1 Introduction

Assuming that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(a_{n},b_{n}\geq0\), \(0<\sum_{n=1}^{\infty}a_{n}^{p}<\infty\), and \(0<\sum_{n=1}^{\infty }b_{n}^{q}<\infty\), we have the following Hardy–Hilbert inequality (cf. [1]):

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

where the constant factor \(\frac{\pi}{\sin(\pi/p)}\) is the best possible. We still have the following Mulholland inequality with the same best possible constant factor \(\frac{\pi}{\sin(\pi/p)}\) (cf. Theorem 343 of [1], replacing \(\frac{a_{m}}{m}\), \(\frac{b_{n}}{n}\) by \(a_{m}\), \(b_{n}\)):

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

Inequalities (1)–(2) are important in the analysis and its applications (cf. [1, 2]).

In 2007, Yang [3] first gave a Hilbert-type integral inequality in the whole plane as follows:

$$\begin{aligned} & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac{f(x)g(y)}{(1+e^{x+y})^{\lambda}}\,dx\,dy \\ &\quad < B \biggl(\frac{\lambda}{2},\frac{\lambda}{2} \biggr) \biggl( \int_{-\infty}^{\infty }e^{-\lambda x}f^{2}(x)\,dx \int_{-\infty}^{\infty}e^{-\lambda y}f^{2}(y)\,dy \biggr)^{\frac{1}{2}}, \end{aligned}$$
(3)

where the constant factor \(B(\frac{\lambda}{2},\frac{\lambda}{2})\) (\(\lambda>0\)) is the best possible. Some new results on inequalities (1)–(3) were provided by [4–24]. In 2016, Yang and Chen [25] gave a more accurate extension of (1) in the whole plane as follows:

$$\begin{aligned} &\sum_{ \vert n \vert =1}^{\infty}\sum _{ \vert m \vert =1}^{\infty}\frac{a_{m}b_{n}}{ ( \vert m-\xi \vert + \vert n-\eta \vert ) ^{\lambda}} \\ &\quad < 2B ( \lambda _{1},\lambda_{2} ) \Biggl[ \sum _{ \vert m \vert =1}^{\infty} \vert m-\xi \vert ^{p(1-\lambda_{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{ \vert n \vert =1}^{\infty} \vert n-\eta \vert ^{q(1-\lambda_{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
(4)

where the constant factor \(2B ( \lambda_{1},\lambda_{2} ) \) (\(0<\lambda_{1},\lambda_{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda \), \(\xi , \eta\in{[} 0,\frac{1}{2}]\)) is the best possible. Another result was provided by Xin et al. [26].

In this paper, by introducing multi-parameters, applying the weight coefficients and Hermite–Hadamard’s inequality, we give a reverse of the extension of Mulholland’s inequality (2) in the whole plane with the best possible constant factor similar to the reverse of (4). The equivalent forms and a few particular cases are also considered.

2 Some lemmas

In the following, we agree that \(0< p<1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\lambda_{1},\lambda_{2}>0\), \(\lambda_{1}+\lambda_{2}=\lambda\leq 1\), \(\alpha,\beta\in{[} \arccos\frac{1}{3},\frac{\pi}{2}]\), \(\xi ,\eta \in{[} 0,\frac{1}{2}]\), and

$$ H_{\gamma}(\lambda_{1}):=\frac{2\pi\csc^{2}\gamma}{\lambda\sin (\frac{\pi\lambda_{1}}{\lambda})}\quad (\gamma= \alpha,\beta). $$
(5)

Remark 1

In view of the conditions that \(\alpha,\beta\in{[} \arccos\frac{1}{3},\frac{\pi}{2}]\), \(\xi,\eta\in{[} 0,\frac{1}{2}]\), it follows that \((\frac{3}{2}\pm\eta)(1\mp\cos\beta)\geq1\) and \((\frac{3}{2}\pm\xi)(1\mp\cos\alpha)\geq1\).

We set the following functions:

$$ A_{\alpha}(\xi,x)= \vert x-\xi \vert +(x-\xi)\cos\alpha, $$

\(B_{\beta}(\eta,y)= \vert y-\eta \vert +(y-\eta)\cos\beta\), and

$$ H(x,y):=\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,x)+\ln^{\lambda }B_{\beta }(\eta,y)}\quad \biggl( \vert x \vert , \vert y \vert > \frac{3}{2} \biggr). $$
(6)

Definition 1

Define two weight coefficients as follows:

$$\begin{aligned} &\omega(\lambda_{2},m):=\sum_{ \vert n \vert =2}^{\infty} \frac{H(m,n)}{B_{\beta}(\eta,n)}\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi ,m)}{\ln^{1-\lambda_{2}}B_{\beta}(\eta,n)}, \quad \vert m \vert \in \mathbf{N} \backslash \{1\}, \end{aligned}$$
(7)
$$\begin{aligned} &\varpi(\lambda_{1},n):=\sum_{ \vert m \vert =2}^{\infty} \frac{H(m,n)}{A_{\alpha}(\xi,m)}\frac{\ln^{\lambda_{2}}B_{\beta}(\eta ,n)}{\ln^{1-\lambda_{1}}A_{\alpha}(\xi,m)}, \quad \vert n \vert \in \mathbf{N} \backslash \{1\}, \end{aligned}$$
(8)

where \(\sum_{ \vert j \vert =2}^{\infty}\cdots =\sum_{j=-2}^{-\infty}\cdots+\sum_{j=2}^{\infty}\cdots\) (\(j=m,n\)).

Lemma 1

(cf. [26])

Suppose that \(g(t)\) (>0) is strictly decreasing in \((1,\infty)\), satisfying \(\int_{1}^{\infty}g(t)\,dt\in \mathbf{R}_{+}\). We have

$$ \int_{2}^{\infty}g(t)\,dt< \sum _{n=2}^{\infty}g(n)< \int_{1}^{\infty}g(t)\,dt. $$
(9)

If \((-1)^{i}g(t)>0\) (\(i=0,1,2\); \(t\in(\frac{3}{2},\infty)\)), \(\int_{\frac {3}{2}}^{\infty}g(t)\,dt\in\mathbf{R}_{+}\), then we have the following Hermite–Hadamard inequality (cf. [27]):

$$ \sum_{n=2}^{\infty}g(n)< \int_{\frac{3}{2}}^{\infty}g(t)\,dt. $$
(10)

Lemma 2

For \(0<\lambda\leq1\), \(0<\lambda_{2}<1\), the following inequalities are valid:

$$ H_{\beta}(\lambda_{1}) \bigl(1-\theta(\lambda _{2},m) \bigr)< \omega(\lambda_{2},m)< H_{\beta}( \lambda_{1}),\quad \vert m \vert \in\mathbf{N}\backslash \{1\}, $$
(11)

where

$$\begin{aligned} \theta(\lambda_{2},m) :=&\frac{\lambda}{\pi}\sin \biggl( \frac{\pi\lambda_{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln[(2+\eta)(1+\cos\beta)]}{\ln A_{\alpha }(\xi,m)}}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{2}}A_{\alpha}(\xi,m)} \biggr) \in(0,1). \end{aligned}$$
(12)

Proof

For \(\vert m \vert \in\mathbf{N}\backslash\{ 1\}\), we set the following functions:

$$\begin{aligned} &H^{(1)}(m,y) :=\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,m)+\ln ^{\lambda }[(y-\eta)(\cos\beta-1)]},\quad y< -\frac{3}{2}, \\ &H^{(2)}(m,y) :=\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,m)+\ln ^{\lambda }[(y-\eta)(\cos\beta+1)]},\quad y>\frac{3}{2}, \end{aligned}$$

wherefrom

$$ H^{(1)}(m,-y)=\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,m)+\ln ^{\lambda }[(y+\eta)(1-\cos\beta)]},\quad y>\frac{3}{2}. $$

We find

$$\begin{aligned} \omega(\lambda_{2},m) =&\sum_{n=-2}^{-\infty} \frac{H^{(1)}(m,n)\ln ^{\lambda_{1}}A_{\alpha}(\xi,m)}{(n-\eta)(\cos\beta-1)\ln ^{1-\lambda _{2}} [ (n-\eta)(\cos\beta-1) ] } \\ &{}+\sum_{n=2}^{\infty}\frac{H^{(2)}(m,n)\ln^{\lambda_{1}}A_{\alpha }(\xi ,m)}{(n-\eta)(1+\cos\beta)\ln^{1-\lambda_{2}} [ (n-\eta)(1+\cos \beta) ] } \\ =&\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi,m)}{1-\cos\beta}\sum_{n=2}^{\infty} \frac{H^{(1)}(m,-n)}{(n+\eta)\ln^{1-\lambda _{2}}[(n+\eta)(1-\cos\beta)]} \\ &{}+\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi,m)}{1+\cos\beta}\sum_{n=2}^{\infty} \frac{H^{(2)}(m,n)}{(n-\eta)\ln^{1-\lambda _{2}}[(n-\eta)(1+\cos\beta)]}. \end{aligned}$$
(13)

In virtue of \(0<\lambda\leq1\), \(0<\lambda_{2}<1\), we find that for \(y>\frac{3}{2}\),

$$\begin{aligned} &\frac{d}{dy}\frac{H^{(i)}(m,(-1)^{i}y)}{(y-(-1)^{i}\eta)\ln ^{1-\lambda _{2}}[(y-(-1)^{i}\eta)(1+(-1)^{i}\cos\beta)]} < 0, \\ &\frac{d^{2}}{dy^{2}}\frac{H^{(i)}(m,(-1)^{i}y)}{(y-(-1)^{i}\eta)\ln ^{1-\lambda_{2}}[(y-(-1)^{i}\eta)(1+(-1)^{i}\cos\beta)]} >0\quad (i=1.2), \end{aligned}$$

it follows that \(\frac{H^{(i)}(m,(-1)^{i}y)}{(y-(-1)^{i}\eta)\ln ^{1-\lambda_{2}}[(y-(-1)^{i}\eta)(1+(-1)^{i}\cos\beta)]}\) (\(i=1.2\)) are strictly decreasing and strictly convex in \((\frac{3}{2},\infty)\). By (10) and (13) we find

$$\begin{aligned} \omega(\lambda_{2},m) < &\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi ,m)}{1-\cos\beta} \int_{3/2}^{\infty}\frac{H^{(1)}(m,-y)\,dy}{(y+\eta)\ln ^{1-\lambda_{2}}[(y+\eta)(1-\cos\beta)]} \\ &{}+\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi,m)}{1+\cos\beta}\int_{3/2}^{\infty}\frac{H^{(2)}(m,y)\,dy}{(y-\eta)\ln^{1-\lambda _{2}}[(y-\eta)(1+\cos\beta)]}. \end{aligned}$$

Setting \(u=\frac{\ln[(y+\eta)(1-\cos\beta)]}{\ln A_{\alpha}(\xi ,m)}\) (\(u=\frac{\ln[(y-\eta)(1+\cos\beta)]}{\ln A_{\alpha}(\xi,m)}\)) in the above first (second) integral, in view of Remark 1, by simplifications, we obtain

$$\begin{aligned} \omega(\lambda_{2},m) < & \biggl( \frac{1}{1-\cos\beta}+ \frac{1}{1+\cos \beta} \biggr) \int_{0}^{\infty}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&\frac{2\csc^{2}\beta}{\lambda} \int_{0}^{\infty}\frac{v^{(\lambda _{2}/\lambda)-1}\,dv}{1+v}= \frac{2\pi\csc^{2}\beta}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}=H_{\beta}(\lambda_{1}). \end{aligned}$$

By (9) and (13), in the same way, we still have

$$\begin{aligned} \omega(\lambda_{2},m) >&\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi ,m)}{1-\cos\beta} \int_{2}^{\infty}\frac{H^{(1)}(m,-y)\,dy}{(y+\eta)\ln ^{1-\lambda_{2}}[(y+\eta)(1-\cos\beta)]} \\ &{}+\frac{\ln^{\lambda_{1}}A_{\alpha}(\xi,m)}{1+\cos\beta}\int_{2}^{\infty}\frac{H^{(2)}(m,y)\,dy}{(y-\eta)\ln^{1-\lambda _{2}}[(y-\eta)(1+\cos\beta)]} \\ \geq& \biggl( \frac{1}{1-\cos\beta}+\frac{1}{1+\cos\beta} \biggr) \int_{\frac{\ln[(2+\eta)(1+\cos\beta)]}{\ln A_{\alpha}(\xi ,m)}}^{\infty}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&H_{\beta}(\lambda_{1})-2\csc^{2}\beta \int_{0}^{\frac{\ln[(2+\eta )(1+\cos\beta)]}{\ln A_{\alpha}(\xi,m)}}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&H_{\beta}(\lambda_{1}) \bigl(1-\theta(\lambda _{2},m) \bigr)>0, \end{aligned}$$

where \(\theta(\lambda_{2},m)\) is indicated by (12). It follows that \(\theta(\lambda_{2},m)<1\) and

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

Hence, (11) and (12) are valid. □

In the same way, for \(0<\lambda\leq1\), \(0<\lambda_{1}<1\), we still have the following.

Lemma 3

The following inequalities are valid:

$$ H_{\alpha}(\lambda_{1}) \bigl(1-\widetilde{\theta}(\lambda _{1},n) \bigr)< \varpi(\lambda_{1},n)< H_{\alpha}( \lambda_{1}), \quad \vert n \vert \in\mathbf{N}\backslash\{1\}, $$
(14)

where

$$\begin{aligned} \widetilde{\theta}(\lambda_{1},n) :=&\frac{\lambda}{\pi}\sin \biggl( \frac{\pi \lambda_{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln[(2+\xi)(1+\cos\alpha)]}{\ln B_{\beta}(\eta,n)}}\frac{u^{\lambda_{1}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{1}}B_{\beta}(\eta,n)} \biggr) \in(0,1). \end{aligned}$$
(15)

Lemma 4

If \((\zeta,\gamma)=(\xi,\alpha)\) (or \((\eta ,\beta)\)), \(\rho>0\), then we have

$$\begin{aligned} h_{\rho}(\zeta,\gamma) :=&\sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{-1-\rho} [ \vert n-\zeta \vert +(n-\zeta)\cos\gamma ] }{ [ \vert n-\zeta \vert +(n-\zeta)\cos\gamma ] } \\ =&\frac{1}{\rho} \bigl(2\csc^{2}\gamma+o(1) \bigr)\quad \bigl( \rho\rightarrow0^{+} \bigr) . \end{aligned}$$
(16)

Proof

Still by (10) we find

$$\begin{aligned} h_{\rho}(\zeta,\gamma) =&\sum_{n=-2}^{-\infty} \frac{\ln^{-1-\rho }[(n-\zeta)(\cos\gamma-1)]}{(n-\zeta)(\cos\gamma-1)} \\ &{}+\sum_{n=2}^{\infty}\frac{\ln^{-1-\rho}[(n-\zeta)(\cos\gamma +1)]}{(n-\zeta)(\cos\gamma+1)} \\ =&\sum_{n=2}^{\infty} \biggl\{ \frac{\ln^{-1-\rho}[(n+\zeta)(1-\cos \gamma)]}{(n+\zeta)(1-\cos\gamma)}+\frac{\ln^{-1-\rho}[(n-\zeta )(\cos \gamma+1)]}{(n-\zeta)(\cos\gamma+1)} \biggr\} \\ \leq& \int_{\frac{3}{2}}^{\infty} \biggl\{ \frac{\ln^{-1-\rho }[(y+\zeta )(1-\cos\gamma)]}{(y+\zeta)(1-\cos\gamma)}+ \frac{\ln^{-1-\rho }[(y-\zeta)(\cos\gamma+1)]}{(y-\zeta)(\cos\gamma+1)} \biggr\} \,dy \\ =&\frac{1}{\rho} \biggl\{ \frac{\ln^{-\rho}[(\frac{3}{2}+\zeta )(1-\cos \gamma)]}{1-\cos\gamma}+\frac{\ln^{-\rho}[(\frac{3}{2}-\zeta )(1+\cos \gamma)]}{1+\cos\gamma} \biggr\} \\ =&\frac{1}{\rho} \biggl\{ \frac{1}{1-\cos\gamma}+\frac {1}{1+\cos\gamma}+o_{1}(1) \biggr\} \quad \bigl(\rho\rightarrow0^{+} \bigr). \end{aligned}$$

By (9), we still can find that

$$\begin{aligned} &h_{\rho}(\zeta,\gamma) \\ &\quad =\sum_{n=2}^{\infty} \biggl\{ \frac{\ln^{-1-\rho}[(n+\zeta)(1-\cos \gamma)]}{(n+\zeta)(1-\cos\gamma)}+\frac{\ln^{-1-\rho}[(n-\zeta )(\cos \gamma+1)]}{(n-\zeta)(\cos\gamma+1)} \biggr\} \\ &\quad \geq \int_{2}^{\infty} \biggl\{ \frac{\ln^{-1-\rho}[(y+\zeta)(1-\cos \gamma)]}{(y+\zeta)(1-\cos\gamma)}+ \frac{\ln^{-1-\rho}[(y-\zeta)(\cos \gamma+1)]}{(y-\zeta)(\cos\gamma+1)} \biggr\} \,dy \\ &\quad = \frac{1}{\rho} \biggl\{ \frac{\ln^{-\rho}[(2+\zeta)(1-\cos \gamma)]}{1-\cos\gamma}+\frac{\ln^{-\rho}[(2-\zeta)(1+\cos\gamma )]}{1+\cos \gamma} \biggr\} \\ &\quad = \frac{1}{\rho} \biggl\{ \frac{1}{1-\cos\gamma}+\frac {1}{1+\cos\gamma}+o_{2}(1) \biggr\} \quad \bigl(\rho\rightarrow0^{+} \bigr). \end{aligned}$$

Hence, we prove that (16) is valid. □

3 Main results and a few particular cases

We also define

$$ H(\lambda_{1}):=H_{\beta}^{1/p}(\lambda _{1})H_{\alpha}^{1/q}(\lambda_{1})= \frac{2\pi\csc^{2/p}\beta\csc^{2/q}\alpha}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}. $$
(17)

Theorem 1

Suppose that \(a_{m},b_{n}\geq0\) (\(\vert m \vert , \vert n \vert \in\mathbf{N}\backslash\{1\}\)) and

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

We have the following reverse equivalent inequalities:

$$\begin{aligned} &\begin{aligned}[b] I:={}&\sum_{ \vert n \vert =2}^{\infty} \sum_{ \vert m \vert =2}^{\infty}\frac{1}{\ln^{\lambda}A_{\alpha}(\xi,m)+\ln^{\lambda }B_{\beta}(\eta,n)}a_{m}b_{n} \\ >{}&H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \end{aligned}$$
(18)
$$\begin{aligned} &\begin{aligned}[b] J :={}& \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{p\lambda _{2}-1}B_{\beta}(\eta,n)}{B_{\beta}(\eta,n)} \Biggl( \sum_{ \vert m \vert =2}^{\infty} \frac{a_{m}}{\ln^{\lambda}A_{\alpha}(\xi ,m)+\ln^{\lambda}B_{\beta}(\eta,n)} \Biggr) ^{p} \Biggr] ^{\frac {1}{p}} \\ >{}&H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{1/p}, \end{aligned} \end{aligned}$$
(19)
$$\begin{aligned} &\begin{aligned}[b] L :={}& \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{\ln^{q\lambda _{1}-1}A_{\alpha}(\xi,m)}{(1-\theta(\lambda_{2},m))^{q-1}A_{\alpha }(\xi ,m)} \\ &{}\times \Biggl( \sum_{ \vert n \vert =2}^{\infty} \frac{1}{\ln ^{\lambda}A_{\alpha}(\xi,m)+\ln^{\lambda}B_{\beta}(\eta,n)}b_{n} \Biggr) ^{q} \Biggr] ^{\frac{1}{q}} \\ >{}&H(\lambda_{1}) \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta ,n))^{1-q}}b_{n}^{q} \Biggr] ^{1/q}. \end{aligned} \end{aligned}$$
(20)

In particular, (i) for \(\alpha=\beta=\frac{\pi}{2}\), \(\xi,\eta\in {}[ 0,\frac{1}{2}]\), setting

$$\begin{aligned} \theta_{1}(\lambda_{2},m) :=&\frac{\lambda}{\pi}\sin \biggl(\frac{\pi\lambda _{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln(2+\eta)}{\ln \vert m-\xi \vert }}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{2}} \vert m-\xi \vert } \biggr) \in(0,1), \end{aligned}$$
(21)

we have the following reverse equivalent inequalities:

$$\begin{aligned} &\begin{aligned}[b] &\sum_{ \vert n \vert =2}^{\infty}\sum _{ \vert m \vert =2}^{\infty}\frac{1}{\ln^{\lambda} \vert m-\xi \vert +\ln ^{\lambda} \vert n-\eta \vert }a_{m}b_{n} \\ &\quad >\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})} \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1-\theta_{1}(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1} \vert m-\xi \vert }{ \vert m-\xi \vert ^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &\qquad {}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1} \vert n-\eta \vert }{ \vert n-\eta \vert ^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned} \end{aligned}$$
(22)
$$\begin{aligned} &\begin{aligned}[b] & \Biggl\{ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{p\lambda _{2}-1} \vert n-\eta \vert }{ \vert n-\eta \vert } \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{a_{m}}{\ln^{\lambda } \vert m-\xi \vert +\ln^{\lambda} \vert n-\eta \vert } \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\ &\quad >\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})} \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1-\theta_{1}(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1} \vert m-\xi \vert }{ \vert m-\xi \vert ^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned} \end{aligned}$$
(23)
$$\begin{aligned} &\begin{aligned}[b] & \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{\ln^{q\lambda _{1}-1} \vert m-\xi \vert }{(1-\theta_{1}(\lambda_{2},m))^{q-1} \vert m-\xi \vert } \Biggl( \sum_{ \vert n \vert =2}^{\infty} \frac{1}{\ln ^{\lambda} \vert m-\xi \vert +\ln^{\lambda} \vert n-\eta \vert }b_{n} \Biggr) ^{q} \Biggr] ^{\frac{1}{q}} \\ &\quad >\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})} \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda _{2})-1} \vert n-\eta \vert }{ \vert n-\eta \vert ^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned} \end{aligned}$$
(24)

(ii) For \(\xi,\eta=0\), \(\alpha,\beta\in{[} \arccos\frac{1}{3},\frac{\pi}{2}]\), setting

$$\begin{aligned} \theta_{2}(\lambda_{2},m) :=&\frac{\lambda}{\pi}\sin \biggl(\frac{\pi\lambda _{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln2(1+\cos\beta)}{\ln( \vert m \vert +m\cos \alpha)}}\frac{u^{\lambda_{2}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{2}}( \vert m \vert +m\cos \alpha)} \biggr) \in(0,1), \end{aligned}$$
(25)

we have the following equivalent inequalities:

$$\begin{aligned} &\begin{aligned}[b] &\sum_{ \vert n \vert =2}^{\infty}\sum _{ \vert m \vert =2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda}( \vert m \vert +m\cos\alpha)+\ln^{\lambda}( \vert n \vert +n\cos\beta)} \\ &\quad >H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta_{2}(\lambda_{2},m) \bigr) \frac{\ln^{p(1-\lambda_{1})-1}( \vert m \vert +m\cos\alpha)}{( \vert m \vert +m\cos\alpha)^{1-p}} a_{m}^{p} \Biggr] ^{1/p} \\ &\qquad {}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}( \vert n \vert +n\cos\beta)}{( \vert n \vert +n\cos\beta)^{1-q}}b_{n}^{q} \Biggr] ^{1/q}, \end{aligned} \end{aligned}$$
(26)
$$\begin{aligned} &\begin{aligned}[b] & \Biggl\{ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{p\lambda _{2}-1}( \vert n \vert +n\cos\beta)}{ \vert n \vert +n\cos\beta} \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{a_{m}}{\ln^{\lambda}( \vert m \vert +m\cos\alpha)+\ln^{\lambda }( \vert n \vert +n\cos\beta)} \Biggr] ^{p} \Biggr\} ^{\frac {1}{p}} \\ &\quad >H(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert =2}^{\infty } \bigl(1-\theta_{2}(\lambda_{2},m) \bigr) \frac{\ln^{p(1-\lambda_{1})-1}( \vert m \vert +m\cos\alpha)}{( \vert m \vert +m\cos\alpha)^{1-p}} a_{m}^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned} \end{aligned}$$
(27)
$$\begin{aligned} &\begin{aligned}[b] & \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{\ln^{q\lambda _{1}-1}( \vert m \vert +m\cos\alpha)}{(1-\theta_{2}(\lambda _{2},m))^{q-1}( \vert m \vert +m\cos \alpha)} \\ &\qquad {}\times \Biggl( \sum_{ \vert n \vert =2}^{\infty} \frac{1}{\ln ^{\lambda}( \vert m \vert +m\cos\alpha)+\ln^{\lambda}( \vert n \vert +n\cos\beta)}b_{n} \Biggr) ^{q} \Biggr] ^{\frac{1}{q}} \\ &\quad > H(\lambda_{1}) \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{q(1-\lambda _{2})-1}( \vert n \vert +n\cos\beta)}{( \vert n \vert +n\cos \beta)^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned} \end{aligned}$$
(28)

Proof

By the reverse Hölder inequality with weight (cf. [28]) and (8), we find

$$\begin{aligned} & \Biggl( \sum_{ \vert m \vert =2}^{\infty}H(m,n)a_{m} \Biggr) ^{p} \\ &\quad = \Biggl\{ \sum_{ \vert m \vert =2}^{\infty}H(m,n) \biggl[ \frac{(A_{\alpha}(\xi,m))^{1/q}\ln^{(1-\lambda_{1})/q}A_{\alpha}(\xi ,m)}{\ln ^{(1-\lambda_{2})/p}B_{\beta}(\eta,n)}a_{m} \biggr] \\ &\qquad {}\times \biggl[ \frac{\ln^{(1-\lambda_{2})/p}B_{\beta}(\eta,n)}{(A_{\alpha}(\xi,m))^{1/q}\ln^{(1-\lambda_{1})/q}A_{\alpha}(\xi,m)} \biggr] \Biggr\} ^{p} \\ &\quad \geq \sum_{ \vert m \vert =2}^{\infty}H(m,n) \frac{(A_{\alpha }(\xi,m))^{p/q}\ln^{(1-\lambda_{1})p/q}A_{\alpha}(\xi,m)}{\ln ^{1-\lambda_{2}}B_{\beta}(\eta,n)}a_{m}^{p} \\ &\qquad {}\times \Biggl[ \sum_{ \vert m \vert =2}^{\infty}H(m,n) \frac{\ln ^{(1-\lambda_{2})q/p}B_{\beta}(\eta,n)}{A_{\alpha}(\xi,m)\ln ^{1-\lambda_{1}}A_{\alpha}(\xi,m)} \Biggr] ^{p-1} \\ &\quad =\frac{(\varpi(\lambda_{1},n))^{p-1}B_{\beta}(\eta,n)}{\ln ^{p\lambda _{2}-1}B_{\beta}(\eta,n)}\sum_{ \vert m \vert =2}^{\infty} \frac{H(m,n)(A_{\alpha}(\xi,m))^{\frac{p}{q}}\ln^{(1-\lambda_{1})\frac {p}{q}}A_{\alpha}(\xi,m)}{B_{\beta}(\eta,n)\ln^{1-\lambda_{2}}B_{\beta }(\eta,n)}a_{m}^{p}. \end{aligned}$$

By (14), in view of \(p-1<0\), it follows that

$$\begin{aligned} J >&H_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert n \vert =2}^{\infty}\sum_{ \vert m \vert =2}^{\infty} \frac{H(m,n)(A_{\alpha}(\xi,m))^{p/q}\ln^{(1-\lambda_{1})p/q}A_{\alpha }(\xi ,m)}{B_{\beta}(\eta,n)\ln^{1-\lambda_{2}}B_{\beta}(\eta ,n)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&H_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert m \vert =2}^{\infty}\sum_{ \vert n \vert =2}^{\infty} \frac{H(m,n)(A_{\alpha}(\xi,m))^{p/q}\ln^{(1-\lambda_{1})p/q}A_{\alpha }(\xi ,m)}{B_{\beta}(\eta,n)\ln^{1-\lambda_{2}}B_{\beta}(\eta ,n)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&H_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert m \vert =2}^{\infty}\omega(\lambda_{2},m) \frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi,m)}{(A_{\alpha}(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$
(29)

By (11) and (17), we have (19).

Using the reverse Hölder inequality again, we have

$$\begin{aligned} I =&\sum_{ \vert n \vert =2}^{\infty} \Biggl[ \frac{\ln^{\lambda _{2}-(1/p)}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1/p}}\sum_{ \vert m \vert =2}^{\infty}H(m,n)a_{m} \Biggr] \biggl[ \frac{\ln ^{(1/p)-\lambda_{2}}B_{\beta}(\eta,n)}{(B_{\beta}(\eta ,n))^{-1/p}}b_{n} \biggr] \\ \geq&J \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
(30)

and then by (19) we have (18).

On the other hand, assuming that (18) is valid, we set

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

and find

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

By (29) it follows that \(J>0\). If \(J=\infty\), then (19) is trivially valid; if \(J<\infty\), then we have

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

Hence (19) is valid, which is equivalent to (18).

We have proved that (18) is valid. Then we set

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

and find

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

If \(L=0\), then (20) is impossible, namely \(L>0\). If \(L=\infty \), then (20) is trivially valid; if \(L<\infty\), then we have

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

Hence, (20) is valid.

On the other hand, assuming that (20) is valid, using the reverse Hölder inequality, we have

$$\begin{aligned} I =&\sum_{ \vert m \vert =2}^{\infty} \biggl[ \frac{\ln ^{(1/q)-\lambda_{1}}A_{\alpha}(\xi,m)}{(1-\theta(\lambda _{2},m))^{-1/p}(A_{\alpha}(\xi,m))^{-1/q}}a_{m} \biggr] \\ &{}\times \Biggl[ \frac{\ln^{\lambda_{1}-(1/q)}A_{\alpha}(\xi,m)}{(1-\theta(\lambda_{2},m))^{1/p}(A_{\alpha}(\xi,m))^{1/q}}\sum _{ \vert n \vert =2}^{\infty}H(m,n)b_{n} \Biggr] \\ \geq& \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1- \theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi,m)}{(A_{\alpha }(\xi,m))^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}L, \end{aligned}$$
(31)

and then by (20) we have (18), which is equivalent to (20).

Therefore, inequalities (18), (19), and (20) are equivalent.

The theorem is proved. □

Theorem 2

With regards to the assumptions of Theorem 1, the constant factor \(H(\lambda_{1})\) in (18), (19), and (20) 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

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

By (16) and (14) we find

$$\begin{aligned} &\begin{aligned} \widetilde{I}_{1} := {}& \Biggl[ \sum _{ \vert m \vert =2}^{\infty } \bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda _{1})-1}A_{\alpha}(\xi ,m)}{(A_{\alpha}(\xi,m))^{1-p}} \widetilde{a}_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1}B_{\beta}(\eta,n)}{(B_{\beta}(\eta,n))^{1-q}}\widetilde{b}_{n}^{q} \Biggr] ^{\frac{1}{q}} \\ ={}& \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \frac{\ln ^{-1-\varepsilon}A_{\alpha}(\xi,m)}{A_{\alpha}(\xi,m)}-\sum_{ \vert m \vert =2}^{\infty} \frac{O(\ln^{-1-(\lambda_{2}+\varepsilon )}A_{\alpha}(\xi,m))}{A_{\alpha}(\xi,m)} \Biggr] ^{\frac{1}{p}} \\ &{}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{-1-\varepsilon}B_{\beta}(\eta,n)}{B_{\beta}(\eta,n)} \Biggr] ^{\frac{1}{q}} \\ ={}&\frac{1}{\varepsilon} \bigl(2\csc^{2}\alpha+o(1)-\varepsilon O(1) \bigr)^{1/p} \bigl(2\csc^{2}\beta+\widetilde{o}(1) \bigr)^{1/q}\quad \bigl(\varepsilon\rightarrow0^{+} \bigr), \end{aligned} \\ &\begin{aligned} \widetilde{I} &:=\sum_{ \vert m \vert =2}^{\infty } \sum_{ \vert n \vert =2}^{\infty}H(m,n)\widetilde{a}_{m}\widetilde{b}_{n} \\ &=\sum_{ \vert n \vert =2}^{\infty}\sum _{ \vert m \vert =2}^{\infty}H(m,n)\frac{\ln^{\widetilde{\lambda}_{1}-1}A_{\alpha }(\xi ,m)\ln^{\widetilde{\lambda}_{2}-\varepsilon-1}B_{\beta}(\eta,n)}{A_{\alpha}(\xi,m)B_{\beta}(\eta,n)} \\ &=\sum_{ \vert n \vert =2}^{\infty}\varpi(\widetilde{ \lambda}_{1},n)\frac{\ln^{-1-\varepsilon}B_{\beta}(\eta,n)}{B_{\beta }(\eta,n)}< H_{\alpha}( \widetilde{\lambda}_{1})\sum_{ \vert n \vert =2}^{\infty} \frac{\ln^{-1-\varepsilon}B_{\beta}(\eta,n)}{B_{\beta }(\eta,n)} \\ &=\frac{1}{\varepsilon}H_{\alpha} \biggl(\lambda_{1}- \frac{\varepsilon}{p} \biggr) \bigl(2\csc^{2}\beta+ \widetilde{o}(1) \bigr). \end{aligned} \end{aligned}$$

If there exists a positive number \(K\geq H(\lambda_{1})\) such that (18) is still valid when replacing \(H(\lambda_{1})\) by K, then, in particular, we have

$$ \varepsilon\widetilde{I}=\varepsilon\sum_{ \vert m \vert =2}^{\infty} \sum_{ \vert n \vert =2}^{\infty}H(m,n)\widetilde{a}_{m} \widetilde{b}_{n}>\varepsilon K\widetilde{I}_{1}. $$

In view of the above 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

$$ \frac{4\pi\csc^{2}\alpha}{\lambda\sin(\frac{\pi\lambda _{1}}{\lambda})}\csc^{2}\beta\geq2K\csc^{2/p}\alpha \csc^{2/q}\beta \quad \bigl(\varepsilon\rightarrow0^{+} \bigr), $$

namely

$$ H(\lambda_{1})=\frac{2\pi}{\lambda\sin(\frac{\pi\lambda _{1}}{\lambda})}\csc^{2/p}\beta\csc ^{2/q}\alpha\geq K. $$

Hence, \(K=H(\lambda_{1})\) is the best possible constant factor in (18).

The constant factor \(H(\lambda_{1})\) in (19) ((20)) is still the best possible. Otherwise we would reach a contradiction by (30) ((31)) that the constant factor in (18) is not the best possible. □

Remark 2

(i) For \(\xi=\eta=0\) in (22), setting

$$\begin{aligned} \widetilde{\theta}_{1}(\lambda_{2},m) :=& \frac{\lambda}{\pi}\sin \biggl(\frac{\pi\lambda_{1}}{\lambda} \biggr) \int_{0}^{\frac{\ln2}{\ln \vert m \vert }}\frac{u^{\lambda _{2}-1}}{1+u^{\lambda}}\,du \\ =&O \biggl( \frac{1}{\ln^{\lambda_{2}} \vert m \vert } \biggr) \in(0,1), \end{aligned}$$

we have the following new inequality with the best possible constant factor \(\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}\):

$$\begin{aligned} &\sum_{ \vert n \vert =2}^{\infty}\sum _{ \vert m \vert =2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda} \vert m \vert +\ln ^{\lambda} \vert n \vert } \\ &\quad >\frac{2\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})} \Biggl[ \sum_{ \vert m \vert =2}^{\infty} \bigl(1-\widetilde{\theta}_{1}(\lambda_{2},m) \bigr)\frac{\ln^{p(1-\lambda_{1})-1} \vert m \vert }{ \vert m \vert ^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &\qquad {}\times \Biggl[ \sum_{ \vert n \vert =2}^{\infty} \frac{\ln ^{q(1-\lambda_{2})-1} \vert n \vert }{ \vert n \vert ^{1-q}}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(32)

It follows that (20) ((22)) is an extension of (29).

(ii) If \(a_{-m}=a_{m}\) and \(b_{-n}=b_{n}\) (\(m,n\in\mathbf{N}\backslash\{1\}\)), setting

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

then (22) reduces to

$$\begin{aligned} &\sum_{n=2}^{\infty}\sum _{m=2}^{\infty} \biggl[ \frac{1}{\ln^{\lambda }(m-\xi)+\ln^{\lambda}(n-\eta)}+ \frac{1}{\ln^{\lambda}(m-\xi)+\ln ^{\lambda}(n+\eta)} \\ &\qquad {} +\frac{1}{\ln^{\lambda}(m+\xi)+\ln^{\lambda}(n-\eta)}+\frac{1 }{\ln^{\lambda}(m+\xi)+\ln^{\lambda}(n+\eta)} \biggr] a_{m}b_{n} \\ &\quad >\frac{2\pi}{\lambda\sin(\pi\lambda_{1}/\lambda)} \Biggl\{ \sum_{m=2}^{\infty} \biggl[ \bigl(1-\theta_{2}(\lambda_{2},m) \bigr) \frac{\ln ^{p(1-\lambda_{1})-1}(m-\xi)}{(m-\xi)^{1-p}} \\ &\qquad {} + \bigl(1-\theta_{3}(\lambda_{2},m) \bigr) \frac{\ln^{p(1-\lambda _{1})-1}(m+\xi)}{(m+\xi)^{1-p}} \biggr] a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &\qquad {}\times \Biggl\{ \sum_{n=2}^{\infty} \biggl[ \frac{\ln^{q(1-\lambda _{2})-1}(n-\eta)}{(n-\eta)^{1-q}}+\frac{\ln^{q(1-\lambda _{2})-1}(n+\eta)}{(n+\eta)^{1-q}} \biggr] b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(33)

In particular, for \(\xi=\eta=0\), \(\lambda=1\), \(\lambda_{1}=\lambda _{2}=\frac{1}{2}\) in (33), setting

$$ \theta(m):=\frac{1}{\pi} \int_{0}^{\frac{\ln2}{\ln m}}\frac{u^{-1/2}}{1+u}\,du=O \biggl( \frac{1}{\ln^{1/2}m} \biggr) \in(0,1), $$

we have the following reverse Mulholland inequality (2) with the best possible constant π:

$$\begin{aligned} \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln mn} >&\pi \Biggl[ \sum_{m=2}^{\infty} \bigl(1-\theta(m) \bigr) \frac{\ln^{\frac{p}{2}-1}m}{m^{1-p}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ &{} \times \Biggl( \sum_{n=2}^{\infty} \frac{\ln^{\frac{q}{2}-1}n}{n^{1-q}}b_{n}^{q} \Biggr) ^{\frac{1}{q}}. \end{aligned}$$
(34)

4 Conclusions

In this paper, by introducing multi-parameters, applying the weight coefficients, and using Hermite–Hadamard’s inequality, we give a reverse of the extension of Mulholland’s inequality in the whole plane with the best possible constant factor in Theorems 1–2. The equivalent forms and a few particular cases are considered. The technique of real analysis is very important as it is the key to proving the reverse equivalent inequalities with the best possible constant factor. The lemmas and theorems provide an extensive account of this type of inequalities.