A double inequality for tanhx

Abstract

In this paper, we prove that, for $x>0$,

$$ \sqrt{1-\exp \biggl(-\frac{x^{2}}{\sqrt{x^{2}+1}} \biggr)}< \tanh x< \sqrt[3]{1-\exp \biggl(- \frac{x^{3}}{\sqrt{x^{3}+1}} \biggr)}. $$

This solves an open problem proposed by Ivády.

1 Introduction

Ivády (see [1, Problem 51]) proposed the following problem: Show that, for $x>0$,

$$ \begin{gathered}\sqrt{1-\exp \biggl(-\frac{x^{2}}{\sqrt{x^{2}+1}} \biggr)}< \tanh x< \sqrt[3]{1- \exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr)}\end{gathered} $$

(1.1)

holds. Subsequently, a solution was presented by the proposer (see [2]).

In [2], the proof of the left-hand side of (1.1) is correct, but the proof of the right-hand side of (1.1) is not correct.

Using the inverse function of tanhx, the second inequality in (1.1) has the equivalent form

$$ \begin{gathered}\frac{1}{2}\ln \biggl(\frac{1+\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}}{1-\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}} \biggr)>x\quad \text{for } x>0.\end{gathered} $$

(1.2)

According to the mean-value theorem, Ivády [2, Eq. (8)] got on $[0, x]$

$$ \begin{aligned}[b]&\frac{1}{2x}\ln \biggl(\frac{1+\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}}{1-\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}} \biggr)\\ &\quad= \frac{\frac{1}{2}\ln (\frac{1+\sqrt[3]{1-\exp (-\frac{x^{3}}{\sqrt{x^{3}+1}} )}}{1-\sqrt[3]{1-\exp (-\frac{x^{3}}{\sqrt{x^{3}+1}} )}} )}{x} =\frac{\eta^{2}(\eta^{3}+2)}{2(\eta^{3}+1)^{3/2} (1-\exp (-\frac{\eta^{3}}{\sqrt{\eta^{3}+1}} ) )^{2/3} (1- (1-\exp (-\frac{\eta^{3}}{\sqrt{\eta^{3}+1}} ) )^{2/3} )}\hspace{-12pt}\end{aligned} $$

(1.3)

for some $0 < \eta< x$, and then proved that, for all $\eta> 0$,

$$\begin{aligned} \frac{\eta^{2}(\eta^{3}+2)}{2(\eta^{3}+1)^{3/2} (1-\exp (-\frac{\eta^{3}}{\sqrt{\eta^{3}+1}} ) )^{2/3} (1- (1-\exp (-\frac{\eta^{3}}{\sqrt{\eta^{3}+1}} ) )^{2/3} )}>1. \end{aligned}$$

We note that (1.3) may be corrected as

$$\begin{aligned} &\frac{1}{2x}\ln \biggl(\frac{1+\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}}{1-\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}} \biggr) \\ &\quad=\frac{\eta^{2}(\eta^{3}+2)\exp (-\frac{\eta^{3}}{\sqrt {\eta^{3}+1}} )}{2(\eta^{3}+1)^{3/2} (1-\exp (-\frac {\eta^{3}}{\sqrt{\eta^{3}+1}} ) )^{2/3} (1- (1-\exp (-\frac{\eta^{3}}{\sqrt{\eta^{3}+1}} ) )^{2/3} )} \end{aligned}$$

for some $0 < \eta< x$.

In this paper, we provide a proof of the right-hand side of (1.1).

The numerical values given in this paper have been calculated via the computer program MAPLE 13.

2 Lemma

Lemma 2.1

Let

$$\begin{aligned} G(x)&=\frac{x^{2}(x^{3}+2)}{2(x^{3}+1)^{3/2}}\exp \biggl(-\frac{x^{3}}{\sqrt {x^{3}+1}} \biggr)- \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{2/3} \\ &\quad+ \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{4/3}. \end{aligned}$$

(2.1)

Then, for $x>0$,

$$\begin{aligned} G(x)>0. \end{aligned}$$

(2.2)

Proof

We split the proof into three cases.

Case 1. $0< x<0.5$.

We first prove the following inequalities:

$$\begin{aligned}& \frac{x^{3}+2}{2(x^{3}+1)^{3/2}}\exp \biggl(-\frac{x^{3}}{\sqrt {x^{3}+1}} \biggr)>1-2x^{3}, \end{aligned}$$

(2.3)

$$\begin{aligned}& \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{2/3}< x^{2}-\frac{2}{3}x^{5}+ \frac{7}{12}x^{8}, \end{aligned}$$

(2.4)

and

$$\begin{aligned} \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{4/3}>x^{4}-\frac{4}{3}x^{7} \end{aligned}$$

(2.5)

for $0< x<0.5$.

The inequality (2.3) can be converted to

$$\begin{aligned} \frac{y}{\sqrt{y+1}}+\ln \biggl(\frac{2(y+1)^{3/2}(1-2y)}{y+2} \biggr)< 0 \quad\text{for } 0< y< 0.125. \end{aligned}$$

We consider the function $f_{1}(y)$ defined, for $0< y<0.125$, by

$$\begin{aligned} f_{1}(y)=\frac{y}{\sqrt{y+1}}+\ln2 +\frac{3}{2}\ln(y+1)+\ln (1-2y)-\ln(y+2). \end{aligned}$$

Differentiation yields

$$\begin{aligned} -2(y+1)f'_{1}(y)=\frac{6y^{2}+19y+4}{(1-2y)(y+2)}-\frac{y+2}{(y+1)^{1/2}}. \end{aligned}$$

By direct computation, we get, for $0< y<0.125$,

$$\begin{aligned} \biggl(\frac{6y^{2}+19y+4}{(1-2y)(y+2)} \biggr)^{2}-\frac {(y+2)^{2}}{y+1}= \frac{y (4y^{4}(2-y)+199y^{3}+597y^{2}+601y+200 )}{(1-2y)^{2}(y+2)^{2}(y+1)}>0. \end{aligned}$$

We then obtain $f'_{1}(y)<0$ for $0< y<0.125$. Hence, $f_{1}(y)$ is strictly decreasing for $0< y<0.125$, and we have

$$\begin{aligned} f_{1}(y)=\frac{y}{\sqrt{y+1}}+\ln \biggl(\frac {2(y+1)^{3/2}(1-2y)}{y+2} \biggr)< f_{1}(0)=0. \end{aligned}$$

The inequality (2.4) can be written for $0< x<0.5$ as

$$\begin{aligned} \frac{x^{3}}{\sqrt{x^{3}+1}}+\ln \biggl(1-x^{3} \biggl(1- \frac {2}{3}x^{3}+\frac{7}{12}x^{6} \biggr)^{3/2} \biggr)< 0. \end{aligned}$$

(2.6)

In order to prove (2.6), it suffices to show that

$$\begin{aligned} f_{2}(y)< 0\quad\text{for } 0< y< 0.125, \end{aligned}$$

where

$$\begin{aligned} f_{2}(y)=\frac{y}{\sqrt{y+1}}+\ln \biggl(1-y \biggl(1- \frac {2}{3}y+\frac{7}{12}y^{2} \biggr)^{3/2} \biggr). \end{aligned}$$

Differentiation yields

$$\begin{aligned} -f'_{2}(y)=\frac{4(3-5y+7y^{2})\sqrt {36-24y+21y^{2}}}{72-(12y-8y^{2}+7y^{3})\sqrt{36-24y+21y^{2}}}-\frac {y+2}{2(y+1)^{3/2}}. \end{aligned}$$

We now prove $f'_{2}(y)<0$ for $0< y<0.125$. It suffices to show that, for $0< y<0.125$,

$$\begin{aligned} \frac{4(3-5y+7y^{2})\sqrt{36-24y+21y^{2}}}{72-(12y-8y^{2}+7y^{3})\sqrt {36-24y+21y^{2}}}>\frac{y+2}{2(y+1)^{3/2}}. \end{aligned}$$

It is not difficult to prove that

$$\begin{aligned} \frac{4(3-5y+7y^{2})\sqrt{36-24y+21y^{2}}}{72-(12y-8y^{2}+7y^{3})\sqrt {36-24y+21y^{2}}}>1-y+\frac{9}{8}y^{2} \end{aligned}$$

and

$$\begin{aligned} \frac{y+2}{2(y+1)^{3/2}}< 1-y+\frac{9}{8}y^{2} \end{aligned}$$

for $0< y<0.125$ (we here omit the proofs). Hence, $f'_{2}(y)<0$ holds for $0< y<0.125$. So, $f_{2}(y)$ is strictly decreasing for $0< y<0.125$, and we have

$$\begin{aligned} f_{2}(y)=\frac{y}{\sqrt{y+1}}+\ln \biggl(1-y \biggl(1- \frac {2}{3}y+\frac{7}{12}y^{2} \biggr)^{3/2} \biggr)< f_{2}(0)=0 \end{aligned}$$

for $0< y<0.125$.

The inequality (2.5) can be written for $0< x<0.5$ as

$$\begin{aligned} \frac{x^{3}}{\sqrt{x^{3}+1}}+\ln \biggl(1-x^{3} \biggl(1- \frac {4}{3}x^{3} \biggr)^{3/4} \biggr)>0. \end{aligned}$$

(2.7)

In order to prove (2.7), it suffices to show that

$$\begin{aligned} f_{3}(y)>0\quad\text{for } 0< y< 0.125, \end{aligned}$$

where

$$\begin{aligned} f_{3}(y)=\frac{y}{\sqrt{y+1}}+\ln \biggl(1-y \biggl(1- \frac {4}{3}y \biggr)^{3/4} \biggr). \end{aligned}$$

Differentiation yields

$$\begin{aligned} f'_{3}(y)=\frac{y+2}{2(y+1)^{3/2}}-\frac{3-7y}{3 (1-\frac {4}{3}y )^{1/4} (1-y(1-\frac{4}{3}y)^{3/4} )}. \end{aligned}$$

We now prove $f'_{3}(y)>0$ for $0< y<0.125$. It suffices to show that, for $0< y<0.125$,

$$\begin{aligned} 1-y \biggl(1-\frac{4}{3}y \biggr)^{3/4}>\frac {2(3-7y)(y+1)^{3/2}}{3(y+2) (1-\frac{4}{3}y )^{1/4}}. \end{aligned}$$

It is not difficult to prove that

$$\begin{aligned} 1-y \biggl(1-\frac{4}{3}y \biggr)^{3/4}>1-y \quad\text{and}\quad \frac{2(3-7y)(y+1)^{3/2}}{3(y+2) (1-\frac{4}{3}y )^{1/4}}< 1-y \end{aligned}$$

for $0< y<0.125$ (we here omit the proofs). Hence, $f'_{3}(y)>0$ holds for $0< y<0.125$. So, $f_{3}(y)$ is strictly increasing for $0< y<0.125$, and we have

$$\begin{aligned} f_{3}(y)=\frac{y}{\sqrt{y+1}}+\ln \biggl(1-y \biggl(1- \frac {4}{3}y \biggr)^{3/4} \biggr)>f_{3}(0)=0 \end{aligned}$$

for $0< y<0.125$.

We then obtain by (2.3), (2.4) and (2.5), for $0< x<0.5$,

$$\begin{aligned} G(x)&>x^{2}\bigl(1-2x^{3}\bigr)- \biggl(x^{2}- \frac{2}{3}x^{5}+\frac{7}{12}x^{8} \biggr)+x^{4}-\frac{4}{3}x^{7} \\ &=\frac{1}{12}x^{4} \bigl(12-16x-16x^{3}-7x^{4} \bigr)>0. \end{aligned}$$

Case 2. $0.5\leq x \leq3$.

Let

$$\begin{aligned} G(x)=G_{1}(x)+G_{2}(x), \end{aligned}$$

where

$$\begin{aligned} G_{1}(x)&=\frac{x^{2}(x^{3}+2)}{2(x^{3}+1)^{3/2}}\exp \biggl(- \frac {x^{3}}{\sqrt{x^{3}+1}} \biggr)- \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt {x^{3}+1}} \biggr) \biggr)^{2/3} \end{aligned}$$

(2.8)

and

$$\begin{aligned} G_{2}(x)&= \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{4/3}. \end{aligned}$$

(2.9)

It is not difficult to prove that

$$\begin{aligned} \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{1/3}< x,\quad x>0 \end{aligned}$$

(2.10)

(we here omit the proof). Differentiating $G_{1}(x)$ and using (2.10), we obtain, for $x>0$,

$$\begin{aligned} &{-}\frac{(x^{3}+1)^{3/2}}{x}\exp \biggl(\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr)G'_{1}(x) \\ &\quad=\frac{-(x^{6}+8)\sqrt{x^{3}+1}+3x^{3}(x^{3}+2)^{2}}{4(x^{3}+1)^{3/2}} +\frac{x(x^{3}+2)}{ (1-\exp (-\frac{x^{3}}{\sqrt {x^{3}+1}} ) )^{1/3}} \\ &\quad>\frac{-(x^{6}+8)\sqrt{x^{3}+1}+3x^{3}(x^{3}+2)^{2}}{4(x^{3}+1)^{3/2}} +x^{3}+2 \\ &\quad=\frac{3x^{3} ((x^{3}+4)\sqrt{x^{3}+1}+x^{6}+4x^{3}+4 )}{4(x^{3}+1)^{3/2}}>0. \end{aligned}$$

Therefore, the function $G_{1}(x)$ is strictly decreasing for $x>0$.

Differentiation yields

$$\begin{aligned} &G'_{2}(x)= \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{1/3}\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \frac {2x^{2}(x^{3}+2)}{(x^{3}+1)^{3/2}}>0. \end{aligned}$$

Therefore, the function $G_{2}(x)$ is strictly increasing for $x>0$.

Let $0.5\leq r \leq x \leq s \leq3$. Since $G_{1}(x)$ is decreasing and $G_{2}(x)$ is increasing, we obtain

$$\begin{aligned} G(x)\geq G_{1}(s)+G_{2}(r)=:\sigma(r,s). \end{aligned}$$

We divide the interval $[0.5, 3]$ into 250 subintervals:

$$\begin{aligned}{} [0.5, 3]=\bigcup_{k=0}^{249} \biggl[0.5+ \frac{k}{100}, 0.5+\frac {k+1}{100} \biggr]\quad\text{for } k = 0, 1, 2, \ldots, 249. \end{aligned}$$

By direct computation we get

$$\begin{aligned} \sigma \biggl(0.5+\frac{k}{100}, 0.5+\frac{k+1}{100} \biggr)>0\quad \text{for } k = 0, 1, 2, \ldots, 249. \end{aligned}$$

Hence,

$$\begin{aligned} G(x)>0 \quad\text{for } x\in \biggl[0.5+\frac{k}{100}, 0.5+ \frac {k+1}{100} \biggr]\text{ and } k = 0, 1, 2, \ldots, 249. \end{aligned}$$

This implies that $G(x)$ is positive for $0.5\leq x \leq3$.

Case 3. $x>3$.

We first prove that, for $x>3$,

$$\begin{aligned} \frac{3x^{3}(x^{3}+2)-(x^{3}+1)^{3/2}}{4\sqrt{x}} \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{1/3}>x^{2}\bigl(x^{3}+2\bigr), \end{aligned}$$

(2.11)

which can be written for $x>3$ as

$$\begin{aligned} \frac{x^{3}}{\sqrt{x^{3}+1}}+\ln \biggl(1- \biggl(\frac {4x^{5/2}(x^{3}+2)}{3x^{3}(x^{3}+2)-(x^{3}+1)^{3/2}} \biggr)^{3} \biggr)>0, \end{aligned}$$

(2.12)

which can be converted to

$$\begin{aligned} \frac{y}{\sqrt{y+1}}+\ln \biggl(1- \biggl(\frac {4y^{5/6}(y+2)}{3y(y+2)-(y+1)^{3/2}} \biggr)^{3} \biggr)>0\quad\text{for } y>27. \end{aligned}$$

(2.13)

It is not difficult to prove that

$$\begin{aligned} \frac{y}{\sqrt{y+1}}>\sqrt{y}-\frac{1}{2\sqrt{y}} \end{aligned}$$

and

$$\begin{aligned} 1- \biggl(\frac{4y^{5/6}(y+2)}{3y(y+2)-(y+1)^{3/2}} \biggr)^{3}>1-\frac {64}{27\sqrt{y}}- \frac{64}{27y}-\frac{128}{81y^{3/2}} \end{aligned}$$

for $y>27$ (we here omit the proofs).

In order to prove (2.13), it suffices to show that

$$\begin{aligned} \sqrt{y}-\frac{1}{2\sqrt{y}}+\ln \biggl(1-\frac{64}{27\sqrt {y}}- \frac{64}{27y}-\frac{128}{81y^{3/2}} \biggr)>0\quad\text{for } y>27. \end{aligned}$$

(2.14)

We consider the function $f_{4}(z)$ defined, for $z>3\sqrt{3}$, by

$$\begin{aligned} f_{4}(z)=z-\frac{1}{2z}+\ln \biggl(1-\frac{64}{27z}- \frac {64}{27z^{2}}-\frac{128}{81z^{3}} \biggr). \end{aligned}$$

Differentiation yields, for $z>3\sqrt{3}$,

$$\begin{aligned} f'_{4}(z)=\frac {162z^{5}-384z^{4}+81z^{3}+320z^{2}+576z-128}{2z^{2}(81z^{3}-192z^{2}-192z-128)}>0. \end{aligned}$$

Hence, $f_{4}(z)$ is strictly increasing for $z>3\sqrt{3}$, and we have

$$\begin{aligned} f_{4}(z)=z-\frac{1}{2z}+\ln \biggl(1-\frac{64}{27z}- \frac {64}{27z^{2}}-\frac{128}{81z^{3}} \biggr)>f_{4}(3\sqrt{3})=4.289 \ldots>0 \end{aligned}$$

for $z>3\sqrt{3}$. Hence, (2.14) holds for $y>27$.

We now prove $G(x)>0$ for $x>3$. It is easy to see that

$$\begin{aligned} \frac{x^{2}(x^{3}+2)}{(x^{3}+1)^{3/2}}>\sqrt{x}\quad\text{for } x>3. \end{aligned}$$

In order to prove $G(x)>0$ for $x>3$, it is enough to prove the following inequality:

$$\begin{aligned} H(x)>0 \quad\text{for } x>3, \end{aligned}$$

where

$$\begin{aligned} H(x)&=\frac{\sqrt{x}}{2}\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr)- \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{2/3} \\ &\quad+ \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{4/3}. \end{aligned}$$

Differentiating $H(x)$ and using (2.11) yield, for $x>3$,

$$\begin{aligned} &{-}\bigl(x^{3}+1\bigr)^{3/2}\exp \biggl(\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr)H'(x) \\ &\quad=\frac{3x^{3}(x^{3}+2)-(x^{3}+1)^{3/2}}{4\sqrt{x}} \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{1/3} \\ &\qquad+x^{2}\bigl(x^{3}+2\bigr)-2x^{2} \bigl(x^{3}+2\bigr) \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt {x^{3}+1}} \biggr) \biggr)^{2/3} \\ &\quad>2x^{2}\bigl(x^{3}+2\bigr) \biggl\{ 1- \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt {x^{3}+1}} \biggr) \biggr)^{2/3} \biggr\} >0. \end{aligned}$$

Therefore, the function $H(x)$ is strictly decreasing for $x\geq3$, and we have

$$ H(x)>\lim_{t\to\infty}H(t)=0 \quad\text{for } x\geq3. $$

Hence, we have $G(x)>0$ for all $x>0$. The proof of Lemma 2.1 is complete. □

3 Proof of the right-hand side of (1.1)

It is sufficient to prove the following inequality:

$$ F(x)=\frac{1}{2}\ln \biggl(\frac{1+\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}}{1-\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}} \biggr)-x>0 \quad\text{for } x>0. $$

(3.1)

Differentiating $F(x)$ and using (2.2), we obtain, for $x>0$,

$$\begin{aligned} & \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{2/3} \biggl[1- \biggl(1-\exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr) \biggr)^{2/3} \biggr]F'(x)=G(x)>0, \end{aligned}$$

where $G(x)$ is given in (2.1). Therefore, $F(x)$ is strictly increasing for $x>0$, and we have

$$ F(x)=\frac{1}{2}\ln \biggl(\frac{1+\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}}{1-\sqrt[3]{1-\exp (-\frac {x^{3}}{\sqrt{x^{3}+1}} )}} \biggr)-x>F(0)=0 $$

for $x>0$. The proof is complete.