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 $$0. Let $$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$$ etc.
It's enough to prove that $$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!