---
abstract: 'In this paper, we present monotonicity results of a function involving to the inverse hyperbolic sine. From these, we derive some inequalities for bounding the inverse hyperbolic sine.'
address:
- 'Research Institute of Mathematical Inequality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China'
- 'School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China'
author:
- Feng Qi
- 'Bai-Ni Guo'
date:
- Commenced on 10 March 2009 and completed on 11 March 2009 in Jiaozuo
- 
title: Monotonicity results and bounds for the inverse hyperbolic sine
---

[^1]

Introduction and main results
=============================

In [@Zhu-New-Arc-Hyper Theorem 1.9 and Theorem 1.10], the following inequalities were established: For $0\le x\le r$ and $r>0$, the double inequality $$\label{zhu-arcsinh-ineq-1}
\frac{(a+1)x}{a+\sqrt{1+x^2}\,}\le\operatorname{arcsinh}x\le \frac{(b+1)x}{b+\sqrt{1+x^2}\,}$$ holds true if and only if $a\le2$ and $$\label{zhu-cond-1}
b\ge\frac{\sqrt{1+r^2}\,\operatorname{arcsinh}r-r}{r-\operatorname{arcsinh}r}.$$

The aim of this paper is to elementarily generalize the inequality  to monotonicity results and to deduce more inequalities.

Our results may be stated as the following theorems.

\[Arc-Hyperbolic-Sine-thm1\] For $\theta\in\mathbb{R}$, let $$\label{f-theta(x)-dfn}
f_\theta(x)=\frac{\bigl(\theta+\sqrt{1+x^2}\,\bigr)\operatorname{arcsinh}x}{x},\quad  x>0.$$

1.  When $\theta\le2$, the function $f_\theta(x)$ is strictly increasing;

2.  When $\theta>2$, the function $f_\theta(x)$ has a unique minimum.

As straightforward consequences of Theorem \[Arc-Hyperbolic-Sine-thm1\], the following inequalities are inferred.

\[Arc-Hyperbolic-Sine-thm2\] Let $r>0$.

1.  For $\theta\le2$, the double inequality $$\label{Arc-Hyperbolic-Sine-ineq1}
    \frac{(1+\theta)x}{\theta+\sqrt{1+x^2}\,}<\operatorname{arcsinh}x\le \frac{\bigl[\bigl(\theta+\sqrt{1+r^2}\,\bigr)(\operatorname{arcsinh}r)/r\bigr]x}{\theta+\sqrt{1+x^2}\,}$$ holds true on $(0,r]$, where the constants $1+\theta$ and $\frac{\left(\theta+\sqrt{1+r^2}\,\right)\operatorname{arcsinh}r}{r}$ in  are the best possible.

2.  For $\theta>2$, the double inequality $$\label{Arc-Hyperbolic-Sine-ineq2}
    \frac{4\bigl(1-1/{\theta^2}\bigr)x}{\theta+\sqrt{1+x^2}\,}\le \operatorname{arcsinh}x \le \frac{\max\bigl\{1+\theta, \bigl(\theta+\sqrt{1+r^2}\,\bigr)(\operatorname{arcsinh}r)/r\bigr\}x} {\theta+\sqrt{1+x^2}\,}.$$

Replacing $\operatorname{arcsinh}x$ by $x$ in  and  yields $$\begin{gathered}
\left.\begin{aligned}
\frac{\bigl[\bigl(\theta+\sqrt{1+r^2}\,\bigr)(\operatorname{arcsinh}r)/r\bigr]\sinh x}{\theta+\cosh x}&\\
\frac{\max\bigl\{1+\theta, \bigl(\theta+\sqrt{1+r^2}\,\bigr)(\operatorname{arcsinh}r)/r\bigr\}\sinh x} {\theta+\cosh x}&
\end{aligned}\right\}>x\\
>\begin{cases}
\dfrac{(1+\theta)\sinh x}{\theta+\cosh x},&\theta\le2\\[0.7em]
\dfrac{4\bigl(1-1/{\theta^2}\bigr)\sinh x}{\theta+\cosh x},&\theta>2
\end{cases}\end{gathered}$$ for $x\in(0,\operatorname{arcsinh}r)$. This can be regarded as Oppenheim type inequalities for the hyperbolic sine and cosine functions. For information on Oppenheim’s double inequality for the sine and cosine functions, please refer to [@Oppeheim-Sin-Cos.tex] and closely related references therein.

It is clear that the left-hand side inequality in  recovers the left-hand side inequality in  while the right-hand side inequalities in  and  do not include each other.

By similar approach to prove our theorems in next section, we can procure similar monotonicity results and inequalities for the inverse hyperbolic cosine.

Proof of theorems
=================

Now we prove our theorems elementarily.

Direct differentiation yields $$\begin{aligned}
f_\theta'(x)&=\frac1{x^2}\biggl(\theta+\frac1{\sqrt{x^2+1}\,}\biggr) \biggl[\frac{x\bigl({\theta}/{\sqrt{x^2+1}\,}+1\bigr)} {\theta+1/{\sqrt{x^2+1}\,}}-\operatorname{arcsinh}x\biggr]\\
&\triangleq\frac1{x^2}\biggl(\theta+\frac1{\sqrt{x^2+1}\,}\biggr)h_{\theta}(x),\\
h_{\theta}'(x)&=\frac{x^2 \bigl(2-\theta^2+\theta\sqrt{x^2+1}\,\bigr)} {\sqrt{x^2+1}\,\bigl(\theta\sqrt{x^2+1}\,+1\bigr)^2}\\
&\triangleq\frac{x^2q_x(\theta)} {\sqrt{x^2+1}\,\bigl(\theta\sqrt{x^2+1}\,+1\bigr)^2}.\end{aligned}$$ The function $q_x(\theta)$ has two zeros $$\theta_1(x)=\frac{\sqrt{1+x^2}\,-\sqrt{9+x^2}\,}2\quad \text{and}\quad \theta_2(x)=\frac{\sqrt{1+x^2}\,+\sqrt{9+x^2}\,}2.$$ They are strictly increasing and have the bounds $-1\le\theta_1(x)<0$ and $\theta_2(x)\ge2$ on $(0,\infty)$. As a result, under the condition $\theta\not\in(-1,0)$,

1.  when $\theta\le-1$, the function $q_x(\theta)$ and the derivative $h_{\theta}'(x)$ are negative, and so the function $h_{\theta}(x)$ is strictly decreasing on $(0,\infty)$;

2.  when $0\le\theta\le2$, the function $q_x(\theta)$ and the derivative $h_{\theta}'(x)$ is positive, and so the function $h_{\theta}(x)$ is strictly increasing on $(0,\infty)$;

3.  when $\theta>2$, the function $q_x(\theta)$ and the derivative $h_{\theta}'(x)$ have a unique zero which is the unique minimum point of $h_{\theta}(x)$.

Furthermore, since $\lim_{x\to\infty}h_{\theta}(x)=\infty$ for $\theta\ge0$ and $\lim_{x\to0^+}h_{\theta}(x)=0$, it follows that

1.  when $\theta\le-1$, the function $h_{\theta}(x)$ is negative, and so the derivative $f_\theta'(x)$ is positive, that is, the function $f_\theta(x)$ is strictly increasing on $(0,\infty)$;

2.  when $0\le\theta\le2$, the function $h_{\theta}(x)$ is positive, and so the derivative $f_\theta'(x)$ is also positive, accordingly, the function $f_\theta(x)$ is strictly increasing on $(0,\infty)$;

3.  when $\theta>2$, the function $h_{\theta}(x)$ and the derivative $f_\theta'(x)$ have a unique zero which is the unique minimum point of the function $f_\theta(x)$ on $(0,\infty)$.

On the other hand, when $\theta\in(-1,0)$, we have $$\begin{aligned}
\bigl[x^2f_\theta'(x)\bigr]'&=\frac{x^2\bigl[\sqrt{x^2+1}\, +(\operatorname{arcsinh}x)/x-\theta\bigr]} {(x^2+1)^{3/2}}>0\end{aligned}$$ which means that the function $x^2f_\theta'(x)$ is strictly increasing on $(0,\infty)$. From the limit $\lim_{x\to0^+}\bigl[x^2f_\theta'(x)\bigr]=0$, it is derived that the function $x^2f_\theta'(x)$ is positive. Hence the function $f_\theta(x)$ is strictly increasing on $(0,\infty)$. The proof of Theorem \[Arc-Hyperbolic-Sine-thm1\] is complete.

Since $\lim_{x\to0^+}f_\theta(x)=1+\theta$, by Theorem \[Arc-Hyperbolic-Sine-thm1\], it is easy to see that $1+\theta<f_\theta(x)\le f_\theta(r)$ on $(0,r]$ for $\theta\le2$. The inequality  is thus proved.

For $\theta>2$, the minimum point $x_0\in(0,\infty)$ satisfies $$\operatorname{arcsinh}x_0=\frac{x_0\bigl({\theta}/{\sqrt{x_0^2+1}\,}+1\bigr)} {\theta+1/{\sqrt{x_0^2+1}\,}}.$$ Therefore, the minimum of the function $f_\theta(x)$ on $(0,\infty)$ equals $$\begin{gathered}
\frac{\bigl(\theta+\sqrt{1+x_0^2}\,\bigr)\operatorname{arcsinh}x_0}{x_0} =\frac{x_0\bigl({\theta}/{\sqrt{x_0^2+1}\,}+1\bigr)} {\theta+1/{\sqrt{x_0^2+1}\,}} \cdot \frac{\bigl(\theta+\sqrt{1+x_0^2}\,\bigr)}{x_0}\\
=\frac{\bigl({\theta}/{\sqrt{x_0^2+1}\,}+1\bigr)\bigl(\theta+\sqrt{1+x_0^2}\,\bigr)} {\theta+1/{\sqrt{x_0^2+1}\,}} =\frac{\bigl(\theta+\sqrt{1+x_0^2}\,\bigr)^2} {\theta{\sqrt{x_0^2+1}\,}+1} \ge4\biggl(1-\frac1{\theta^2}\biggr).\end{gathered}$$ From this, it is obtained that $$4\biggl(1-\frac1{\theta^2}\biggr)\le f_\theta(x)\le \max\Bigl\{\lim_{x\to0^+}f_\theta(x),f_\theta(r)\Bigr\}$$ for $x\in(0,r]$, which implies the inequality . The proof of Theorem \[Arc-Hyperbolic-Sine-thm2\] is thus completed.

[9]{}

F. Qi and B.-N. Guo, *A concise proof of Oppenheim’s double inequality relating to the cosine and sine functions*, Avaliable online at <http://arxiv.org/abs/0902.2511>.

L. Zhu, *New inequalities of Shafer-Fink type for arc-hyperbolic sine*, J. Inequal. Appl. **2008** (2008), Article ID 368275, 5 pages; Available online at <http://dx.doi.org/10.1155/2008/368275>.

[^1]: This paper was typeset using -LaTeX
