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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. $x\arctan(x) \leqslant x^2$.

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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.

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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!

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