TITLE: Prove inequality $\tan(x) \arctan(x) \geqslant x^2$
QUESTION [5 upvotes]: 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. $x\arctan(x) \leqslant x^2$.

REPLY [4 votes]: It's enough   to prove that $f(x)>0,$     where
$$f(x)=\arctan{x}-\frac{x^2}{\tan{x}}$$ and $x\in\left(0,\frac{\pi}{2}\right).$
Indeed, by AM-GM $$f'(x)=\frac{1}{1+x^2}-\frac{2x\tan{x}-\frac{x^2}{\cos^2x}}{\tan^2x}=\frac{1}{1+x^2}+\frac{x^2}{\sin^2x}-\frac{2x\cos{x}}{\sin{x}}\geq$$
$$\geq\frac{2x}{\sqrt{1+x^2}\sin{x}}-\frac{2x\cos{x}}{\sin{x}}>\frac{2x}{\sqrt{1+\tan^2x}\sin{x}}-\frac{2x\cos{x}}{\sin{x}}=0.$$
Id est, $$f(x)>\lim_{x\rightarrow0^+}f(x)=0$$ and we are done!