Prove : $\arctan(x)>\tanh(x)$ for $0< x\leq 1$

Question

It's a classical inequality that I want to propose :

Claim:

Let $0< x\leq 1$ then we have : $$\arctan(x)>\tanh(x)$$

My proof :

We introduce the function :

$$f(x)=\arctan(x)-\tanh(x)$$

We differentiate :

$$f'(x)=\frac{1}{(x^2+1)}+\frac{-4}{(e^{-x} + e^{x})^2}$$

Remains to show :

$$\frac{(e^{-x} + e^{x})^2}{4}\geq x^2+1\quad \quad (0)$$

Or :

$$-\frac{1}{2}+\frac{e^{-2x}}{4}+\frac{e^{2x}}{4}-x^2\geq 0$$

We introduce the function :

$$g(x)=-\frac{1}{2}+\frac{e^{-2x}}{4}+\frac{e^{2x}}{4}-x^2$$

We differentiate twice :

$$g''(x)=e^{-2x}+e^{2x}-2\geq 0$$

We deduce that the first derivative of $g(x)$ is increasing and $g'(0)=0$ so $g(x)$ is increasing but $g(0)=0$ so we establish the inequality $(0)$ . We conclude that the function $f(x)$ is increasing and $f(0)=0$ wich prove the claim .

My question :

Have you one or more alternative proof (using integration by example)?

Thanks in advance !

Erik

Answer 1

Note $\sinh t-t>0$ and

\begin{align} \arctan x-\tanh x & =\int_0^x \left(\frac{1}{t^2+1}-\frac{1}{\cosh^2 t}\right)dt\\ &= \int_0^x \frac{(\sinh t +t)(\sinh t-t)}{(t^2+1)\cosh^2 t}dt>0 \end{align} Thus, $\arctan x>\tanh x$.

Answer 2

For all $u$, $$ \sin^2(u)\le u^2\implies\cos^2(u)\ge1-u^2\implies\sec^2(u)\left(1-u^2\right)\le1\tag1 $$ Set $u=\tanh(x)$: $$\newcommand{\sech}{\operatorname{sech}} \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\tan(\tanh(x)) &=\sec^2(\tanh(x))\sech^2(x)\tag{2a}\\ &=\sec^2(\tanh(x))\left(1-\tanh^2(x)\right)\tag{2b}\\[3pt] &\le1\tag{2c} \end{align} $$ Thus, for $x\ge0$, the Mean Value Theorem says $$ \tan(\tanh(x))\le x\tag3 $$ Therefore, because $\arctan(x)$ is increasing, $$ \tanh(x)\le\arctan(x)\tag4 $$