frac{{\beta }^{2}s}{12{(m^{\alpha }+1)}^s}-\frac{{\beta }^{2}s}{12{(m^{\alpha }+1)}^{s+1}}\\ \hspace{1em}\hspace{1em}\hspace{1em}-\frac{{\beta }^{2}s}{720}[\frac{(s+1)(s+2){\beta }^2}{{(m^{\alpha }+1)}^{s+2}}+\frac{\beta (s+1)(5-\beta -2\beta s_2)}{{(m^{\alpha }+1)}^{s+1}}+\frac{(2-\beta s_2)(3-\beta s_2)}{{(m^{\alpha }+1)}^s}].\end{array}$$

Hence, we have $h(m)>\frac{1}{{(m^{\alpha }+1)}^s}{h}_1+\frac{\lambda }{{(m^{\alpha }+1)}^{s+1}}{h}_2+\frac{s(s+1)}{{(m^{\alpha }+1)}^{s+2}}{h}_3,$ where

$$h_1:=\frac{1}{s_2}-\frac{\beta }{2}-\frac{\beta -{\beta }^2{s}_2}{12}-\frac{{\beta }^{2}s(2-\beta s_2)(3-\beta s_2)}{720},$$

$$h_2:=\frac{1}{s_2(s_2+1)}-\frac{{\beta }^2}{12}-\frac{{\beta }^3(s+1)(5-\beta -2\beta s_2)}{720},$$

and $h_3:=\frac{1}{s_2(s_2+1)(s_2+2)}-\frac{{\beta }^4(s+2)}{720}.$ We can find

$$h_1\ge \frac{1}{s_2}-\frac{\beta }{2}-\frac{\beta -{\beta }^2{s}_2}{12}-\frac{s{\beta }^2(2-\beta s_2)(3-\beta s_2)}{720}=\frac{g(s_2)}{720s_2},$$

where the function $g(\sigma )(\sigma \in (0,\frac{2}{\beta }])$ is indicated by

$$g(\sigma ):=720-(420\beta +6s{\beta }^2)\sigma +(60{\beta }^2+5s{\beta }^3){\sigma }^2-s{\beta }^4{\sigma }^3.$$

For $\beta \in (0,1], s\in (0,6]$, we obtain

$$\begin{array}{ll}{g}^{\prime }(\sigma )& =-(420\beta +6s{\beta }^2)+2(60{\beta }^2+5s{\beta }^3)\sigma -3{\beta }^4{\sigma }^2\\ & \le -420\beta -6s{\beta }^2+2(60{\beta }^2+5s{\beta }^3)\frac{2}{\beta }=(14s\beta -180)\beta <0,\end{array}$$

and then it follows that $h_1\ge \frac{g(s_2)}{720 s_2}\ge \frac{g(2/\beta )}{720 s_2}=\frac{1}{6 s_2}>0$. For $s_2\in (0,\frac{2}{\beta }]$, we still find

$$h_2>\frac{{\beta }^2}{6}-\frac{{\beta }^2}{12}-\frac{5(s+1){\beta }^2}{720}=(\frac{1}{12}-\frac{s+1}{140}){\beta }^2>0,$$

and $h_3\ge (\frac{1}{24}-\frac{s+2}{720}){\beta }^3>0(s\in (0,6])$.

Hence, it follows that $h(m)>0$. Setting $t=m^{\frac{\alpha }{\beta }}{u}^{\frac{1}{\beta }},$ we have

$$\begin{array}{l}{\varpi }_s(s_2,m)=m^{\alpha (s-s_2)}{\displaystyle \sum _{n=1}^{\infty }g(m,n)}<m^{\alpha (s-s_2)}{\displaystyle {\int }_0^{\infty }g(m,t)}dt\\ \hspace{1em}\hspace{1em}\hspace{1em}=\beta m^{\alpha (s-s_2)}{\displaystyle {\int }_0^{\infty }\frac{t^{\beta s_2-1}dt}{{(m^{\alpha }+t^{\beta })}^s}}={\displaystyle {\int }_0^{\infty }\frac{u^{s_2-1}du}{{(1+u)}^s}}=B(s_2,s-s_2)=k_s(s_2).\end{array}$$

On the other hand, in view of (11), we find

$$\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }g(m,n)={\displaystyle {\int }_1^{\infty }g(m,t)dt+\frac{1}{2}}}g(m,1)+{\displaystyle {\int }_1^{\infty }{P}_1(t)g^{\prime }(m,t)dt}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\displaystyle {\int }_1^{\infty }g(m,t)dt+H(m),}\end{array}$$

where we set $H(m):=\frac{1}{2}g(m,1)+{\displaystyle {\int }_1^{\infty }{P}_1(t)g^{\prime }(m,t)dt}$.

We find $\frac{1}{2}g(m,1)=\frac{\beta }{2{(m^{\alpha }+1)}^s}$, and

$$g^{\prime }(m,t)=-\frac{\beta (\beta s-\beta s_2+1)t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^s}+\frac{{\beta }^{2}s{m}^{\alpha }{t}^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}}.$$

For $s_2\in (0,\frac{2}{\beta }]\cap (0,s), 0<s\le 6$, by (7), we find

$$-\beta (\beta s-\beta s_2+1){\displaystyle {\int }_1^{\infty }{P}_1(t)\frac{t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^s}}dt>0,$$

and

$${\beta }^2{m}^{\alpha }s{\displaystyle {\int }_1^{\infty }{P}_1(t)\frac{t^{\beta s_2-2}}{{(m^{\alpha }+t^{\beta })}^{s+1}}}dt>-\frac{{\beta }^2{m}^{\alpha }s}{12{(m^{\alpha }+1)}^{s+1}}>-\frac{{\beta }^{2}s}{12{(m^{\alpha }+1)}^s}.$$

Hence, it follows that

$$H(m)>\frac{\beta }{2{(m^{\alpha }+1)}^s}-\frac{{\beta }^{2}s}{12{(m^{\alpha }+1)}^s}\ge \frac{\beta }{2{(m^{\alpha }+1)}^s}-\frac{6\beta }{12{(m^{\alpha }+1)}^s}=0.$$

Then, we have

$$\begin{array}{l}{\varpi }_s({\lambda }_2,m)=m^{\alpha (s-s_2)}{\displaystyle \sum _{n=1}^{\infty }g(m,n)>m^{\alpha (s-s_2)}{\displaystyle {\int }_1^{\infty }g(m,t)dt}}\\ \hspace{1em}\hspace{1em}\hspace{1em}=m^{\alpha (s-s_2)}{\displaystyle {\int }_0^{\infty }g(m,t)dt}-m^{\alpha (s-s_2)}{\displaystyle {\int }_0^{1}g(m,t)dt}\\ \hspace{1em}\hspace{1em}\hspace{1em}=k_s(s_2)\left[1-\frac{1}{k_s(s_2)}{\displaystyle {\int }_0^{\frac{1}{m^{\alpha }}}\frac{u^{s_2-1}}{{(1+u)}^s}du}\right]>0,\end{array}$$

where we indicate that $O(\frac{1}{m^{\alpha s_2}})=\frac{1}{k_s(s_2)}{\displaystyle {\int }_0^{\frac{1}{m^{\alpha }}}\frac{u^{s_2-1}}{{(1+u)}^s}du},$ satisfying

$$0<{\displaystyle {\int }_0^{\frac{1}{m^{\alpha }}}\frac{u^{s_2-1}}{{(1+u)}^s}du}<{\displaystyle {\int }_0^{\frac{1}{m^{\alpha }}}{u}^{s_2-1}du}=\frac{1}{s_2{m}^{\alpha s_2}}.$$

Therefore, inequalities (14) are valid.

The lemma is proved. □

In view of Lemma 3, the key inequality is obtained as follows:

Lemma 4.

We have the reverse inequality, as follows:

$$\begin{array}{c}{I}_{\lambda }:={\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{a_m{b}_n}{{(m^{\alpha }+n^{\beta })}^{\lambda }}}}>{(\frac{1}{\beta }{k}_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(\frac{1}{\alpha }{k}_{\lambda }({\lambda }_1))}^{\frac{1}{q}}\\ \times {\left[{\displaystyle \sum _{m=1}^{\infty }(1-O(\frac{1}{m^{\alpha {\lambda }_2}}))m^{p(1-\alpha {\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-\beta {\widehat{\lambda }}_2)-1}}{b}_n^q\right]}^{\frac{1}{q}}.\end{array}$$

(15)

Proof. 

By the symmetry, for $s_1\in (0,\frac{2}{\alpha }]\cap (0,s),s\in (0,6],$ we obtain the inequalities of the next weight coefficient, as follows:

$$\begin{array}{c}0<k_s(s_1)(1-O(\frac{1}{n^{\beta s_1}}))\\ <{\omega }_s(s_1,n):=n^{\beta (s-s_1)}{\displaystyle \sum _{m=1}^{\infty }\frac{\alpha m^{\alpha s_1-1}}{{(m^{\alpha }+n^{\beta })}^s}}<k_s(s_1)(n\in \mathrm{N}),\end{array}$$

(16)

where we indicate $O(\frac{1}{n^{\beta s_1}}):=\frac{1}{k_s(s_1)}{\displaystyle {\int }_0^{\frac{1}{n^{\beta }}}\frac{u^{s_1-1}}{{(1+u)}^s}}du>0$.

Using the reverse Hölder’s inequality (cf. [31]), it follows that

$$\begin{array}{l}{I}_{\lambda }={\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{1}{{(m^{\alpha }+n^{\beta })}^{\lambda }}}}\left[\frac{m^{\alpha (1-{\lambda }_1)/q}{(\beta n^{\beta -1})}^{1/p}}{n^{\beta (1-{\lambda }_2)/p}{(\alpha m^{\alpha -1})}^{1/q}}{a}_m\right]\left[\frac{n^{\beta (1-{\lambda }_2)/p}{(\alpha m^{\alpha -1})}^{1/q}}{m^{\alpha (1-{\lambda }_1)/q}{(\beta n^{\beta -1})}^{1/p}}{b}_n\right]\\ \hspace{1em}\ge {\left[{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{\beta }{{(m^{\alpha }+n^{\beta })}^{\lambda }}}}\frac{m^{\alpha (1-{\lambda }_1)(p-1)}{n}^{\beta -1}{a}_m^p}{n^{\beta (1-{\lambda }_2)}{(\alpha m^{\alpha -1})}^{p-1}}\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{\alpha }{{(m^{\alpha }+n^{\beta })}^{\lambda }}}}\frac{n^{\beta (1-{\lambda }_2)(q-1)}{m}^{\alpha -1}{b}_n^q}{m^{\alpha (1-{\lambda }_1)}{(\beta n^{\beta -1})}^{q-1}}\right]}^{\frac{1}{q}}\\ \hspace{1em}={(\frac{1}{\alpha })}^{\frac{1}{q}}{(\frac{1}{\beta })}^{\frac{1}{p}}{\left[{\displaystyle \sum _{m=1}^{\infty }}{\varpi }_{\lambda }({\lambda }_2,m)m^{p(1-\alpha {\widehat{\lambda }}_1)-1}{a}_m^p\right]}^{\frac{1}{p}}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times {\left[{\displaystyle \sum _{n=1}^{\infty }{\omega }_{\lambda }({\lambda }_1,n)}{n}^{q(1-\beta {\widehat{\lambda }}_2)-1}{b}_n^q\right]}^{\frac{1}{q}}.\end{array}$$

By (14), (16) and (5) (for $s=\lambda ,s_i={\lambda }_i(i=1,2)$), since $0<p<1(q<0)$ and the assumptions, we obtain (15).

The lemma is proved. □

3. Main Results

By Lemma 1 and Lemma 4, the following theorem follows:

Theorem 1.

The following reverse inequality with two partial sums and parameters is valid:

$$\begin{array}{c}I:={\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{m^{\alpha -1}{n}^{\beta -1}}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}}}{A}_m{B}_n>\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)\alpha \beta }{(\frac{1}{\beta }{k}_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(\frac{1}{\alpha }{k}_{\lambda }({\lambda }_1))}^{\frac{1}{q}}\\ \hspace{1em}\times {\left[{\displaystyle \sum _{m=1}^{\infty }(1-O(\frac{1}{m^{\alpha {\lambda }_2}}))m^{p(1-\alpha {\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-\beta {\widehat{\lambda }}_2)-1}}{b}_n^q\right]}^{\frac{1}{q}}.\end{array}$$

(17)

In particular, for ${\lambda }_1+{\lambda }_2=\lambda$, we have

$$0<{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1-\alpha {\lambda }_1)-1}}{a}_m^p<\infty , 0<{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-\beta {\lambda }_2)-1}}{b}_n^q<\infty ,$$

as well as:

$$\begin{array}{c}{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{m^{\alpha -1}{n}^{\beta -1}{A}_m{B}_n}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}>}}{\left(\frac{1}{\alpha }\right)}^{1+\frac{1}{q}}{\left(\frac{1}{\beta }\right)}^{1+\frac{1}{p}}\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)}B({\lambda }_1,{\lambda }_2)\\ \times {\left[{\displaystyle \sum _{m=1}^{\infty }(1-O(\frac{1}{m^{\alpha {\lambda }_2}}))m^{p(1-\alpha {\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-\beta {\lambda }_2)-1}}{b}_n^q\right]}^{\frac{1}{q}}.\end{array}$$

(18)

Proof. 

Based on the expression as follows

$$\frac{1}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}=\frac{1}{\Gamma (\lambda +2)}{\displaystyle {\int }_0^{\infty }{t}^{(\lambda +2)-1}}{e}^{-(m^{\alpha }+n^{\beta })t}dt,$$

by (6) and (7), we have

$$\begin{array}{l}I=\frac{1}{\Gamma (\lambda +2)}{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }(m^{\alpha -1}{A}_m)(n^{\beta -1}{B}_n){\displaystyle {\int }_0^{\infty }{t}^{\lambda +1}}{e}^{-(m^{\alpha }+n^{\beta })t}dt}}\\ =\frac{1}{\Gamma (\lambda +2)}{\displaystyle {\int }_0^{\infty }{t}^{\lambda +1}}\left({\displaystyle \sum _{m=1}^{\infty }{e}^{-m^{\alpha }t}{m}^{\alpha -1}{A}_m}\right)\left({\displaystyle \sum _{n=1}^{\infty }{e}^{-n^{\beta }t}{n}^{\beta -1}{B}_n}\right)dt\\ \ge \frac{1}{\Gamma (\lambda +2)}{\displaystyle {\int }_0^{\infty }{t}^{\lambda +1}}\left(\frac{1}{t\alpha }{\displaystyle \sum _{m=1}^{\infty }{e}^{-m^{\alpha }t}{a}_m}\right)\left(\frac{1}{t\beta }{\displaystyle \sum _{n=1}^{\infty }{e}^{-n^{\beta }t}{b}_n}\right)dt\\ =\frac{1}{\Gamma (\lambda +2)\alpha \beta }{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }{a}_m{b}_n}}{\displaystyle {\int }_0^{\infty }{t}^{\lambda -1}}{e}^{-(m^{\alpha }+n^{\beta })t}dt=\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)\alpha \beta }{I}_{\lambda }.\end{array}$$

Then, by (15), inequality (17) follows.

The theorem is proved. □

In the following two theorems, we provide a few equivalent conditions on (17).

Theorem 2.

Assume that ${\lambda }_1\in (0,\frac{2}{\alpha }-1]\cap (0,\lambda ), {\lambda }_2\in (0,\frac{2}{\beta })\cap (0,\lambda ). \lambda \in (0,4].$ If ${\lambda }_1+{\lambda }_2=\lambda ,$ then $\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)\alpha \beta }$${(\frac{1}{\beta }{k}_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(\frac{1}{\alpha }{k}_{\lambda }({\lambda }_1))}^{\frac{1}{q}}$ in (17) is the best possible. constant factor.

Proof. 

We now prove that the constant factor $\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)}{\left(\frac{1}{\alpha }\right)}^{1+\frac{1}{q}}{\left(\frac{1}{\beta }\right)}^{1+\frac{1}{p}}B({\lambda }_1,{\lambda }_2)$ in (18) is the best possible. For any $0<\epsilon <\min \{p{\lambda }_1,\left|q\right|(\frac{2}{\beta }-{\lambda }_2)\}$, we set

$${\tilde{a}}_m:=m^{\alpha ({\lambda }_1-\frac{\epsilon }{p})-1}, {\tilde{b}}_n:=n^{\beta ({\lambda }_2-\frac{\epsilon }{q})-1}(m,n\in \mathrm{N}).$$

Since $0<{\lambda }_1-\frac{\epsilon }{p}\le \frac{2}{\alpha }-1, 0<\alpha ({\lambda }_1-\frac{\epsilon }{p})\le 2-\alpha <2,$ by (2.2.24) (cf. [5]), we have

$$\begin{array}{l}{\tilde{A}}_m:={\displaystyle \sum _{i=1}^m{\tilde{a}}_i}={\displaystyle \sum _{i=1}^m{i}^{\alpha ({\lambda }_1-\frac{\epsilon }{p})-1}}={\displaystyle {\int }_1^m{t}^{\alpha ({\lambda }_1-\frac{\epsilon }{p})-1}}dt\\ \hspace{1em}\hspace{1em} +\frac{1}{2}[m^{\alpha ({\lambda }_1-\frac{\epsilon }{p})-1}+1]+\frac{{\epsilon }_0}{12}[\alpha ({\lambda }_1-\frac{\epsilon }{p})-1][m^{\alpha ({\lambda }_1-\frac{\epsilon }{p})-2}-1]\\ \hspace{1em} =\frac{1}{\alpha ({\lambda }_1-\frac{\epsilon }{p})}(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p})}+c_1+O_1(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p})-1})({\epsilon }_0\in (0,1);m\in \mathrm{N},\mathrm{m}\to \infty ).\end{array}$$

In the same way, for $0<\beta ({\lambda }_2-\frac{\epsilon }{q})<2,$ we have

$${\tilde{B}}_n:={\displaystyle \sum _{k=1}^n{\tilde{b}}_k}=\frac{1}{\beta ({\lambda }_2-\frac{\epsilon }{q})}(n^{\beta ({\lambda }_2-\frac{\epsilon }{q})}+c_2+O_2(n^{\beta ({\lambda }_2-\frac{\epsilon }{q})-1}))(n\in \mathrm{N};n\to \infty ).$$

We observe that ${\tilde{A}}_m=o(e^{t{m}^{\alpha }}),{\tilde{B}}_n=o(e^{t{n}^{\beta }})(t>0;m,n\to \infty )$.

If there exists a constant $M\ge {\left(\frac{1}{\alpha }\right)}^{1+\frac{1}{q}}{\left(\frac{1}{\beta }\right)}^{1+\frac{1}{p}}\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)}B({\lambda }_1,{\lambda }_2)$, such that (18) is valid when we replace ${\left(\frac{1}{\alpha }\right)}^{1+\frac{1}{q}}{\left(\frac{1}{\beta }\right)}^{1+\frac{1}{p}}\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)}B({\lambda }_1,{\lambda }_2)$ by $M$, then in particular, we have

$$\begin{array}{l}\tilde{I}:={\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{m^{\alpha -1}{n}^{\beta -1}}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}{\tilde{A}}_m{\tilde{B}}_n}}\\ \hspace{1em} >M{\left[{\displaystyle \sum _{m=1}^{\infty }\left(1-O(\frac{1}{m^{\alpha {\lambda }_2}})\right)m^{p(1-\alpha {\lambda }_1)-1}}{\tilde{a}}_m^p\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-\beta {\lambda }_2)-1}}{\tilde{b}}_n^q\right]}^{\frac{1}{q}}.\end{array}$$

(19)

By (19) and using the decreasingness property of series, it follows that

$$\begin{array}{l}\tilde{I}>M{\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{-\alpha \epsilon -1}-{\displaystyle \sum _{m=1}^{\infty }{m}^{-\alpha \epsilon -1}}O(\frac{1}{m^{\epsilon {\lambda }_2}})}\right]}^{\frac{1}{p}}{\left(1+{\displaystyle \sum _{n=2}^{\infty }{n}^{-\beta \epsilon -1}}\right)}^{\frac{1}{q}}\\ \hspace{1em}>M{\left({\displaystyle {\int }_1^{\infty }{x}^{-\alpha \epsilon -1}}dx-O(1)\right)}^{\frac{1}{p}}{\left(1+{\displaystyle {\int }_1^{\infty }{y}^{-\beta \epsilon -1}}dy\right)}^{\frac{1}{q}}\\ \hspace{1em}=\frac{M}{\epsilon }{\left(\frac{1}{\alpha }-\epsilon O(1)\right)}^{\frac{1}{p}}{\left(\epsilon +\frac{1}{\beta }\right)}^{\frac{1}{q}}.\end{array}$$

We still find that

$$\begin{array}{l}\tilde{I}<\frac{1}{\alpha ({\lambda }_1-\frac{\epsilon }{p})}\frac{1}{\beta ({\lambda }_2-\frac{\epsilon }{q})}{\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{1}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}\left[m^{\alpha -1}\left(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p})}+\left|c_1\right|+|O_1(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p})-1})|\right)\right]}}\\ \hspace{1em}\hspace{1em}\times \left[n^{\beta -1}\left(n^{\beta ({\lambda }_2-\frac{\epsilon }{q})}+\left|c_2\right|+|O_2(n^{\beta ({\lambda }_2-\frac{\epsilon }{q})-1})|\right)\right]=\frac{1}{\alpha ({\lambda }_1-\frac{\epsilon }{p})}\frac{1}{\beta ({\lambda }_2-\frac{\epsilon }{q})}\\ \hspace{1em}\times {\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{1}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}\left[m^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-1}+\left|c_1\right|m^{\alpha -1}+|O_1(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-2})|\right]}}\\ \hspace{1em}\hspace{1em}\times \left[n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)-1}+\left|c_2\right|n^{\beta -1}+|O_2(n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)-2})|\right]\\ \hspace{1em}=\frac{1}{\alpha ({\lambda }_1-\frac{\epsilon }{p})}\frac{1}{\beta ({\lambda }_2-\frac{\epsilon }{q})}(I_0+I_1),\end{array}$$

where we indicate that $I_0:={\displaystyle \sum _{n=1}^{\infty }\left[n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)}-{\displaystyle \sum _{m=1}^{\infty }\frac{m^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-1}}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}}\right]},$ and

$$\begin{array}{l}{I}_1:={\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{1}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}}}[(\left|c_1\right|m^{\alpha -1}+|O_1(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-2}))n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)-1}|\\ \hspace{1em}\hspace{1em}+(\left|c_1\right|m^{\alpha -1}+|O_1(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-2}))(\left|c_2\right|n^{\beta -1}+|O_2(n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)-2})|)\\ \hspace{1em}\hspace{1em}+m^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-1}(\left|c_2\right|n^{\beta -1}+|O_2(n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)-2})|)]\\ \hspace{1em}\hspace{1em}\le {\displaystyle \sum _{n=1}^{\infty }\frac{n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)-1}}{{(n^{\beta })}^{{\lambda }_2+1}}{\displaystyle \sum _{m=1}^{\infty }\frac{1}{{(m^{\alpha })}^{{\lambda }_1+1}}(\left|c_1\right|m^{\alpha -1}+|O_1(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-2})|)}}\\ \hspace{1em}\hspace{1em}+{\displaystyle \sum _{n=1}^{\infty }\frac{(\left|c_2\right|n^{\beta -1}+|O_2(n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)-2})|)}{{(n^{\beta })}^{{\lambda }_2+1}}{\displaystyle \sum _{m=1}^{\infty }\frac{1}{{(m^{\alpha })}^{{\lambda }_1+1}}}}(\left|c_1\right|m^{\alpha -1}+|O_1(m^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-2}|)\\ \hspace{1em}\hspace{1em}+{\displaystyle \sum _{n=1}^{\infty }\frac{(\left|c_2\right|n^{\beta -1}+|O_2(n^{\beta ({\lambda }_2-\frac{\epsilon }{q}+1)-2})|)}{{(n^{\beta })}^{{\lambda }_2+1}}{\displaystyle \sum _{m=1}^{\infty }\frac{1}{{(m^{\alpha })}^{{\lambda }_1+1}}}}{m}^{\alpha ({\lambda }_1-\frac{\epsilon }{p}+1)-1}\le M_1<\infty .\end{array}$$

By (14), for $s=\lambda +2\in (0,6],s_1={\lambda }_1+1-\frac{\epsilon }{p}$ $(\in (0,\frac{2}{\alpha }]\cap (0,\lambda +2)),$ we have

$$\begin{array}{l}{I}_0=\frac{1}{\alpha }{\displaystyle \sum _{n=1}^{\infty }\left[n^{\beta ({\lambda }_2+1+\frac{\epsilon }{p})}{\displaystyle \sum _{m=1}^{\infty }\frac{\alpha m^{\alpha ({\lambda }_1+1-\frac{\epsilon }{p})-1}}{{(m+n)}^{\lambda +2}}}\right]}{n}^{-\beta \epsilon -1}\\ \hspace{1em}=\frac{1}{\alpha }{\displaystyle \sum _{n=1}^{\infty }{\omega }_{\lambda +2}({\lambda }_1+1-\frac{\epsilon }{p},n)n^{-\beta \epsilon -1}}\\ \hspace{1em}<\frac{1}{\alpha }{k}_{\lambda +2}({\lambda }_1+1-\frac{\epsilon }{p})\left(1+{\displaystyle \sum _{n=2}^{\infty }{n}^{-\beta \epsilon -1}}\right)\\ \hspace{1em}<\frac{1}{\alpha }{k}_{\lambda +2}({\lambda }_1+1-\frac{\epsilon }{p})\left(1+{\displaystyle {\int }_1^{\infty }{y}^{-\beta \epsilon -1}}dy\right)\\ \hspace{1em}=\frac{1}{\epsilon }\frac{1}{\alpha \beta }B({\lambda }_1+1-\frac{\epsilon }{p},{\lambda }_2+1+\frac{\epsilon }{p})(1+\beta \epsilon ).\end{array}$$

Based on the above results, we have

$$\frac{1}{{\alpha }^2({\lambda }_1-\frac{\epsilon }{p})}\frac{1}{{\beta }^2({\lambda }_2-\frac{\epsilon }{q})}[B({\lambda }_1+1-\frac{\epsilon }{p},{\lambda }_2+1+\frac{\epsilon }{p})(1+\beta \epsilon )+\epsilon M_1]>\epsilon \tilde{I}>M{\left(\frac{1}{\alpha }-\epsilon O(1)\right)}^{\frac{1}{p}}{\left(\epsilon +\frac{1}{\beta }\right)}^{\frac{1}{q}}.$$

Setting $\epsilon \to 0^{+}$ in the above inequality, in virtue of the continuity of the beta function, we find

$${(\frac{1}{\alpha })}^{1+\frac{1}{q}}{(\frac{1}{\beta })}^{1+\frac{1}{p}}\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)}B({\lambda }_1,{\lambda }_2)={(\frac{1}{\alpha })}^{1+\frac{1}{q}}{(\frac{1}{\beta })}^{1+\frac{1}{p}}\frac{1}{{\lambda }_1{\lambda }_2}B({\lambda }_1+1,{\lambda }_2+1)\ge M.$$

Hence, $M={\left(\frac{1}{\alpha }\right)}^{1+\frac{1}{q}}{\left(\frac{1}{\beta }\right)}^{1+\frac{1}{p}}\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)}B({\lambda }_1,{\lambda }_2)$ is the best possible constant factor in (18).

The theorem is proved. □

Theorem 3.

Suppose that ${\lambda }_1\in (0,\frac{2}{\alpha }]\cap (0,\lambda ), {\lambda }_2\in (0,\frac{2}{\beta }]\cap (0,\lambda ). \lambda \in (0,6]$. If the constant factor $\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)\alpha \beta }$${(\frac{1}{\beta }{k}_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(\frac{1}{\alpha }{k}_{\lambda }({\lambda }_1))}^{\frac{1}{q}}$ in (17) is the best possible, then for

$$\lambda -{\lambda }_1-{\lambda }_2\in (-p{\lambda }_1,p(\lambda -{\lambda }_1))\cap [q(\frac{2}{\beta }-{\lambda }_2),p(\frac{2}{\alpha }-{\lambda }_1)],$$

we have ${\lambda }_1+{\lambda }_2=\lambda .$

Proof. 

For ${\widehat{\lambda }}_1=\frac{\lambda -{\lambda }_2}{p}+\frac{{\lambda }_1}{q}=\frac{\lambda -{\lambda }_1-{\lambda }_2}{p}+{\lambda }_1, {\widehat{\lambda }}_2=\frac{\lambda -{\lambda }_1}{q}+\frac{{\lambda }_2}{p}=\frac{\lambda -{\lambda }_1-{\lambda }_2}{q}+{\lambda }_2$, we find ${\widehat{\lambda }}_1+{\widehat{\lambda }}_2=\lambda .$ For $\lambda -{\lambda }_1-{\lambda }_2\in (-p{\lambda }_1,p(\lambda -{\lambda }_1)),$ we have ${\widehat{\lambda }}_1\in (0,\lambda ),$ and then ${\widehat{\lambda }}_2=\lambda -{\widehat{\lambda }}_1\in (0,\lambda );$ for $\lambda -{\lambda }_1-{\lambda }_2\in [q(\frac{2}{\beta }-{\lambda }_2),p(\frac{2}{\alpha }-{\lambda }_1)],$ we still have ${\widehat{\lambda }}_1\le \frac{2}{\alpha },$ ${\widehat{\lambda }}_2\le \frac{2}{\beta }.$ Then, for ${\lambda }_1+{\lambda }_2=\lambda$ in (17), substitution of ${\lambda }_i={\widehat{\lambda }}_i (i=1,2),$ we still have

$$\begin{array}{c}{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{A_m{B}_n}{{(m^{\alpha }+n^{\beta })}^{\lambda +2}}>}}{(\frac{1}{\alpha })}^{1+\frac{1}{q}}{(\frac{1}{\beta })}^{1+\frac{1}{p}}\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)}B({\widehat{\lambda }}_1,{\widehat{\lambda }}_2)\\ \times {\left[{\displaystyle \sum _{m=1}^{\infty }(1-O(\frac{1}{m^{\alpha {\widehat{\widehat{\lambda }}}_2}}))m^{p(1-\alpha {\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-\beta {\widehat{\lambda }}_2)-1}}{b}_n^q\right]}^{\frac{1}{q}}.\end{array}$$

(20)

By using the reverse Hölder’s inequality (cf. [31]), we still obtain

$$\begin{array}{l}\hspace{1em}B({\widehat{\lambda }}_1,{\widehat{\lambda }}_2)=k_{\lambda }(\frac{\lambda -{\lambda }_2}{p}+\frac{{\lambda }_1}{q})\\ ={\displaystyle {\int }_0^{\infty }\frac{1}{{(1+u)}^{\lambda }}}{u}^{\frac{\lambda -{\lambda }_2}{p}+\frac{{\lambda }_1}{q}-1}du={\displaystyle {\int }_0^{\infty }\frac{1}{{(1+u)}^{\lambda }}}\left(u^{\frac{\lambda -{\lambda }_2-1}{p}}\right)\left(u^{\frac{{\lambda }_1-1}{q}}\right)du\\ \ge {\left[{\displaystyle {\int }_0^{\infty }\frac{1}{{(1+u)}^{\lambda }}}{u}^{\lambda -{\lambda }_2-1}du\right]}^{\frac{1}{p}}{\left[{\displaystyle {\int }_0^{\infty }\frac{1}{{(1+u)}^{\lambda }}}{u}^{{\lambda }_1-1}du\right]}^{\frac{1}{q}}\\ ={\left[{\displaystyle {\int }_0^{\infty }\frac{1}{{(1+v)}^{\lambda }}}{v}^{{\lambda }_2-1}dv\right]}^{\frac{1}{p}}{\left[{\displaystyle {\int }_0^{\infty }\frac{1}{{(1+u)}^{\lambda }}}{u}^{{\lambda }_1-1}du\right]}^{\frac{1}{q}}\\ ={(k_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(k_{\lambda }({\lambda }_1))}^{\frac{1}{q}}.\end{array}$$

(21)

If $\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)\alpha \beta }$${(\frac{1}{\beta }{k}_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(\frac{1}{\alpha }{k}_{\lambda }({\lambda }_1))}^{\frac{1}{q}}$ in (17) is the best possible constant factor, then compare it with the constant factors in (17) and (20), and we have the following inequality:

$$\begin{array}{c}\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)\alpha \beta }{(\frac{1}{\beta }{k}_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(\frac{1}{\alpha }{k}_{\lambda }({\lambda }_1))}^{\frac{1}{q}}\\ \ge {(\frac{1}{\alpha })}^{1+\frac{1}{q}}{(\frac{1}{\beta })}^{1+\frac{1}{p}}\frac{\Gamma (\lambda )}{\Gamma (\lambda +2)}B({\widehat{\lambda }}_1,{\widehat{\lambda }}_2)(\in {\mathrm{R}}_{+}),\end{array}$$

namely, $B({\widehat{\lambda }}_1,{\widehat{\lambda }}_2)\le$${(k_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(k_{\lambda }({\lambda }_1))}^{\frac{1}{q}}$. Then, by (21), we have

$$B({\widehat{\lambda }}_1,{\widehat{\lambda }}_{21})={(k_{\lambda }({\lambda }_2))}^{\frac{1}{p}}{(k_{\lambda }({\lambda }_1))}^{\frac{1}{q}},$$

which follows that (21) protains the form of equality.

We observe that (21) protains the form of equality if and only if there exist $A$ and $B$, such that they are not both zero and (cf. [31]) $A{u}^{\lambda -{\lambda }_2-1}=B{u}^{{\lambda }_1-1}a.e.$ in ${\mathrm{R}}_{+}$. Assume that $A\ne 0$. It follows that $u^{\lambda -{\lambda }_2-{\lambda }_1}=\frac{B}{A}a.e.$ in ${\mathrm{R}}_{+}$, and then $\lambda -{\lambda }_2-{\lambda }_1=0$. Hence, we have ${\lambda }_1+{\lambda }_2=\lambda$.

The theorem is proved. □

Remark 1.

(i) For $\alpha =\beta =1,\lambda \in (0,4],{\lambda }_1\in (0,1]\cap (0,\lambda ),{\lambda }_2\in (0,2)\cap (0,\lambda )$ in (18), we have the following reverse inequality with $\frac{1}{\lambda (\lambda +1)}B({\lambda }_1,{\lambda }_2)$ as the best possible constant factor:

$$\begin{array}{c}{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{A_m{B}_n}{{(m+n)}^{\lambda +2}}>}}\frac{1}{\lambda (\lambda +1)}B({\lambda }_1,{\lambda }_2)\\ \times {\left[{\displaystyle \sum _{m=1}^{\infty }(1-O(\frac{1}{m^{{\lambda }_2}}))m^{p(1-{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-{\lambda }_2)-1}}{b}_n^q\right]}^{\frac{1}{q}}.\end{array}$$

(22)

(ii) For $\alpha =\beta =\frac{1}{2},\lambda \in (0,4],{\lambda }_1\in (0,3]\cap (0,\lambda ),{\lambda }_2\in (0,\lambda )$ in (18), we have the following reverse inequality with $\frac{8}{\lambda (\lambda +1)}B({\lambda }_1,{\lambda }_2)$ as the best possible constant factor:

$$\begin{array}{c}{\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{A_m{B}_n}{{(\sqrt{m}+\sqrt{n})}^{\lambda +2}\sqrt{mn}}>}}\frac{8}{\lambda (\lambda +1)}B({\lambda }_1,{\lambda }_2)\\ \times {\left[{\displaystyle \sum _{m=1}^{\infty }(1-O(\frac{1}{m^{{\lambda }_{2/2}}}))m^{p(1-\frac{{\lambda }_1}{2})-1}}{a}_m^p\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-\frac{{\lambda }_2}{2})-1}}{b}_n^q\right]}^{\frac{1}{q}}.\end{array}$$

(23)

4. Conclusions

In this article, by using the techniques of real analysis, the way of weight coefficients and the idea of introduced parameters, applying the mid-value theorem. We estimate some lemmas and obtain a new reverse extended inequality (2) with two partial sums and multi-parameters in Theorem 1. We consider a few equivalent statements of the best possible constant factor related to several parameters in Theorems 2 and 3. We also deduce some inequalities for setting particular parameters in Remark 1. The theorems and lemmas in this paper provide a useful extensive account of this type of inequality. Further studies should be using the idea of this article to build some other kinds of Hilbert-type inequalities with partial sums and parameters.