# A double inequality for tanhx

## Abstract

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

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

This solves an open problem proposed by Ivády.

## Introduction

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

$$\begin{gathered}\sqrt{1-\exp \biggl(-\frac{x^{2}}{\sqrt{x^{2}+1}} \biggr)}< \tanh x< \sqrt{1- \exp \biggl(-\frac{x^{3}}{\sqrt{x^{3}+1}} \biggr)}\end{gathered}$$
(1.1)

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

In , 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

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

## 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. □

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

It is sufficient to prove the following inequality:

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

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

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

## References

1. 1.

Cakić, N.P., Merkle, M.J. (eds.): Problem section. Publ. Elektroteh. Fak. Univ. Beogr., Mat. 14, 111–114 (2003). http://pefmath2.etf.rs/files/123/problem_section.pdf

2. 2.

Cakić, N.P. (ed.): Problem section. Publ. Elektroteh. Fak. Univ. Beogr., Mat. 16, 146–155 (2005). http://pefmath2.etf.rs/files/125/problem_section.pdf

Not applicable.

## Author information

Both authors contributed equally to this work. Both authors read and approved the final manuscript.

Correspondence to Bo Zhang.

## Ethics declarations

### Competing interests

The authors declare that they have no competing interests. 