Prove inequality $\tan(x) \arctan(x) \geqslant x^2$ Prove that for $x\in \left( - \frac{\pi} {2},\,\frac{\pi}{2}\right)$ the following inequality holds
$$\tan(x) \arctan(x) \geqslant x^2.$$
I have tried proving that function $f(x) := \tan(x) \arctan(x) - x^2 \geqslant 0$ by using derivatives but it gets really messy and I couldn't make it to the end. I also tried by using inequality $\tan(x) \geqslant x$ on the positive part of the interval but this is too weak estimation and gives opposite result i.e. $x\arctan(x) \leqslant x^2$.

It's enough to prove this for $0<x<\pi/2$. Let $f(x)=(\tan x)/x$. Then $f$ is increasing
on $(0,\pi/2)$. To prove this, for instance $f(x)$ has nonnegative Maclaurin coefficients.
Let $x\in(0,\pi/2)$, and let $y=\arctan x$. Then $x=\tan y\ge y$ as $f(y)=(\tan y)/y\ge1$.
Therefore $g(y)\le g(x)$, that is
$$\frac{\tan y}y\le\frac{\tan x}x$$
or
$$\frac{x}{\arctan x}\le\frac{\tan x}x$$
etc.
