Bounds for the Neuman–Sándor Mean in Terms of the Arithmetic and Contra-Harmonic Means

Abstract

In this paper, the authors provide several sharp upper and lower bounds for the Neuman–Sándor mean in terms of the arithmetic and contra-harmonic means, and present some new sharp inequalities involving hyperbolic sine function and hyperbolic cosine function.

1. Introduction

In the literature, the quantities

$$\begin{array}{c}A(s,t)=\frac{s+t}{2},G(s,t)=\sqrt{st},H(s,t)=\frac{2st}{s+t},\\ \begin{array}{cccc}\overline{C}(s,t)\hfill & =\frac{2(s^2+st+t^2)}{3(s+t)},\hfill & C(s,t)\hfill & =\frac{s^2+t^2}{s+t},\hfill \\ S(s,t)\hfill & =\sqrt{\frac{s^2+t^2}{2}},\hfill & M_p(s,t)\hfill & =\{\begin{array}{cc}{\left(\frac{s^p+t^p}{2}\right)}^{1/p},\hfill & p\ne 0;\hfill \\ \sqrt{st},\hfill & p=0\hfill \end{array}\hfill \end{array}\end{array}$$

are called in [1,2,3], for example, the arithmetic mean, geometric mean, harmonic mean, centroidal mean, contra-harmonic mean, root-square mean, and the power mean of order p of two positive numbers s and t, respectively.

For $s,t>0$ with $s\ne t$, the first Seiffert means $P(s,t)$, the second Seiffert means $T(s,t)$, and Neuman–Sándor mean $M(s,t)$ are, respectively, defined [4,5,6] by

$$\begin{array}{cccccc}P(s,t)\hfill & =\frac{s-t}{4arctan\left(\sqrt{\frac{s}{t}}\right)-\pi },\hfill & T(s,t)\hfill & =\frac{s-t}{2arctan\frac{s-t}{s+t}},\hfill & M(s,t)\hfill & =\frac{s-t}{2arsinh\frac{s-t}{s+t}},\hfill \end{array}$$

where $arsinhv=ln\left(v+\sqrt{v^2+1}\right)$ is the inverse hyperbolic sine function.

The first Seiffert mean $P(s,t)$ can be rewritten [6] (Equation (2.4]) as

$$P(s,t)=\frac{s-t}{2arcsin\frac{s-t}{s+t}}.$$

A chain of inequalities

$$G(s,t)<L_{-1}(s,t)<P(s,t)<A(s,t)<M(s,t)<T(s,t)<Q(s,t)$$

were given in [6], where

$$\begin{array}{cc}{L}_p(s,t)\hfill & =\{\begin{array}{cc}{\left[\frac{t^{p+1}-s^{p+1}}{(p+1)(t-s)}\right]}^{1/p},\hfill & p\ne -1,0;\hfill \\ \frac{1}{e}{\left(\frac{t^t}{s^s}\right)}^{1/(t-s)},\hfill & p=0;\hfill \\ \frac{t-s}{lnt-lns},\hfill & p=-1\hfill \end{array}\hfill \end{array}$$

is the p-th generalized logarithmic mean of s and t with $s\ne t$.

In [6,7], three double inequalities

$$\begin{array}{c}A(s,t)<M(s,t)<T(s,t),P(s,t)<M(s,t)<T^2(s,t),\end{array}$$

and

$$A(s,t)T(s,t)<M^2(s,t)<\frac{A^2(s,t)+T^2(s,t)}{2}$$

were established for $s,t>0$ with $s\ne t$.

For $0<s,t<\frac{1}{2}$ with $s\ne t$, the inequalities

$$\begin{array}{c}\frac{G(s,t)}{G(1-s,1-t)}<\frac{L_{-1}(s,t)}{L_{-1}(1-s,1-t)}<\frac{P(s,t)}{P(1-s,1-t)}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}<\frac{A(s,t)}{A(1-s,1-t)}<\frac{M(s,t)}{M(1-s,1-t)}<\frac{T(s,t)}{T(1-s,1-t)}\hfill \end{array}$$

of Ky Fan type were presented in [6] (Proposition 2.2).

In [8], Li and their two coauthors showed that the double inequality

$$L_{p_0}(s,t)<M(s,t)<L_2(s,t)$$

holds for all $s,t>0$ with $s\ne t$ and for $p_0=1.843\dots$, where $p_0$ is the unique solution of the equation ${(p+1)}^{1/p}=2ln\left(1+\sqrt{2}\right)$.

In [9], Neuman proved that the double inequalities

$$\alpha Q(s,t)+(1-\alpha )A(s,t)<M(s,t)<\beta Q(s,t)+(1-\beta )A(s,t)$$

and

$$\lambda C(s,t)+(1-\lambda )A(s,t)<M(s,t)<\mu C(s,t)+(1-\mu )A(s,t)$$

hold for all $s,t>0$ with $s\ne t$ if and only if

$$\alpha \le \frac{1-ln\left(1+\sqrt{2}\right)}{\left(\sqrt{2}-1\right)ln\left(1+\sqrt{2}\right)}=0.3249\dots ,\beta \ge \frac{1}{3}$$

and

$$\lambda \le \frac{1-ln\left(1+\sqrt{2}\right)}{ln\left(1+\sqrt{2}\right)}=0.1345\dots ,\mu \ge \frac{1}{6}.$$

In [10], (Theorems 1.1 to 1.3), it was found that the double inequalities

$${\alpha }_{1}H(s,t)+(1-{\alpha }_1)Q(s,t)<M(s,t)<{\beta }_{1}H(s,t)+(1-{\beta }_1)Q(s,t),$$

$${\alpha }_{2}G(s,t)+(1-{\alpha }_2)Q(s,t)<M(s,t)<{\beta }_{2}G(s,t)+(1-{\beta }_2)Q(s,t),$$

and

$${\alpha }_{3}H(s,t)+(1-{\alpha }_3)C(s,t)<M(s,t)<{\beta }_{3}H(s,t)+(1-{\beta }_3)C(s,t)$$

hold for all $s,t>0$ with $s\ne t$ if and only if

$$\begin{array}{c}{\alpha }_1\ge \frac{2}{9}=0.2222\dots ,{\beta }_1\le 1-\frac{1}{\sqrt{2}ln\left(1+\sqrt{2}\right)}=0.1977\dots ,\\ {\alpha }_2\ge \frac{1}{3}=0.3333\dots ,{\beta }_2\le 1-\frac{1}{\sqrt{2}ln\left(1+\sqrt{2}\right)}=0.1977\dots ,\end{array}$$

and

$${\alpha }_3\ge 1-\frac{1}{2ln\left(1+\sqrt{2}\right)}=0.4327\dots ,{\beta }_3\le \frac{5}{12}=0.4166\dots$$

In 2017, Chen and their two coauthors [11] established bounds for Neuman–Sándor mean $M(s,t)$ in terms of the convex combination of the logarithmic mean and the second Seiffert mean $T(s,t)$. In 2022, Wang and Yin [12] obtained bounds for the reciprocals of the Neuman–Sándor mean $M(s,t)$.

In [13], it was showed that the double inequality

$$\frac{\alpha }{A(s,t)}+\frac{1-\alpha }{\overline{C}(s,t)}<\frac{1}{TD(s,t)}<\frac{\beta }{A(s,t)}+\frac{1-\beta }{\overline{C}(s,t)}$$

(1)

holds for all $s,t>0$ with $s\ne t$ if and only if $\alpha \le \pi -3$ and $\beta \ge \frac{1}{4}$, where $TD(s,t)$ is the Toader mean introduced in [14] by

$$TD(s,t)=\frac{2}{\pi }{\int }_0^{\pi /2}\sqrt{s^2{cos}^2\varphi +t^2{sin}^2\varphi }d\varphi .$$

In this paper, motivated by the double inequality (1), we will aim to find out the largest values ${\alpha }_1,{\alpha }_2$, and ${\alpha }_3$ and the smallest values ${\beta }_1,{\beta }_2$, and ${\beta }_3$ such that the double inequalities

$$\begin{array}{c}\frac{{\alpha }_1}{C(s,t)}+\frac{1-{\alpha }_1}{A(s,t)}<\frac{1}{M(s,t)}<\frac{{\beta }_1}{C(s,t)}+\frac{1-{\beta }_1}{A(s,t)},\end{array}$$

(2)

$$\begin{array}{c}\frac{{\alpha }_2}{C^2(s,t)}+\frac{1-{\alpha }_2}{A^2(s,t)}<\frac{1}{M^2(s,t)}<\frac{{\beta }_2}{C^2(s,t)}+\frac{1-{\beta }_2}{A^2(s,t)},\end{array}$$

(3)

and

$${\alpha }_3{C}^2(s,t)+(1-{\alpha }_3)A^2(s,t)<M^2(s,t)<{\beta }_3{C}^2(s,t)+(1-{\beta }_3)A^2(s,t)$$

(4)

hold for all positive real numbers s and t with $s\ne t$.

2. Lemmas

To attain our main purposes, we need the following lemmas.

Lemma 1 

([15] (Theorem 1.25)). For $-\infty <s<t<\infty$, let $f,g:[s,t]\to \mathbb{R}$ be continuous on $[s,t]$, differentiable on $(s,t)$, and $g^{\prime }\left(v\right)\ne 0$ on $(s,t)$. If $\frac{f^{\prime }\left(v\right)}{g^{\prime }\left(v\right)}$ is (strictly) increasing (or (strictly) decreasing, respectively) on $(s,t)$, so are the functions

$$\frac{f\left(v\right)-f\left(s\right)}{g\left(v\right)-g\left(s\right)}\mathrm{and}\frac{f\left(v\right)-f\left(t\right)}{g\left(v\right)-g\left(t\right)}.$$

Lemma 2 

([16] (Lemma 1.1)). Suppose that the power series $f\left(v\right)={\sum }_{\ell =0}^{\infty }{u}_{\ell }{v}^{\ell }$ and $g\left(v\right)={\sum }_{\ell =0}^{\infty }{w}_{\ell }{v}^{\ell }$ have the convergent radius $r>0$ and $w_{\ell }>0$ for all $\ell \in \mathbb{N}=\{0,1,2,\dots \}$. Let $h\left(v\right)=\frac{f\left(v\right)}{g\left(v\right)}$. Then the following statements are true.

(1)

If the sequence ${\left\{\frac{u_{\ell }}{w_{\ell }}\right\}}_{\ell =0}^{\infty }$ is (strictly) increasing (or decreasing, respectively), then $h\left(v\right)$ is also (strictly) increasing (or decreasing, respectively) on $(0,r)$.

(2)

If the sequence ${\left\{\frac{u_{\ell }}{w_{\ell }}\right\}}_{\ell =0}^{\infty }$ is (strictly) increasing (or decreasing resepctively) for $0<\ell \le {\ell }_0$ and (strictly) decreasing (or increasing resepctively) for $\ell >{\ell }_0$, then there exists $x_0\in (0,r)$ such that $h\left(v\right)$ is (strictly) increasing (decreasing) on $(0,x_0)$ and (strictly) decreasing (or increasing resepctively) on $(x_0,r)$.

Lemma 3. 

Let

$$h_1\left(v\right)=\frac{2vsinhv+coshv-1}{3{sinh}^{2}v}.$$

Then $h_1\left(v\right)$ is strictly decreasing on $(0,\infty )$ with ${lim}_{v\to 0^{+}}{h}_1\left(v\right)=\frac{5}{6}$ and ${lim}_{v\to \infty }{h}_1\left(v\right)=0$.

Proof. 

Let

$$f_1\left(v\right)=2vsinhv+coshv-1\mathrm{and}{f}_2\left(v\right)=3{sinh}^{2}v=\frac{3}{2}[cosh\left(2v\right)-1].$$

Using the power series

$$sinhv=\sum _{\ell =0}^{\infty }\frac{v^{2\ell +1}}{(2\ell +1)!}\mathrm{and}coshv=\sum _{\ell =0}^{\infty }\frac{v^{2\ell }}{(2\ell )!},$$

(5)

we can express the functions $f_1\left(v\right)$ and $f_2\left(v\right)$ as

$$f_1\left(v\right)=\sum _{\ell =0}^{\infty }\frac{2(2\ell +2)!+(2\ell +1)!}{(2\ell +1)!(2\ell +2)!}{v}^{2\ell +2}\mathrm{and}{f}_2\left(v\right)=\frac{3}{2}\sum _{\ell =0}^{\infty }\frac{2^{2\ell +2}{v}^{2\ell +2}}{(2\ell +2)!}.$$

Hence, we have

$$h_1\left(v\right)=\frac{{\sum }_{\ell =0}^{\infty }{u}_{\ell }{v}^{2\ell +2}}{{\sum }_{\ell =0}^{\infty }{w}_{\ell }{v}^{2\ell +2}},$$

(6)

where $u_{\ell }=\frac{2(2\ell +2)!+(2\ell +1)!}{(2\ell +1)!(2\ell +2)!}$ and $\hspace{1em}\hspace{1em}{w}_{\ell }=\frac{3\times 2^{2\ell +1}}{(2\ell +2)!}$.

Let $c_{\ell }=\frac{u_{\ell }}{w_{\ell }}$. Then

$$c_{\ell }=\frac{2(2\ell +2)!+(2\ell +1)!}{3(2\ell +1)!2^{2\ell +1}}\mathrm{and}{c}_{\ell +1}-c_{\ell }=-\frac{4(3\ell +4)(2\ell +2)!+3(2\ell +3)!}{3(2\ell +3)!2^{2\ell +3}}<0.$$

As a result, by Lemma 2, it follows that the function $h_1\left(v\right)$ is strictly decreasing on $(0,\infty )$. From (6), it is easy to see that ${lim}_{v\to 0^{+}}{h}_1\left(v\right)=\frac{u_0}{w_0}=\frac{5}{6}$.

Using the L’Hospital rule leads to ${lim}_{v\to \infty }{h}_1\left(v\right)=0$ immediately. The proof of Lemma 3 is complete.  □

Lemma 4. 

Let

$$h_2\left(v\right)=\frac{\left({sinh}^{2}v-v^2\right){cosh}^{4}v}{\left({cosh}^{2}v+1\right){sinh}^{4}v}.$$

Then $h_2\left(v\right)$ is strictly increasing on $v\in (0,\infty )$ and has the limit ${lim}_{v\to 0^{+}}{h}_2\left(v\right)=\frac{1}{6}$ and ${lim}_{v\to \infty }{h}_2\left(v\right)=1$.

Proof. 

Let

$$f_3\left(v\right)=\left({sinh}^{2}v-v^2\right){cosh}^{4}v\mathrm{and}{f}_4\left(v\right)=\left({cosh}^{2}v+1\right){sinh}^{4}v.$$

Since

$$\begin{array}{cc}\hfill f_3^{\prime }\left(v\right)& =2\left(sinhv+3{sinh}^{3}v-vcoshv-2v^{2}sinhv\right){cosh}^{3}v\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill f_4^{\prime }\left(v\right)& =2\left(3{cosh}^{2}v+1\right){sinh}^{3}vcoshv,\hfill \end{array}$$

we obtain

$$\begin{array}{cc}\hfill \frac{f_3^{\prime }\left(v\right)}{f_4^{\prime }\left(v\right)}& =\frac{\left(sinhv+3{sinh}^{3}v-vcoshv-2v^{2}sinhv\right){cosh}^{2}v}{\left(3{cosh}^{2}v+1\right){sinh}^{3}v}\hfill \\ \hfill & =\frac{{cosh}^{2}v}{3{cosh}^{2}v+1}\frac{sinhv+3{sinh}^{3}v-vcoshv-2v^{2}sinhv}{{sinh}^{3}v}\hfill \\ \hfill & =\frac{1}{3+\frac{1}{{cosh}^{2}v}}\left(3+\frac{sinhv-vcoshv-2v^{2}sinhv}{{sinh}^{3}v}\right)\hfill \\ \hfill & =\frac{1}{3+\frac{1}{{cosh}^{2}v}}[3+g\left(v\right)],\hfill \end{array}$$

where

$$g\left(v\right)=\frac{sinhv-vcoshv-2v^{2}sinhv}{{sinh}^{3}v}.$$

By using the identity that $sinh\left(3v\right)=3sinhv+4{sinh}^{3}v$, we arrive at

$$g\left(v\right)=4\frac{sinhv-vcoshv-2v^{2}sinhv}{sinh\left(3v\right)-3sinhv}\triangleq 4\frac{g_1\left(v\right)}{g_2\left(v\right)},$$

where $g_1\left(v\right)=sinhv-vcoshv-2v^{2}sinhv$ and $g_2\left(v\right)=sinh\left(3v\right)-3sinhv$.

Straightforward computation gives

$$\begin{array}{cccc}{g}_1^{\prime }\left(v\right)\hfill & =-\left(5vsinhv+2v^{2}coshv\right),\hfill & g_2^{\prime }\left(v\right)\hfill & =3[cosh(3v)-coshv],\hfill \\ g_1^{″}\left(v\right)\hfill & =-\left(5sinhv+9vcoshv+2v^{2}sinhv\right),\hfill & g_2^{″}\left(v\right)\hfill & =3[3sinh(3v)-sinhv],\hfill \end{array}$$

and

$$g_1\left(0^{+}\right)=g_2\left(0^{+}\right)=g_1^{\prime }\left(0^{+}\right)=g_2^{\prime }\left(0^{+}\right)=g_1^{″}\left(0^{+}\right)=g_2^{″}\left(0^{+}\right)=0.$$

Consequently, we obtain

$$\frac{g_1^{″}\left(v\right)}{g_2^{″}\left(v\right)}=-\frac{5sinhv+9vcoshv+2v^{2}sinhv}{3[3sinh(3v)-sinhv]}\triangleq -\frac{1}{3}\frac{g_3\left(v\right)}{g_4\left(v\right)}$$

Using the power series of $sinhv$ and $coshv$, we deduce

$$\begin{array}{cc}{g}_3\left(v\right)\hfill & =5\sum _{\ell =0}^{\infty }\frac{v^{2\ell +1}}{(2\ell +1)!}+9\sum _{\ell =0}^{\infty }\frac{v^{2\ell +1}}{(2\ell )!}+2\sum _{\ell =0}^{\infty }\frac{v^{2\ell +3}}{(2\ell +1)!}\hfill \\ \hfill & =14v+\sum _{\ell =1}^{\infty }\left[\frac{5}{(2\ell +1)!}+\frac{9}{(2\ell )!}+\frac{2}{(2\ell -1)!}\right]v^{2\ell +1}\hfill \\ \hfill & =14v+\sum _{\ell =1}^{\infty }\left[\frac{(4\ell +7)(2\ell )!+9(2\ell +1)!}{(2\ell )!(2\ell +1)!}\right]v^{2\ell +1}\hfill \end{array}$$

and

$$g_4\left(v\right)=3\sum _{\ell =0}^{\infty }\frac{{\left(3v\right)}^{2\ell +1}}{(2\ell +1)!}-\sum _{\ell =0}^{\infty }\frac{v^{2\ell +1}}{(2\ell +1)!}=\sum _{\ell =0}^{\infty }\left[\frac{3^{2\ell +2}-1}{(2\ell +1)!}\right]v^{2\ell +1}.$$

Therefore, we find

$$\frac{g_3\left(v\right)}{g_4\left(v\right)}=\frac{{\sum }_{\ell =0}^{\infty }{u}_{\ell }{v}^{2\ell +1}}{{\sum }_{\ell =0}^{\infty }{w}_{\ell }{v}^{2\ell +1}},$$

where

$$u_{\ell }=\{\begin{array}{cc}14,\hfill & \ell =0;\\ \frac{(4\ell +7)(2\ell )!+9(2\ell +1)!}{(2\ell )!(2\ell +1)!},\hfill & \ell \ge 1\end{array}\mathrm{and}{w}_{\ell }=\{\begin{array}{cc}8>0,\hfill & \ell =0;\\ \frac{3^{2\ell +2}-1}{(2\ell +1)!},\hfill & \ell \ge 1.\end{array}$$

Let $c_{\ell }=\frac{u_{\ell }}{w_{\ell }}$. Then

$$c_{\ell }=\{\begin{array}{cc}\frac{7}{4},\hfill & \ell =0;\\ \frac{(4\ell +7)(2\ell )!+9(2\ell +1)!}{\left(3^{2\ell +2}-1\right)(2\ell )!},\hfill & \ell \ge 1.\end{array}$$

When $\ell =0$, we have $c_1-c_0=-\frac{51}{40}<0$. When $\ell \ge 1$, it follows that

$$\begin{array}{cc}{c}_{\ell +1}-c_{\ell }\hfill & =\frac{(4\ell +11)(2\ell +2)!+9(2\ell +3)!}{\left(3^{2\ell +4}-1\right)(2\ell +2)!}-\frac{(4\ell +7)(2\ell )!+9(2\ell +1)!}{\left(3^{2\ell +2}-1\right)(2\ell )!}\hfill \\ \hfill & =\frac{1}{\left(3^{2\ell +2}-1\right)\left(3^{2\ell +4}-1\right)(2\ell +2)!}\left\{\right[(4\ell +11)(2\ell +2)!\hfill \\ \hfill & +9(2\ell +3)!]\left(3^{2\ell +2}-1\right)-[(4\ell +7)(2\ell )!+9(2\ell +1)!](2\ell +2)\left(3^{2\ell +4}-1\right)\}\hfill \\ \hfill & =\frac{1}{\left(3^{2\ell +2}-1\right)\left(3^{2\ell +4}-1\right)(2\ell +2)!}\left\{\right[(4\ell +11)(2\ell +2)!\hfill \\ \hfill & +9(2\ell +3)!]3^{2\ell +2}-[(4\ell +7)(2\ell )!+9(2\ell +1)!](2\ell +2)3^{2\ell +4}\hfill \\ \hfill & +(2\ell +2)\left[\right(4\ell +7\left)\right(2\ell )!+9(2\ell +1)!]-\left[\right(4\ell +11\left)\right(2\ell +2)!+9(2\ell +3)!]\}\hfill \\ \hfill & =-\frac{1}{\left(3^{2\ell +2}-1\right)\left(3^{2\ell +4}-1\right)(2\ell +2)!}\left\{\right[(8\ell +13)(2\ell +2)!\hfill \\ \hfill & +9(16\ell +15)(2\ell +1)!]3^{2\ell +2}+9(2\ell +1)!+4(2\ell +2)!\}\hfill \\ \hfill & <0.\hfill \end{array}$$

By Lemma 2, it follows that the function $\frac{g_3\left(v\right)}{g_4\left(v\right)}$ is strictly decreasing on $(0,\infty )$, so the function $\frac{g_1^{″}\left(v\right)}{g_2^{″}\left(v\right)}$ is strictly increasing on $(0,\infty )$. Applying Lemma 1, it follows that the function $g\left(v\right)$ is strictly increasing on $(0,\infty )$. By the L’Hospital rule, we have

$$\underset{v\to 0^{+}}{lim}g\left(v\right)=-\frac{7}{3}\mathrm{and}\underset{v\to \infty }{lim}g\left(v\right)=0.$$

It is common knowledge that the function $coshv$ is strictly increasing on $(0,\infty )$. Hence, the function $\frac{1}{3+\frac{1}{{cosh}^{2}v}}$ is strictly increasing on $(0,\infty )$. Therefore, the function $h_2\left(v\right)$ is strictly increasing on $(0,\infty )$ with the limits

$$\underset{v\to 0}{lim}{h}_2\left(v\right)=\frac{1}{6}\mathrm{and}\underset{v\to \infty }{lim}{h}_2\left(v\right)=1.$$

The proof of Lemma 3 is complete.  □

Lemma 5. 

Let

$$h_3\left(v\right)=\frac{2v{cosh}^{2}v}{sinhv}.$$

Then $h_3\left(v\right)$ is strictly increasing on $(0,\infty )$ and has the limit ${lim}_{v\to 0^{+}}{h}_3\left(v\right)=2$.

Proof. 

Let $k_1\left(v\right)=2v{cosh}^{2}v=vcosh\left(2v\right)+v$ and $k_2\left(v\right)=sinhv$. By Equation (5), we have

$$k_1\left(v\right)=2v+\sum _{\ell =1}^{\infty }\frac{2^{2\ell }}{(2\ell )!}{v}^{2\ell +1}\mathrm{and}{k}_2\left(v\right)=\sum _{\ell =0}^{\infty }\frac{v^{2\ell +1}}{(2\ell +1)!}.$$

Hence,

$$h_3\left(v\right)=\frac{2v+{\sum }_{\ell =1}^{\infty }{u}_{\ell }{v}^{2\ell +1}}{{\sum }_{\ell =0}^{\infty }{w}_{\ell }{v}^{2\ell +1}},$$

(7)

where

$$\begin{array}{c}{u}_{\ell }=\{\begin{array}{cc}2,\hfill & \ell =0;\hfill \\ \frac{2^{2\ell }}{(2\ell )!},\hfill & \ell \ge 1\hfill \end{array}\hfill \end{array}\mathrm{and}{w}_{\ell }=\frac{1}{2\ell +1!}.$$

Let $c_{\ell }=\frac{u_{\ell }}{w_{\ell }}$. Then

$$c_{\ell }=\{\begin{array}{cc}2,\hfill & \ell =0;\\ \frac{(2\ell +1)!2^{2\ell }}{(2\ell )!},\hfill & \ell \ge 1\end{array}\mathrm{and}{c}_{\ell +1}-c_{\ell }=\{\begin{array}{cc}10,\hfill & \ell =0;\\ \frac{(3\ell +5)(2\ell +1)!2^{2\ell +1}}{(2\ell +2)!}>0,\hfill & \ell \ge 1.\end{array}$$

Thus, by Lemma 2, it follows that the function $h_3\left(v\right)$ is strictly increasing on $(0,\infty )$. From (7), it is easy to see that ${lim}_{v\to 0^{+}}{h}_3\left(v\right)=\frac{u_0}{w_0}=2$. The proof of Lemma 5 is complete.  □

3. Bounds for Neuman–Sándor Mean

Now we are in a position to state and prove our main results.

Theorem 1. 

For $s,t>0$ with $s\ne t$, the double inequality (2) holds if and only if

$${\alpha }_1\ge 2\left[1-ln\left(1+\sqrt{2}\right)\right]=0.237253\dots \mathrm{and}{\beta }_1\le \frac{1}{6}.$$

Proof. 

Without loss of generality, we assume that $s>t>0$. Let $q=\frac{s-t}{s+t}$. Then $q\in (0,1)$ and

$$\frac{\frac{1}{M(s,t)}-\frac{1}{A(s,t)}}{\frac{1}{C(s,t)}-\frac{1}{A(s,t)}}=\frac{\frac{arsinhq}{q}-1}{\frac{1}{1+q^2}-1}.$$

Let $q=sinh\varphi$. Then $\varphi \in \left(0,ln\left(1+\sqrt{2}\right)\right)$ and

$$\frac{\frac{1}{M(s,t)}-\frac{1}{A(s,t)}}{\frac{1}{C(s,t)}-\frac{1}{A(s,t)}}=\frac{\frac{\varphi }{sinh\varphi }-1}{\frac{1}{{cosh}^2\varphi }-1}=\frac{(sinh\varphi -\varphi ){cosh}^2\varphi }{{sinh}^3\varphi }\triangleq F\left(\varphi \right)=\frac{k_1\left(\varphi \right)}{k_2\left(\varphi \right)}.$$

Let

$$k_1\left(\varphi \right)=(sinh\varphi -\varphi ){cosh}^2\varphi \mathrm{and}{k}_2\left(\varphi \right)={sinh}^3\varphi .$$

Then elaborated computations lead to $k_1\left(0^{+}\right)=k_2\left(0^{+}\right)=0$ and

$$\frac{k_1^{\prime }\left(\varphi \right)}{k_2^{\prime }\left(\varphi \right)}=\frac{2(sinh\varphi -\varphi )sinh\varphi +(cosh\varphi -1)cosh\varphi }{3{sinh}^2\varphi }=1-\frac{2\varphi sinh\varphi +cosh\varphi -1}{3{sinh}^2\varphi }.$$

Combining this with Lemmas 1 and 3 reveals that the function $F\left(\varphi \right)$ is strictly increasing on $\left(0,ln\left(1+\sqrt{2}\right)\right)$. Moreover, it is easy to compute the limits

$$\underset{\varphi \to 0^{+}}{lim}F\left(\varphi \right)=\frac{1}{6}\mathrm{and}\underset{\varphi \to ln{(1+\sqrt{2})}^{-}}{lim}F\left(\varphi \right)=2-2ln\left(1+\sqrt{2}\right).$$

The proof of Theorem 1 is thus complete.  □

Corollary 1. 

For all $\varphi \in \left(0,ln\left(1+\sqrt{2}\right)\right)$, the double inequality

$$1-{\beta }_1\left(1-\frac{1}{{cosh}^2\varphi }\right)<\frac{\varphi }{sinh\varphi }<1-{\alpha }_1\left(1-\frac{1}{{cosh}^2\varphi }\right)$$

(8)

holds if and only if

$${\alpha }_1\le \frac{1}{6}\mathrm{and}{\beta }_1\ge 2\left[1-ln\left(1+\sqrt{2}\right)\right]=0.237253\dots .$$

Theorem 2. 

For $s,t>0$ with $s\ne t$, the double inequality (3) holds if and only if

$${\alpha }_2\ge \frac{4}{3}[1-{ln}^2\left(1+\sqrt{2}\right)]=0.297574\dots \mathrm{and}{\beta }_2\le \frac{1}{6}.$$

Proof. 

Without loss of generality, we assume that $s>t>0$. Let $q=\frac{s-t}{s+t}$. Then $q\in (0,1)$ and

$$\frac{\frac{1}{M^2(s,t)}-\frac{1}{A^2(s,t)}}{\frac{1}{C^2(s,t)}-\frac{1}{A^2(s,t)}}=\frac{\frac{{arsinh}^{2}q}{q^2}-1}{\frac{1}{{(1+q^2)}^2}-1}.$$

Let $q=sinh\varphi$. Then $\varphi \in \left(0,ln\left(1+\sqrt{2}\right)\right)$ and

$$\frac{\frac{1}{M^2(s,t)}-\frac{1}{A^2(s,t)}}{\frac{1}{C^2(s,t)}-\frac{1}{A^2(s,t)}}=\frac{\frac{{\varphi }^2}{{sinh}^2\varphi }-1}{\frac{1}{{cosh}^4\varphi }-1}=\frac{({sinh}^2\varphi -{\varphi }^2){cosh}^4\varphi }{({cosh}^2\varphi +1){sinh}^4\varphi }\triangleq H\left(\varphi \right).$$

By Lemma 4, it is easy to show that $H\left(\varphi \right)$ is strictly increasing on $\left(0,ln\left(1+\sqrt{2}\right)\right)$. Moreover, the limits

$$\underset{\varphi \to 0^{+}}{lim}H\left(\varphi \right)=\frac{1}{6}\mathrm{and}\underset{\varphi \to ln{(1+\sqrt{2})}^{-}}{lim}H\left(\varphi \right)=\frac{4}{3}\left[1-{ln}^2\left(1+\sqrt{2}\right)\right]$$

can be computed readily. The double inequality (3) is thus proved.  □

Corollary 2. 

For all $\varphi \in \left(0,ln\left(1+\sqrt{2}\right)\right)$, the double inequality

$$1-{\beta }_2\left(1-\frac{1}{{cosh}^4\varphi }\right)<{\left(\frac{\varphi }{sinh\varphi }\right)}^2<1-{\alpha }_2\left(1-\frac{1}{{cosh}^4\varphi }\right)$$

(9)

holds if and only if

$${\alpha }_2\le \frac{1}{6}\mathrm{and}{\beta }_2\ge \frac{4}{3}\left[1-{ln}^2\left(1+\sqrt{2}\right)\right]=0.297574\dots .$$

Theorem 3. 

For $s,t>0$ with $s\ne t$, the double inequality (4) holds if and only if

$${\alpha }_3\le \frac{1-{ln}^2\left(1+\sqrt{2}\right)}{3{ln}^2\left(1+\sqrt{2}\right)}=0.095767\dots \mathrm{and}{\beta }_3\ge \frac{1}{6}.$$

Proof. 

Without loss of generality, we assume that $s>t>0$. Let $q=\frac{s-t}{s+t}$. Then $q\in (0,1)$ and

$$\frac{M^2(s,t)-A^2(s,t)}{C^2(s,t)-A^2(s,t)}=\frac{\frac{q^2}{{arsinh}^{2}q}-1}{{(1+q^2)}^2-1}.$$

Let $q=sinh\varphi$. Then $\varphi \in \left(0,ln\left(1+\sqrt{2}\right)\right)$ and

$$\frac{M^2(s,t)-A^2(s,t)}{C^2(s,t)-A^2(s,t)}=\frac{\frac{{sinh}^2\varphi }{{\varphi }^2}-1}{{cosh}^4\varphi -1}\triangleq G\left(\varphi \right)=\frac{k_1\left(\varphi \right)}{k_2\left(\varphi \right)},$$

where

$$k_1\left(\varphi \right)=\frac{{sinh}^2\varphi }{{\varphi }^2}-1\mathrm{and}{k}_2\left(\varphi \right)={cosh}^4\varphi -1.$$

Then $k_1\left(0^{+}\right)=k_2\left(0^{+}\right)=0$ and

$$\frac{k_1^{\prime }\left(\varphi \right)}{k_2^{\prime }\left(\varphi \right)}=\frac{\varphi cosh\varphi -sinh\varphi }{2{\varphi }^3{cosh}^3\varphi }.$$

Denote

$$k_3\left(\varphi \right)=\varphi cosh\varphi -sinh\varphi \mathrm{and}{k}_4\left(\varphi \right)=2{\varphi }^3{cosh}^3\varphi ,$$

it is easy to obtain $k_3\left(0^{+}\right)=k_4\left(0^{+}\right)=0$ and

$$\frac{k_4^{\prime }\left(\varphi \right)}{k_3^{\prime }\left(\varphi \right)}=\frac{6\varphi {cosh}^2\varphi }{sinh\varphi }+6{\varphi }^2{cosh}^2\varphi .$$

(10)

Since the function $v^2{cosh}^{2}v$ is strictly increasing on $(0,\infty )$, by Lemma 5, we see that the ratio in (10) is strictly increasing and $\frac{k_3^{\prime }\left(\varphi \right)}{k_4^{\prime }\left(\varphi \right)}$ is strictly decreasing on $\left(0,ln\left(1+\sqrt{2}\right)\right)$. Consequently, from Lemma 1, it follows that $G\left(\varphi \right)$ is strictly decreasing on $\left(0,ln\left(1+\sqrt{2}\right)\right)$.

The limits

$$\underset{\varphi \to 0^{+}}{lim}G\left(\varphi \right)=\frac{1}{6}\mathrm{and}\underset{\varphi \to ln{(1+\sqrt{2})}^{-}}{lim}G\left(\varphi \right)=\frac{1-{ln}^2\left(1+\sqrt{2}\right)}{3{ln}^2\left(1+\sqrt{2}\right)}$$

can be computed easily. The proof of Theorem 3 is thus complete.  □

Corollary 3. 

For all $\varphi \in \left(0,ln\left(1+\sqrt{2}\right)\right)$, the double inequality

$$1+{\alpha }_3\left({cosh}^4\varphi -1\right)<{\left(\frac{sinh\varphi }{\varphi }\right)}^2<1+{\beta }_3\left({cosh}^4\varphi -1\right)$$

(11)

holds if and only if

$${\alpha }_3\le \frac{1-{ln}^2\left(1+\sqrt{2}\right)}{3{ln}^2\left(1+\sqrt{2}\right)}=0.095767\dots \mathit{and}{\beta }_3\ge \frac{1}{6}.$$

4. A Double Inequality

From Lemma 5, we can deduce

$$\frac{sinhv}{v}<{cosh}^{2}v\mathrm{and}\frac{sinhv}{v}>\frac{{tanh}^{2}x}{v^2}$$

(12)

for $v\in (0,\infty )$. The inequality

$${\left(\frac{sinhv}{v}\right)}^3>coshv$$

(13)

for $v\in (0,\infty )$ can be found and has been applied in [17] (p. 65), [18] (p. 300), [19] (pp. 279, 3.6.9), and [20] (p. 260). In [21], (Lemma 3), Zhu recovered the fact stated in [19] (pp. 279, 3.6.9) that the exponent 3 in the inequality (13) is the least possible, that is, the inequality

$${\left(\frac{sinhv}{v}\right)}^p>coshv$$

(14)

for $x>0$ holds if and only if $p\le 3$.

Inspired by (12) and (14), we find out the following double inequality.

Theorem 4. 

The inequality

$${cosh}^{\alpha }v<\frac{sinhv}{v}<{cosh}^{\beta }v$$

(15)

for $v\ne 0$ holds if and only if $\alpha \le \frac{1}{3}$ and $\beta \ge 1$.

Proof. 

Let

$$h\left(v\right)=\frac{lnsinhv-lnv}{lncoshv}\triangleq \frac{f_1\left(v\right)}{f_2\left(v\right)}.$$

Direct calculation yields

$$\frac{f_1^{\prime }\left(v\right)}{f_2^{\prime }\left(v\right)}=\frac{v{cosh}^{2}v-sinhvcoshv}{v{sinh}^{2}v}=\frac{vcosh\left(2v\right)+v-sinh\left(2v\right)}{vcosh\left(2v\right)-v}\triangleq \frac{f_3\left(v\right)}{f_4\left(v\right)}.$$

Using the power series of $sinhv$ and $coshv$, we obtain

$$\begin{array}{c}{f}_3\left(v\right)=v+v\sum _{\ell =0}^{\infty }\frac{{\left(2v\right)}^{2\ell }}{(2\ell )!}-\sum _{\ell =0}^{\infty }\frac{{\left(2v\right)}^{2\ell +1}}{(2\ell +1)!}=\sum _{\ell =1}^{\infty }\left[\frac{2^{2\ell }}{(2\ell )!}-\frac{2^{2\ell +1}}{(2\ell +1)!}\right]v^{2\ell +1}\\ =\sum _{\ell =0}^{\infty }\frac{(2\ell +1)2^{2\ell +2}}{(2\ell +3)!}{v}^{2\ell +3}\triangleq \sum _{\ell =0}^{\infty }{u}_{\ell }{v}^{2\ell +3}\end{array}$$

and

$$f_4\left(v\right)=v\sum _{\ell =0}^{\infty }\frac{{\left(2v\right)}^{2\ell }}{(2\ell )!}-v=\sum _{\ell =1}^{\infty }\frac{2^{2\ell }}{(2\ell )!}{v}^{2\ell +1}=\sum _{\ell =0}^{\infty }\frac{2^{2\ell +2}}{(2\ell +2)!}{v}^{2\ell +3}\triangleq \sum _{\ell =0}^{\infty }{w}_{\ell }{v}^{2\ell +3},$$

where

$$u_{\ell }=\frac{(2\ell +1)2^{2\ell +2}}{(2\ell +3)!}\mathrm{and}{w}_{\ell }=\frac{2^{2\ell +2}}{(2\ell +2)!}.$$

When setting $c_{\ell }=\frac{u_{\ell }}{w_{\ell }}$, we obtain

$$c_{\ell }=\frac{2\ell +1}{2\ell +3}=1-\frac{2}{2\ell +3}$$

is increasing on $\ell \in \mathbb{N}$. Therefore, by Lemma 2, the ratio $\frac{f_3\left(v\right)}{f_4\left(v\right)}$ is increasing on $(0,\infty )$. Using Lemma 1, we obtain that

$$h\left(v\right)=\frac{f_1\left(v\right)}{f_2\left(v\right)}=\frac{f_1\left(v\right)-f_1\left(0^{+}\right)}{f_2\left(v\right)-f_2\left(0^{+}\right)}$$

is increasing on $(0,\infty )$.

Moreover, the limits ${lim}_{v\to 0^{+}}{h}_1=\frac{1}{3}$ and ${lim}_{v\to \infty }{h}_1=1$ are obvious. The proof of Lemma 4 is thus complete.  □

5. A Remark

For $v,r\in \mathbb{R}$, we have

$${\left(\frac{sinhv}{v}\right)}^r=1+\sum _{m=1}^{\infty }\left[\sum _{k=1}^{2m}\frac{{(-r)}_k}{k!}\sum _{j=1}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k}{j}\right)\frac{T(2m+j,j)}{\left(\genfrac{}{}{0pt}{}{2m+j}{j}\right)}\right]\frac{{\left(2v\right)}^{2m}}{\left(2m\right)!},$$

(16)

where the rising factorial ${\left(r\right)}_k$ is defined by

$${\left(r\right)}_k=\prod _{\ell =0}^{k-1}(r+\ell )=\left\{\begin{array}{cc}r(r+1)\cdots (r+k-1),\hfill & k\ge 1\hfill \\ 1,\hfill & k=0\hfill \end{array}\right.$$

and $T(2m+j,j)$ is called central factorial numbers of the second kind and can be computed by

$$T(n,\ell )=\frac{1}{\ell !}\sum _{j=0}^{\ell }{(-1)}^j\left(\genfrac{}{}{0pt}{}{\ell }{j}\right){\left(\frac{\ell }{2}-j\right)}^n.$$

for $n\ge \ell \ge 0$.

The series expansion (16) was recently derived in [22] (Corollary 4.1).

Can one find bounds of the function ${\left(\frac{sinhv}{v}\right)}^r$ for $v,r\in \mathbb{R}\backslash \left\{0\right\}$?

6. Conclusions

In this paper, we found out the largest values ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_3$ and the smallest values ${\beta }_1$, ${\beta }_2$, ${\beta }_3$ such that the double inequalities (2), (3), and (4) hold for all positive real number $s,t>0$ with $s\ne t$. Moreover, we presented some new sharp inequalities (8), (9), (11), and (15) involving the hyperbolic sine function $sinh\varphi$ and the hyperbolic cosine function $cosh\varphi$.