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
Best Answer
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, $\arctan x>\tanh x$.