1 Introduction

If \(p>1,\frac{1}{p}+\frac{1}{q}=1\), \(a_{m},b_{n}\geq 0\), \(0<\sum_{m=1}^{\infty }a_{m}^{p}<\infty \), \(0<\sum_{n=1}^{\infty }b_{n}^{q}<\infty \), then we have the following Hardy–Hilbert inequality:

$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{ \pi }{\sin (\frac{\pi }{p})}\Biggl( \sum_{m=1}^{\infty }a_{m}^{p} \Biggr)^{ \frac{1}{p}}\Biggl(\sum_{n=1}^{\infty }b_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(1)

with the best possible constant factor \(\frac{\pi }{\sin (\pi /p)}\) [1]. A more accurate form of (1) with the same best possible constant factor was given in [2, Theorem 323]:

$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n-1}< \frac{ \pi }{\sin (\frac{\pi }{p})}\Biggl( \sum_{m=1}^{\infty }a_{m}^{p} \Biggr)^{ \frac{1}{p}}\Biggl(\sum_{n=1}^{\infty }b_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(2)

Inequalities (1) and (2) played an important role in analysis and its applications (see [2–4]).

In 2011, Yang [5] gave the following an extension of (2): If \(0<\lambda_{1}\), \(\lambda_{2}\leq 1\), \(\lambda_{1}+\lambda_{2}=\lambda \), \(a_{m},b_{n}\geq 0\),

$$\begin{aligned}& \Vert a \Vert _{p,\varphi } = \Biggl\{ \sum_{m=1}^{\infty }(m- \alpha )^{p(1-\lambda_{1})-1}a _{m}^{p}\Biggr\} ^{\frac{1}{p}}\in (0, \infty ), \\& \Vert b \Vert _{q,\psi } = \Biggl\{ \sum_{n=1}^{\infty }(n- \alpha )^{q(1-\lambda_{2})-1}b _{n}^{q}\Biggr\} ^{\frac{1}{q}}\in (0, \infty ), \end{aligned}$$

then

$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{(m+n-2 \alpha )^{\lambda }}< B(\lambda_{1}, \lambda_{2})\Vert a \Vert _{p,\varphi }\Vert b \Vert _{q, \psi } \quad \biggl(0\leq \alpha \leq \frac{1}{2}\biggr), $$
(3)

where the constant factor \(B(\lambda_{1},\lambda_{2})\) is the best possible, and \(B(u,v)\) is the beta function defined as (see [6])

$$ B(u,v):= \int_{0}^{\infty }\frac{1}{(1+t)^{u+v}}t^{u-1}\,dt\quad (u,v>0). $$
(4)

For \(\lambda =1\), \(\lambda_{1}=\frac{1}{q}\), \(\lambda_{2}=\frac{1}{p}\), and \(\alpha =\frac{1}{2}\), (3) reduces to (2). Some other results related to (1)–(3) were provided in [7–24]. In 2016–17, a few extensions of (1)–(3) with some reverses in the whole plane were obtained in [25–27].

In this paper, using weight coefficients, a complex integral formula, and Hermite–Hadamard’s inequality, we give the following extension of the reverse of (1) in the whole plane: If \(0< p<1\) (\(q<0\)), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\lambda_{1}\), \(\lambda_{2}<1\), \(\lambda_{1}+ \lambda_{2}=\lambda \leq 1\), \(\xi ,\eta \in [ 0,\frac{1}{2}],a _{m},b_{n}\geq 0\),

$$\begin{aligned}& 0< \sum_{\vert m \vert =1}^{\infty }\vert m-\xi \vert ^{p(1-\lambda_{1})-1}a_{m}^{p}< \infty , \\& 0< \sum _{\vert n \vert =1}^{\infty }\vert n-\eta \vert ^{q(1-\lambda_{2})-1}b_{n}^{q}< \infty , \end{aligned}$$

then setting

$$\begin{aligned} \theta_{1}(\lambda_{2},m) &:=\frac{\lambda \sin (\frac{\pi \lambda _{1}}{\lambda })}{\pi } \int_{\frac{\vert m-\xi \vert }{1+\eta }}^{\infty }\frac{u ^{\lambda_{1}-1}}{u^{\lambda }+1}\,du \\ &=O \biggl( \frac{1}{\vert m-\xi \vert ^{\lambda_{2}}} \biggr) \in (0,1),\quad \vert m \vert \in \mathbf{N}, \end{aligned}$$
(5)

we have the following reverse Hilbert-type inequality in the whole plane:

$$\begin{aligned}& \sum_{\vert n \vert =1}^{\infty }\sum _{\vert m \vert =1}^{\infty }\frac{a_{m}b_{n}}{ \vert m-\xi \vert ^{\lambda }+ \vert n-\eta \vert ^{\lambda }} \\& \quad > \frac{2\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })} \Biggl[ \sum_{\vert m \vert =1}^{\infty } \bigl(1-\theta_{1}(\lambda_{2},m)\bigr)\vert m-\xi \vert ^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\& \quad \quad {}\times \Biggl[ \sum_{\vert n \vert =1}^{\infty }\vert n- \eta \vert ^{q(1-\lambda_{2})-1}b _{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(6)

Moreover, we prove an extended inequality of (6) with multiparameters and a best possible constant factor. We also consider equivalent forms and a few particular cases.

2 Some lemmas and an example

Lemma 1

Let C be the set of complex numbers, \(\mathbf{C}_{\infty }=\mathbf{C}\cup \{\infty \}\), and let \(z_{k}\in \mathbf{C}\backslash \{z\mid \operatorname{Re}z\geq 0, \operatorname{Im}z=0\}\) (\(k=1,2,\ldots ,n\)) be different points. Suppose that a function \(f(z)\) is analytic in \(\mathbf{C}_{ \infty }\) except for \(z_{i}\) (\(i=1,2,\ldots ,n\)) and that \(z=\infty \) is a zero point of \(f(z)\) of order not less than 1. Then, for \(\alpha \in \mathbf{R}\), we have

$$ \int_{0}^{\infty }f(x)x^{\alpha -1}\,dx= \frac{2\pi i}{1-e^{2\pi \alpha i}}\sum_{k=1}^{n} \operatorname{Re}s\bigl[f(z)z^{\alpha -1},z_{k}\bigr], $$
(7)

where \(0<\operatorname{Im}\ln z=\arg z<2\pi \). In particular, if \(z_{k}\) (\(k=1,\ldots ,n\)) are all poles of order 1, then setting \(\varphi_{k}(z)=(z-z _{k})f(z)\) (\(\varphi_{k}(z_{k})\neq 0\)), we have

$$ \int_{0}^{\infty }f(x)x^{\alpha -1}\,dx= \frac{\pi }{\sin \pi \alpha } \sum_{k=1}^{n}(-z_{k})^{\alpha -1} \varphi_{k}(z_{k}). $$
(8)

Proof

By [28] (p. 118) we have (7). We find

$$\begin{aligned} 1-e^{2\pi \alpha i} &=1-\cos 2\pi \alpha -i\sin 2\pi \alpha \\ &=-2i\sin \pi \alpha (\cos \pi \alpha +i\sin \pi \alpha ) \\ &=-2ie^{i \pi \alpha } \sin \pi \alpha . \end{aligned}$$

In particular, since \(f(z)z^{\alpha -1}=\frac{1}{z-z_{k}}(\varphi_{k}(z)z ^{\alpha -1})\), it is obvious that

$$ \operatorname{Re}s\bigl[f(z)z^{\alpha -1},-z_{k} \bigr]=z_{k}{}^{\alpha -1}\varphi_{k}(z _{k})=-e^{i\pi \alpha }(-z_{k})^{\alpha -1} \varphi_{k}(z_{k}). $$

Then by (7) we obtain (8). □

Example 1

For \(s\in \mathbf{N=\{}1,2,\ldots \}\), \(c_{s}\geq \cdots \geq c_{1}>0, \varepsilon >0\), \(\lambda_{1},\lambda_{2}>0\), \(\lambda_{1}+\lambda_{2}=s \lambda \), we define the function

$$ k_{s\lambda }(x,y)=\frac{1}{\prod_{k=1}^{s}(x^{\lambda }+c_{k}y^{ \lambda })} $$

and constants \(\widetilde{c}_{k}=c_{k}+(k-1)\varepsilon \) (\(k=1,\ldots ,s\)).

Since \(\widetilde{c}_{s}>\cdots >\widetilde{c}_{1}=c_{1}>0\), by (8) we find

$$\begin{aligned} \widetilde{k}_{s}(\lambda_{1}) &:=\int_{0}^{\infty }\frac{1}{\prod_{k=1}^{s}(t^{\lambda }+\widetilde{c}_{k})}t^{\lambda_{1}-1}\,dt \\ &\overset{u=t^{\lambda /s}}{=}\frac{1}{\lambda } \int_{0}^{\infty }\frac{1}{ \prod_{k=1}^{s}(u+\widetilde{c}_{k})}u^{\frac{\lambda_{1}}{\lambda }-1}\,du \\ &=\frac{\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })}\sum_{k=1}^{s} \widetilde{c}_{k}^{\frac{\lambda_{1}}{\lambda }-1} \prod_{j=1(j\neq k)}^{s} \frac{1}{\widetilde{c}_{j}-\widetilde{c}_{k}} \in \mathbf{R}_{+}. \end{aligned}$$

Since

$$\begin{aligned} 0 &< \widetilde{k}_{s}(\lambda_{1})=\frac{1}{\lambda } \int_{0}^{ \infty }\prod_{k=1}^{s} \frac{1}{u+\widetilde{c}_{k}}u^{\frac{\lambda _{1}}{\lambda }-1}\,du \\ &\leq \frac{1}{\lambda } \int_{0}^{\infty }\frac{1}{(u+c_{1})^{s}}u ^{\frac{\lambda_{1}}{\lambda }-1}\,du \\ &\overset{u=c_{1}v}{=}\frac{1}{\lambda c_{1}^{\lambda_{2}/\lambda }} \int_{0}^{\infty }\frac{1}{(v+1)^{s}}v^{\frac{\lambda_{1}}{\lambda }-1}\,dv \\ &=\frac{1}{\lambda c_{1}^{\lambda_{2}/\lambda }}B\biggl(\frac{\lambda_{1}}{ \lambda },\frac{\lambda_{2}}{\lambda }\biggr)\in \mathbf{R}_{+}, \end{aligned}$$

it follows that

$$\begin{aligned} k_{s}(\lambda_{1}) &=\lim_{\varepsilon \rightarrow 0^{+}} \widetilde{k}_{s}(\lambda_{1}) \\ &=\frac{\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })}\sum_{k=1}^{s}c_{k}^{\frac{\lambda_{1}}{\lambda }-1} \prod_{j=1(j\neq k)} ^{s}\frac{1}{c_{j}-c_{k}}\in \mathbf{R}_{+}. \end{aligned}$$
(9)

In particular, for \(s=1\), we obtain

$$ k_{1}(\lambda_{1})=\frac{1}{\lambda } \int_{0}^{\infty }\frac{u^{( \lambda_{1}/\lambda )-1}}{u+c_{1}}\,du= \frac{\pi }{\lambda c_{1}^{ \lambda_{2}/\lambda }\sin (\frac{\pi \lambda_{1}}{\lambda })}; $$
(10)

for \(c_{s}=\cdots =c_{1}\), we have

$$ k(\lambda_{1}):= \int_{0}^{\infty }\frac{t^{\lambda_{1}-1}}{(t^{\lambda }+c_{1})^{s}}\,dt= \frac{1}{\lambda c_{1}^{\lambda_{2}/\lambda }}B\biggl(\frac{ \lambda_{1}}{\lambda },\frac{\lambda_{2}}{\lambda }\biggr). $$
(11)

We further assume that \(s\in \mathbf{N}\), \(c_{s}\geq \cdots \geq c_{1}>0\), \(\alpha ,\beta \in (0,\pi )\), \(\xi ,\eta \in [ 0,\frac{1}{2}]\), \(0<\lambda_{1},\lambda_{2},\lambda \leq 1,\lambda_{1}+\lambda_{2}=s \lambda \) (\(s\geq 2\)); \(0<\lambda_{1}\), \(\lambda_{2}<1\), \(0<\lambda_{1}+\lambda_{2}=\lambda \leq 1\) (\(s=1\)). For \(\vert t \vert >\frac{1}{2}\), we set

$$ A_{\zeta ,\theta }(t):=\vert t-\zeta \vert +(t-\zeta )\cos \theta $$

\(((\zeta ,\theta ,t)=(\xi ,\alpha ,x)\mbox{ or }(\eta ,\beta ,y))\) and

$$\begin{aligned} k(x,y) &:=k_{s\lambda }\bigl(A_{\xi ,\alpha }(x),A_{\eta ,\beta }(y)\bigr) \\ &=\frac{1}{\prod_{k=1}^{s}(A_{\xi ,\alpha }^{\lambda }(x)+c_{k}A_{ \eta ,\beta }^{\lambda }(y))}. \end{aligned}$$

Definition 1

Define the following weight coefficients:

$$\begin{aligned}& \omega (\lambda_{2},m) : =\sum_{\vert n \vert =1}^{\infty }k(m,n) \frac{A_{ \xi ,\alpha }^{\lambda_{1}}(m)}{A_{\eta ,\beta }^{1-\lambda_{2}}(n)},\quad \vert m \vert \in \mathbf{N}, \end{aligned}$$
(12)
$$\begin{aligned}& \varpi (\lambda_{1},n) : =\sum_{\vert m \vert =1}^{\infty }k(m,n) \frac{A_{ \eta ,\beta }^{\lambda_{2}}(n)}{A_{\xi ,\alpha }^{1-\lambda_{1}}(m)},\quad \vert n \vert \in \mathbf{N}, \end{aligned}$$
(13)

where \(\sum_{\vert j \vert =1}^{\infty }\cdots =\sum_{j=-1}^{-\infty }+\cdots +\sum_{j=1}^{\infty }\cdots\) (\(j=m,n\)).

Lemma 2

With regards to the above agreement, replacing \(0<\lambda_{1}\leq 1\) (\(0<\lambda_{1}<1\)) by \(\lambda_{1}>0\) and setting

$$ h_{\beta }(\lambda_{1}):=2k_{s}( \lambda_{1})\csc^{2}\beta , $$

we still have

$$ h_{\beta }(\lambda_{1}) \bigl(1-\theta (\lambda_{2},m) \bigr)< \omega (\lambda_{2},m)< h _{\beta }(\lambda_{1}),\quad \vert m \vert \in \mathbf{N,} $$
(14)

where

$$\begin{aligned} \theta (\lambda_{2},m) &:=\frac{1}{k_{s}(\lambda_{1})} \int_{\frac{A_{\xi ,\alpha }(m)}{(1+\eta )(1+\cos \beta )}}^{\infty }\frac{u ^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u^{\lambda }+c_{k})}\,du \\ &=O \biggl( \frac{1}{A_{\xi ,\alpha }^{\lambda_{2}}(m)} \biggr) \in (0,1),\quad \vert m \vert \in \mathbf{N}. \end{aligned}$$
(15)

Proof

For \(\vert x \vert >\frac{1}{2}\), we set

$$\begin{aligned}& k^{(1)}(x,y) : =\frac{1}{\prod_{k=1}^{s}\{A_{\xi ,\alpha }^{\lambda }(x)+c_{k}[(y-\eta )(\cos \beta -1)]^{\lambda }\}}, \\& \quad y < -\frac{1}{2}, \\& k^{(2)}(x,y) : =\frac{1}{\prod_{k=1}^{s}\{A_{\xi ,\alpha }^{\lambda }(x)+c_{k}[(y-\eta )(1+\cos \beta )]^{\lambda }\}}, \\& \quad y > \frac{1}{2}, \end{aligned}$$

wherefrom, for \(y>\frac{1}{2}\),

$$ k^{(1)}(x,-y)=\frac{1}{\prod_{k=1}^{s}\{A_{\xi ,\alpha }^{\lambda }(x)+c _{k}[(y+\eta )(1-\cos \beta )]^{\lambda }\}}. $$

We find

$$\begin{aligned}& \omega (\lambda_{2},m)=\sum_{n=-1}^{-\infty }k^{(1)}(m,n) \frac{A_{ \xi ,\alpha }^{\lambda_{1}}(m)}{[(n-\eta )(\cos \beta -1)]^{1-\lambda _{2}}} \\& \hphantom{\omega (\lambda_{2},m)=}{}+\sum_{n=1}^{\infty }k^{(2)}(m,n) \frac{A_{\xi ,\alpha }^{\lambda _{1}}(m)}{[(n-\eta )(1+\cos \beta )]^{1-\lambda_{2}}} \\& \hphantom{\omega (\lambda_{2},m)}{} = \frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1-\cos \beta )^{1-\lambda _{2}}}\sum_{n=1}^{\infty } \frac{k^{(1)}(m,-n)}{(n+\eta )^{1-\lambda _{2}}} \\& \hphantom{\omega (\lambda_{2},m)=}{} +\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1+\cos \beta )^{1-\lambda _{2}}}\sum_{n=1}^{\infty } \frac{k^{(2)}(m,n)}{(n-\eta )^{1-\lambda _{2}}}. \end{aligned}$$
(16)

 □

It is evident that, for fixed \(m\in \mathbf{N}\), \(0<\lambda_{2}\leq 1\), \(0< \lambda \leq 1\), both \(\frac{k^{(1)}(m,-y)}{(y+\eta )^{1-\lambda_{2}}}\) and \(\frac{k^{(2)}(m,y)}{(y- \eta )^{1-\lambda_{2}}}\) are strictly decreasing and strictly convex with respect to \(y\in (\frac{1}{2},\infty )\) and satisfy

$$\begin{aligned}& \frac{k^{(i)}(m,(-1)^{i}y)}{[y-(-1)^{i}\eta ]^{1-\lambda_{2}}}>0, \\& \frac{d}{dy}\frac{k^{(i)}(m,(-1)^{i}y)}{[y-(-1)^{i}\eta ]^{1-\lambda _{2}}}< 0 \end{aligned}$$

and

$$ \frac{d^{2}}{dy^{2}}\frac{k^{(i)}(m,(-1)^{i}y)}{[y-(-1)^{i}\eta ]^{1- \lambda_{2}}}>0\quad (i=1,2). $$

By Hermite–Hadamard’s inequality (see [29]) we find

$$\begin{aligned} \omega (\lambda_{2},m) < &\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1- \cos \beta )^{1-\lambda_{2}}} \int_{\frac{1}{2}}^{\infty }\frac{k^{(1)}(m,-y)}{(y+ \eta )^{1-\lambda_{2}}}\,dy \\ &{}+\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1+\cos \beta )^{1- \lambda_{2}}} \int_{\frac{1}{2}}^{\infty }\frac{k^{(2)}(m,y)}{(y- \eta )^{1-\lambda_{2}}}\,dy. \end{aligned}$$

Setting \(u=\frac{A_{\xi ,\alpha }(m)}{(y+\eta )(1-\cos \beta )}\) (\(\frac{A_{\xi ,\alpha }(m)}{(y-\eta )(1+\cos \beta )}\)) in the first (second) integral, by simplification we find

$$\begin{aligned} \omega (\lambda_{2},m) &< \biggl(\frac{1}{1-\cos \beta }+ \frac{1}{1+\cos \beta }\biggr) \int_{0}^{\infty }\frac{u^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u ^{\lambda }+c_{k})}\,du \\ &=2k_{s}(\lambda_{1})\csc^{2}\beta =h_{\beta }(\lambda_{1}). \end{aligned}$$

Since both \(\frac{k^{(1)}(m,-y)}{(y+\eta )^{1-\lambda_{2}}}\) and \(\frac{k^{(2)}(m,y)}{(y-\eta )^{1-\lambda_{2}}}\) are strictly decreasing, we still have

$$\begin{aligned} \omega (\lambda_{2},m) &>\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1- \cos \beta )^{1-\lambda_{2}}} \int_{1}^{\infty }\frac{k^{(1)}(m,-y)}{(y+ \eta )^{1-\lambda_{2}}}\,dy \\ &\quad {}+\frac{A_{\xi ,\alpha }^{\lambda_{1}}(m)}{(1+\cos \beta )^{1- \lambda_{2}}} \int_{1}^{\infty }\frac{k^{(2)}(m,y)}{(y-\eta )^{1- \lambda_{2}}}\,dy \\ &=\frac{1}{1-\cos \beta } \int_{0}^{\frac{A_{\xi ,\alpha }(m)}{(1+ \eta )(1-\cos \beta )}}\frac{u^{\lambda_{1}-1}\,du}{\prod_{k=1}^{s}(u ^{\lambda }+c_{k})} \\ &\quad {}+\frac{1}{1+\cos \beta } \int_{0}^{\frac{A_{\xi ,\alpha }(m)}{(1- \eta )(1+\cos \beta )}}\frac{u^{\lambda_{1}-1}\,du}{\prod_{k=1}^{s}(u ^{\lambda }+c_{k})} \\ &\geq 2\csc^{2}\beta \int_{0}^{\frac{A_{\xi ,\alpha }(m)}{(1+\eta )(1+ \cos \beta )}}\frac{u^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u^{\lambda }+c _{k})}\,du \\ &=h_{\beta }(\lambda_{2}) \bigl(1-\theta ( \lambda_{2},m)\bigr)>0, \end{aligned}$$

where \(\theta (\lambda_{2},m)(<1)\) is indicated by (15). We obtain

$$\begin{aligned} 0 &< \theta (\lambda_{2},m)< \frac{1}{k_{s}(\lambda_{1})} \int_{\frac{A_{\xi ,\alpha }(m)}{(1+\eta )(1+\cos \beta )}}^{\infty }\frac{u ^{\lambda_{1}-1}}{u^{s\lambda }}\,du \\ &=\frac{1}{k_{s}(\lambda_{1})} \int_{\frac{A_{\xi ,\alpha }(m)}{(1+\eta )(1+\cos \beta )}}^{\infty }u ^{-\lambda_{2}-1}\,du \\ &=\frac{1}{\lambda_{2}k_{s}(\lambda_{1})} \biggl[ \frac{(1+\eta )(1+ \cos \beta )}{A_{\xi ,\alpha }(m)} \biggr] ^{\lambda_{2}}. \end{aligned}$$

Then we have (14) and estimate (15).  □

In the same way, we have

Lemma 3

With regards to the above agreement, replacing \(0<\lambda_{2}\leq 1\) (\(0<\lambda_{2}<1\)) by \(\lambda_{2}>0\), for

$$ h_{\alpha }(\lambda_{1})=2k_{s}( \lambda_{1})\csc^{2}\alpha , $$

we still have

$$ h_{\alpha }(\lambda_{1}) \bigl(1-\vartheta ( \lambda_{1},n)\bigr)< \varpi (\lambda_{1},n)< h _{\alpha }(\lambda_{1}),\quad \vert n \vert \in \mathbf{N,} $$
(17)

where

$$\begin{aligned} \vartheta (\lambda_{1},n) &:=\frac{1}{k_{s}(\lambda_{1})} \int_{\frac{A_{\eta ,\beta }(n)}{(1+\xi )(1+\cos \alpha)}}^{\infty }\frac{u^{\lambda_{2}-1}}{\prod_{k=1}^{s}(u^{\lambda }+c_{k})}\,du \\ &=O \biggl( \frac{1}{A_{\eta ,\beta }^{\lambda_{1}}(n)} \biggr) \in (0,1),\quad \vert n \vert \in \mathbf{N}. \end{aligned}$$
(18)

Lemma 4

If \(\zeta \in {}[ 0,\frac{1}{2}],\theta \in (0,\pi )\), \((\zeta , \theta )=(\xi ,\alpha )\) (or \((\eta ,\beta )\)), then, for \(\rho >0\),

$$\begin{aligned} H_{\rho }(\zeta ,\theta )&:=\sum_{\vert k \vert =1}^{\infty } \frac{1}{A_{\zeta , \theta }^{1+\rho }(k)}=\frac{1+o(1)}{\rho } \\ &\quad {}\times \biggl[ \frac{1}{(1+\cos \theta )^{1+\rho }}+\frac{1}{(1-\cos \theta )^{1+\rho }} \biggr] \quad \bigl(\rho \rightarrow 0^{+}\bigr). \end{aligned}$$
(19)

Proof

We find

$$\begin{aligned} H_{\rho }(\zeta ,\theta ) &=\sum_{k=-1}^{-\infty } \frac{1}{[(k- \zeta )(\cos \theta -1)]^{1+\rho }} \\ &\quad {}+\sum_{k=1}^{\infty }\frac{1}{[(k-\zeta )(\cos \theta +1)]^{1+ \rho }} \\ &=\frac{1}{(1-\cos \theta )^{1+\rho }}\sum_{k=1}^{\infty } \frac{1}{(k+ \zeta )^{1+\rho }} \\ &\quad {}+\frac{1}{(1+\cos \theta )^{1+\rho }}\sum_{k=1}^{\infty } \frac{1}{(k- \zeta )^{1+\rho }}. \end{aligned}$$

For \(a=\frac{1}{(1-\zeta )^{1+\rho }}>0\), by Hermite–Hadamard’s inequality we have

$$\begin{aligned} H_{\rho }(\zeta ,\theta ) &\leq \biggl[ \frac{1}{(1-\cos \theta )^{1+ \rho }}+ \frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] \\ &\quad {}\times \Biggl[ a+\sum_{k=2}^{\infty } \frac{1}{(k-\zeta )^{1+\rho }} \Biggr] \\ &< \biggl[ \frac{1}{(1-\cos \theta )^{1+\rho }}+\frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] \\ &\quad {}\times \biggl[ a+ \int_{\frac{3}{2}}^{\infty }\frac{1}{(y-\zeta )^{1+ \rho }}\,dy \biggr] \\ &=\frac{a\rho +(\frac{3}{2}-\zeta )^{-\rho }}{\rho } \biggl[ \frac{1}{(1- \cos \theta )^{1+\rho }}+\frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] . \end{aligned}$$

We still obtain

$$\begin{aligned}& H_{\rho }(\zeta ,\theta )\geq \biggl[ \frac{1}{(1-\cos \theta )^{1+ \rho }}+ \frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] \sum_{k=1}^{\infty }\frac{1}{(k+\zeta )^{1+\rho }} \\& \hphantom{ H_{\rho }(\zeta ,\theta )}{}> \biggl[ \frac{1}{(1-\cos \theta )^{1+\rho }}+\frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] \int_{1}^{\infty }\frac{dy}{(y+\zeta )^{1+ \rho }} \\& \hphantom{ H_{\rho }(\zeta ,\theta )}{}= \frac{(1+\zeta )^{-\rho }}{\rho } \biggl[ \frac{1}{(1-\cos \theta )^{1+ \rho }}+\frac{1}{(1+\cos \theta )^{1+\rho }} \biggr] . \end{aligned}$$

Hence we have (19). □

3 Main results and some particular cases

Theorem 5

Suppose that \(0< p<1\) (\(q<0\)), \(\frac{1}{p}+\frac{1}{q}=1\),

$$ K_{\alpha ,\beta }(\lambda_{1}):=h_{\beta }^{\frac{1}{p}}( \lambda_{1})h _{\alpha }^{\frac{1}{q}}(\lambda_{1})=2k_{s}( \lambda_{1}) \csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha , $$
(20)

\(a_{m},b_{n}\geq 0\) \((\vert m\vert ,\vert n\vert \in \mathbf{N})\), and

$$\begin{aligned}& 0< \sum_{\vert m \vert =1}^{\infty }A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p}< \infty , \\& 0< \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda _{2})-1}(n)b_{n}^{q}< \infty . \end{aligned}$$

We have the following reverse equivalent inequalities:

$$\begin{aligned}& I:=\sum_{\vert n \vert =1}^{\infty }\sum _{\vert m \vert =1}^{\infty }\frac{1}{\prod_{k=1} ^{s}(A_{\xi ,\alpha }^{\lambda }(m)+c_{k}A_{\eta ,\beta }^{\lambda }(n))}a _{m}b_{n} \\& \hphantom{I}{}>K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\bigl(1- \theta (\lambda_{2},m) \bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
(21)
$$\begin{aligned}& J : = \Biggl\{ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{p\lambda_{2}-1}(n) \Biggl[ \sum_{\vert m \vert =1}^{\infty }\frac{a_{m}}{\prod_{k=1}^{s}(A_{\xi , \alpha }^{\lambda }(m)+c_{k}A_{\eta ,\beta }^{\lambda }(n))} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\& \hphantom{J}{}> K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\bigl(1- \theta (\lambda_{2},m) \bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned}$$
(22)
$$\begin{aligned}& L : = \Biggl\{ \sum_{\vert m\vert =1}^{\infty }\frac{A_{ \xi ,\alpha }^{q\lambda_{1}-1}(m)}{(1-\theta (\lambda_{2},m))^{q-1}} \Biggl[ \sum_{\vert n\vert =1}^{\infty } \frac{1}{ \prod_{k=1}^{s}(A_{\xi ,\alpha }^{\lambda }(m)+c_{k}A_{\eta ,\beta } ^{\lambda }(n))}b_{n} \Biggr] ^{q} \Biggr\} ^{\frac{1}{q}} \\& \hphantom{L}{}>K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert n\vert =1} ^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(23)

In particular, for \(s=c_{1}=1\), \(\alpha =\beta =\frac{\pi }{2}\) (\(0<\lambda_{1}\), \(\lambda_{2}<1\), \(\lambda_{1}+\lambda_{2}=\lambda \leq 1\)), (21) reduces to (6); and (22) and (23) reduce to the equivalent forms of (6) as follows:

$$\begin{aligned}& \Biggl\{ \sum_{\vert n \vert =1}^{\infty }\vert n-\eta \vert ^{p\lambda_{2}-1} \Biggl( \sum_{\vert m \vert =1}^{\infty } \frac{a_{m}}{\vert m-\xi \vert ^{\lambda }+\vert n-\eta \vert ^{ \lambda }} \Biggr) ^{p} \Biggr\} ^{\frac{1}{p}} \\& \quad > \frac{2\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })} \Biggl[ \sum_{\vert m \vert =1}^{\infty } \bigl(1-\theta_{1}(\lambda_{2},m)\bigr)\vert m-\xi \vert ^{p(1- \lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned}$$
(24)
$$\begin{aligned}& \Biggl[ \sum_{\vert m\vert =1}^{\infty } \frac{\vert m-\xi \vert ^{q \lambda_{1}-1}}{(1-\theta_{1}(\lambda_{2},m))^{q-1}} \Biggl( \sum_{\vert n\vert =1}^{\infty } \frac{b_{n}}{\vert m-\xi \vert ^{ \lambda }+\vert n-\eta \vert ^{\lambda }} \Biggr) ^{q} \Biggr] ^{\frac{1}{q}} \\& \quad > \frac{2\pi }{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda })} \Biggl[ \sum_{\vert n\vert =1}^{\infty } \vert n-\eta \vert ^{q(1-\lambda_{2})-1}b _{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(25)

Proof

By the reverse Hölder inequality with weight (see [29]) and (12) we find

$$\begin{aligned}& \Biggl( \sum_{\vert m \vert =1}^{\infty }k(m,n)a_{m} \Biggr) ^{p} \\& \quad = \Biggl[ \sum_{\vert m \vert =1}^{\infty }k(m,n) \frac{A_{\xi ,\alpha }^{(1- \lambda_{1})/q}(m)a_{m}}{A_{\eta ,\beta }^{(1-\lambda_{2})/p}(n)}\frac{A _{\eta ,\beta }^{(1-\lambda_{2})/p}(n)}{A_{\xi ,\alpha }^{(1-\lambda _{1})/q}(m)} \Biggr] ^{p} \\& \quad \geq \sum_{\vert m \vert =1}^{\infty }h(m,n) \frac{A_{\xi ,\alpha }^{(1-\lambda_{1})p/q}(m)}{A _{\eta ,\beta }^{1-\lambda_{2}}(n)}a_{m}^{p} \Biggl[ \sum _{\vert m \vert =1}^{ \infty }k(m,n)\frac{A_{\eta ,\beta }^{(1-\lambda_{2})q/p}(n)}{A_{ \xi ,\alpha }^{1-\lambda_{1}}(m)} \Biggr] ^{p-1} \\& \quad = \frac{(\varpi (\lambda_{1},n))^{p-1}}{A_{\eta ,\beta }^{p\lambda_{2}-1}(n)} \sum_{\vert m \vert =1}^{\infty }k(m,n) \frac{A_{\xi ,\alpha }^{(1-\lambda_{1})p/q}(m)}{A _{\eta ,\beta }^{1-\lambda_{2}}(n)}a_{m}^{p}. \end{aligned}$$

By (17), in view of \(p-1<0\), we have

$$\begin{aligned} J >&h_{\alpha }^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum _{\vert n \vert =1}^{ \infty }\sum _{\vert m \vert =1}^{\infty }k(m,n)\frac{A_{\xi ,\alpha }^{(1-\lambda _{1})p/q}(m)}{A_{\eta ,\beta }^{1-\lambda_{2}}(n)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&h_{\alpha }^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\sum_{\vert n \vert =1}^{\infty }k(m,n) \frac{A_{\xi ,\alpha }^{(1-\lambda_{1})p/q}(m)}{A _{\eta ,\beta }^{1-\lambda_{2}}(n)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&h_{\alpha }^{\frac{1}{q}}(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\omega (\lambda_{2},m)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$
(26)

Then by (14) we have (22).

By Hölder’s inequality (see [29]) we have

$$\begin{aligned} I = \sum_{\vert n \vert =1}^{\infty } \Biggl[ A_{\eta ,\beta }^{\lambda_{2}- \frac{1}{p}}(n)\sum_{\vert m \vert =1}^{\infty }k(m,n)a_{m} \Biggr] A_{\eta , \beta }^{\frac{1}{p}-\lambda_{2}}(n)b_{n} \geq J \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(27)

Then by (22) we have (21).

On the other hand, assuming that (21) is valid, we set

$$ b_{n}:=A_{\eta ,\beta }^{p\lambda_{2}-1}(n) \Biggl( \sum _{\vert m \vert =1}^{ \infty }k(m,n)a_{m} \Biggr) ^{p-1},\quad \vert n \vert \in \mathbf{N,} $$

and then

$$ J= \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{p}}. $$

By (26) we find \(J>0\). If \(J=\infty \), then (22) is evidently valid; if \(J<\infty \), then by (21) we have

$$\begin{aligned}& \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b_{n} ^{q} \\& \quad=J^{p}=I >K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum_{\vert m \vert =1}^{\infty }\bigl(1- \theta (\lambda_{2},m)\bigr)A_{ \xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{q}}, \\& J = \Biggl[ \sum_{\vert n \vert =1}^{\infty }A_{\eta ,\beta }^{q(1-\lambda_{2})-1}(n)b _{n}^{q} \Biggr] ^{\frac{1}{p}} > K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert m \vert =1}^{\infty }\bigl(1- \theta (\lambda_{2},m) \bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m} ^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned}$$

namely, (22) follows, which is equivalent to (21).

We have proved that (21) is valid. Then we set

$$ a_{m}:= \frac{A_{\xi ,\alpha }^{q\lambda_{1}-1}(m)}{(1-\theta ( \lambda_{2},m))^{q-1}} \Biggl( \sum _{\vert n\vert =1}^{ \infty }k(m,n)b_{n} \Biggr) ^{q-1},\quad \vert m\vert \in \mathbf{N,} $$

and find

$$ L= \Biggl[ \sum_{\vert m\vert =1}^{\infty }\bigl(1-\theta ( \lambda_{2},m)\bigr)A _{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \Biggr] ^{\frac{1}{q}}. $$

If \(L=0\), then (23) is impossible, so that \(L>0\). If \(L=\infty \), then (23) is trivially valid; if \(L<\infty \), then we have

$$\begin{aligned}& \sum_{\vert m\vert =1}^{\infty }\bigl(1-\theta ( \lambda_{2},m)\bigr)A _{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \\& \quad =L^{q}=I>K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert m\vert =1} ^{\infty }\bigl(1-\theta (\lambda_{2},m) \bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a _{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{\vert n\vert =1}^{\infty }A_{\eta , \beta }^{q(1-\lambda_{2})-1}(n)b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \\& L = \Biggl[ \sum_{\vert m\vert =1}^{\infty }\bigl(1-\theta ( \lambda_{2},m)\bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \Biggr] ^{\frac{1}{q}} > K_{\alpha ,\beta }(\lambda_{1}) \Biggl[ \sum _{\vert n\vert =1}^{\infty }A_{\eta ,\beta }^{q(1- \lambda_{2})-1}(n)b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$

thats is, (23) follows.

On the other-hand, assuming that (23) is valid, using the reverse Hölder inequality, we have

$$\begin{aligned} I =&\sum_{\vert m\vert =1}^{\infty } \biggl[ \frac{A_{ \xi ,\alpha }^{(1/q)-\lambda_{1}}(m)}{(1-\theta (\lambda_{2},m))^{-1/p}}a _{m} \biggr] \\ &\times \Biggl[ \frac{A_{\xi ,\alpha }^{\lambda_{1}-(1/q)}(m)}{(1- \theta (\lambda_{2},m))^{1/p}}\sum_{\vert n\vert =1}^{ \infty }k(m,n)b_{n} \Biggr] \\ \geq & \Biggl[ \sum_{\vert m\vert =1}^{\infty }\bigl(1- \theta ( \lambda_{2},m)\bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m)a_{m}^{p} \Biggr] ^{\frac{1}{p}}L, \end{aligned}$$
(28)

and then by (23) we have (21), which is equivalent to (23).

Therefore, inequalities (21), (22), and (23) are equivalent. □

Theorem 6

With regards to the assumptions of Theorem 5, the constant factor\(K _{\alpha ,\beta }(\lambda_{1})\) in (21), (22), and (23) is the best possible.

Proof

For \(0<\varepsilon <p\lambda_{1}\), we set \(\widetilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon }{p}\) (\(\in (0,1)\)), \(\widetilde{\lambda }_{2}=\lambda_{2}+\frac{\varepsilon }{p}\) (>0), and

$$\begin{aligned}& \widetilde{a}_{m}:=A_{\xi ,\alpha }^{\lambda_{1}-(\varepsilon /p)-1}(m)=A _{\xi ,\alpha }^{\widetilde{\lambda }_{1}-1}(m)\quad \bigl(\vert m\vert \in \mathbf{N}\bigr), \\& \widetilde{b}_{n}:=A_{\eta ,\beta }^{\lambda_{2}-(\varepsilon /q)-1}(n)=A_{\eta ,\beta }^{\widetilde{\lambda }_{2}-\varepsilon -1}(n)\quad \bigl(\vert n\vert \in \mathbf{N}\bigr). \end{aligned}$$

By (19) and (17) we find

$$\begin{aligned}& \widetilde{I}_{1}:= \Biggl[ \sum_{\vert m\vert =1}^{\infty } \bigl(1- \theta (\lambda_{2},m)\bigr)A_{\xi ,\alpha }^{p(1-\lambda_{1})-1}(m) \widetilde{a}_{m}^{p} \Biggr] ^{\frac{1}{p}}\Biggl[ \sum_{\vert n\vert =1}^{\infty }A_{\eta , \beta }^{q(1-\lambda_{2})-1}(n) \widetilde{b}_{n}^{q} \Biggr] ^{ \frac{1}{q}} \\& \hphantom{\widetilde{I}_{1}} = \Biggl[ \sum_{\vert m\vert =1}^{\infty }A_{\xi ,\alpha } ^{-1-\varepsilon }(m)-\sum_{\vert m\vert =1}^{\infty }O\bigl(A _{\xi ,\alpha }^{-1-(\lambda_{2}+\varepsilon )}(m)\bigr) \Biggr] ^{ \frac{1}{p}} \Biggl( \sum _{\vert n\vert =1}^{\infty }A_{ \eta ,\beta }^{-1-\varepsilon }(n) \Biggr) ^{\frac{1}{q}} \\& \hphantom{\widetilde{I}_{1}} =\frac{1}{\varepsilon }\bigl(2\csc^{2}\alpha +o(1)-\varepsilon O(1) \bigr)^{ \frac{1}{p}}\bigl(2\csc^{2}\beta +\widetilde{o}(1) \bigr)^{\frac{1}{q}}\quad \bigl( \varepsilon \rightarrow 0^{+}\bigr), \\& \widetilde{I}:=\sum_{\vert m\vert =1}^{\infty } \sum _{\vert n\vert =1}^{\infty }k(m,n)\widetilde{a}_{m} \widetilde{b}_{n} \\& \hphantom{\widetilde{I}}=\sum_{\vert n\vert =1}^{\infty } \sum_{\vert m\vert =1}^{\infty }k(m,n)A_{\xi ,\alpha }^{ \widetilde{\lambda }_{1}-1}(m)A_{\eta ,\beta }^{\widetilde{\lambda }_{2}-\varepsilon -1}(n) \\& \hphantom{\widetilde{I}} = \sum_{\vert n\vert =1}^{\infty }\varpi ( \widetilde{ \lambda }_{1},n)A_{\eta ,\beta }^{-1-\varepsilon }(n)< h _{\alpha }( \widetilde{\lambda }_{1})\sum_{\vert n\vert =2} ^{\infty }A_{\eta ,\beta }^{-1-\varepsilon }(n) \\& \hphantom{\widetilde{I}} = \frac{1}{\varepsilon }h_{\alpha }\biggl(\lambda_{1}- \frac{\varepsilon }{p}\biggr) \bigl(2\csc^{2}\beta +\widetilde{o}(1) \bigr). \end{aligned}$$

If there exists a positive number \(K\geq K_{\alpha ,\beta }(\lambda _{1})\) such that (21) is still valid when replacing \(K_{\alpha ,\beta }(\lambda_{1})\) by K, then, in particular, we have

$$ \varepsilon \widetilde{I}=\varepsilon \sum_{\vert m\vert =1} ^{\infty }\sum_{\vert n\vert =1}^{\infty }k(m,n) \widetilde{a}_{m}\widetilde{b}_{n}>\varepsilon K \widetilde{I}_{1}. $$

In view of the preceding results, it follows that

$$\begin{aligned}& h_{\alpha }\biggl(\lambda_{1}-\frac{\varepsilon }{p}\biggr) \bigl(2\csc^{2}\beta + \widetilde{o}(1)\bigr) >K\cdot \bigl(2\csc^{2}\alpha +o(1)-\varepsilon O(1) \bigr)^{1/p}\bigl(2\csc^{2} \beta +\widetilde{o}(1) \bigr)^{1/q}, \end{aligned}$$

and then

$$ 4k_{s}(\lambda_{1})\csc^{2}\alpha\csc^{2}\beta \geq 2K\csc^{2/p} \alpha \csc^{2/q}\beta \quad \bigl(\varepsilon \rightarrow 0^{+}\bigr), $$

namely,

$$ K_{\alpha ,\beta }(\lambda_{1})=2k_{s}(\lambda_{1})\csc^{2/p}\beta \csc^{2/q}\alpha \geq K. $$

Hence \(K=K_{\alpha ,\beta }(\lambda_{1})\) is the best possible constant factor in (21).

The constant factor \(K_{\alpha ,\beta }(\lambda_{1})\) in (22) ((23)) is still the best possible. Otherwise, we would reach a contradiction by (27) ((28)) that the constant factor in (21) is not the best possible. □

Remark 1

(i) For \(\xi =\eta =0\) and \(\alpha =\beta =\frac{\pi }{2}\) in (21), setting

$$\begin{aligned} \widetilde{\theta }(\lambda_{2},m) &:=\frac{1}{k_{s}(\lambda_{1})} \int_{\vert m \vert }^{\infty }\frac{u^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u^{ \lambda }+c_{k})}\,du \\ &=O \biggl( \frac{1}{\vert m \vert ^{\lambda_{2}}} \biggr) \in (0,1),\quad \vert m \vert \in \mathbf{N}, \end{aligned}$$

we have the following new reverse inequality with the best possible constant factor \(2k_{s}(\lambda_{1})\):

$$\begin{aligned}& \sum_{\vert n \vert =1}^{\infty }\sum _{\vert m \vert =1}^{\infty }\frac{1}{\prod_{k=1} ^{s}(\vert m \vert ^{\lambda }+c_{k}\vert n \vert ^{\lambda })}a_{m}b_{n} \\& \quad >2k_{s}( \lambda _{1}) \Biggl[ \sum_{\vert m \vert =1}^{\infty }\bigl(1- \widetilde{\theta }(\lambda_{2},m)\bigr)\vert m \vert ^{p(1- \lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum _{\vert n \vert =1}^{ \infty }\vert n \vert ^{q(1-\lambda_{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(29)

It follows that (21) is an extension of (29).

(ii) If \(a_{-m}=a_{m}\) and \(b_{-n}=b_{n}\) (\(m,n\in \mathbf{N}\)), then for

$$\begin{aligned} \widetilde{\theta }(\lambda_{2},m) =&\frac{1}{k_{s}(\lambda_{1})} \int _{m}^{\infty }\frac{u^{\lambda_{1}-1}}{\prod_{k=1}^{s}(u^{\lambda }+c _{k})}\,du \\ =&O \biggl( \frac{1}{m^{\lambda_{2}}} \biggr) \in (0,1),\quad m\in \mathbf{N}, \end{aligned}$$

(29) reduces to the following reverse Hilbert-type inequality:

$$\begin{aligned}& \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{1}{\prod_{k=1}^{s}(m ^{\lambda }+c_{k}n^{\lambda })}a_{m}b_{n} \\& \quad > k_{s}(\lambda_{1}) \Biggl[ \sum _{m=1}^{\infty }\bigl(1- \widetilde{\theta }( \lambda_{2},m)\bigr)m^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty }n^{q(1-\lambda_{2})-1}b_{n} ^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(30)

(iii) If \(a_{-m}=a_{m}\) and \(b_{-n}=b_{n}\) (\(m,n \in \mathbf{N}\)), then setting

$$\begin{aligned} \theta_{2}(\lambda_{2},m) :=&\frac{\lambda \sin (\frac{\pi \lambda _{1}}{\lambda })}{\pi } \int_{\frac{m-\xi }{1+\eta }}^{\infty }\frac{u ^{\lambda_{1}-1}}{u^{\lambda }+1}\,du \\ =&O \biggl( \frac{1}{(m-\xi )^{\lambda_{2}}} \biggr) \in (0,1),\quad m\in \mathbf{N}, \\ \theta_{3}(\lambda_{2},m) :=&\frac{\lambda \sin (\frac{\pi \lambda _{1}}{\lambda })}{\pi } \int_{\frac{m+\xi }{1+\eta }}^{\infty }\frac{u ^{\lambda_{1}-1}}{u^{\lambda }+1}\,du \\ =&O \biggl( \frac{1}{(m+\xi )^{\lambda_{2}}} \biggr) \in (0,1),\quad m\in \mathbf{N}, \end{aligned}$$

(6) reduces to

$$\begin{aligned}& \sum_{n=1}^{\infty }\sum _{m=1}^{\infty } \biggl[ \frac{1}{(m-\xi )^{ \lambda }+(n-\eta )^{\lambda }}+ \frac{1}{(m-\xi )^{\lambda }+(n+ \eta )^{\lambda }} \\& \quad \quad {}+\frac{1}{(m+\xi )^{\lambda }+(n-\eta )^{\lambda }}+\frac{1}{(m+ \xi )^{\lambda }+(n+\eta )^{\lambda }} \biggr] a_{m}b_{n} \\& \quad > \frac{2\pi }{\lambda \sin (\pi \lambda_{1}/\lambda )} \Biggl\{ \sum_{m=1}^{\infty } \bigl[ \bigl(1-\theta_{2}(\lambda_{2},m)\bigr) (m-\xi )^{p(1- \lambda_{1})-1} \\& \quad \quad {} +\bigl(1-\theta_{3}(\lambda_{2},m) \bigr) (m+\xi )^{p(1-\lambda_{1})-1} \bigr] a _{m}^{p} \Biggr\} ^{\frac{1}{p}} \\& \quad \quad {}\times \Biggl\{ \sum_{n=1}^{\infty } \bigl[ (n- \eta )^{q(1-\lambda_{2})-1}+(n+ \eta )^{q(1-\lambda_{2})-1} \bigr] b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(31)

In particular, for \(\xi =\eta =0\), \(\lambda =1\), \(\lambda_{1}=\lambda_{2}=\frac{1}{2}\) in (31) (or for \(s=\lambda =c_{1}=1\), \(\lambda_{1}=\lambda_{2}=\frac{1}{2}\) in (30)), setting

$$ \theta (m):=\frac{1}{\pi } \int_{m}^{\infty }\frac{u^{-1/2}}{u+1}\,du=O \biggl( \frac{1}{m^{1/2}} \biggr) \in (0,1),\quad m\in \mathbf{N}, $$

we have the following reverse Hardy–Hilbert inequality with the best possible constant π:

$$\begin{aligned}& \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n} > \pi \Biggl[ \sum_{m=1}^{\infty }\bigl(1-\theta (m)\bigr)m^{\frac{p}{2}-1}a _{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty }n^{ \frac{q}{2}-1}b_{n}^{q} \Biggr) ^{\frac{1}{q}}. \end{aligned}$$

Hence (21) is an extended reverse Hardy–Hilbert’s inequality in the whole plane.

4 Conclusions

In this paper, using the weight coefficients, a complex integral formula, and Hermite–Hadamard’s inequality, we give an extended reverse Hardy–Hilbert’s inequality in the whole plane with multiparameters and a best possible constant factor (Theorems 5 and 6). We consider equivalent forms and a few particular cases. The technique of real analysis is very important, which is the key to prove the reverse equivalent inequalities with the best possible constant factor. The lemmas and theorems provide an extensive account of this type inequalities.