<s>[INST] Inequality involving the tangent function How can one show that
$$x \tan x > \frac{4x - \pi}{\pi - 2x},$$
for $x \in \left(0,\frac{\pi}{2}\right)$. 
Clearly, for $x \in \left(0,\frac{\pi}{4}\right)$ as $4x - \pi < 0$ and $\pi - 2x > 0$ the term $\frac{4x - \pi}{\pi - 2x}$ is negative while $x \tan x$ is always positive, so one need only show the inequality is true for $x \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$. 
 [/INST] For $0<x\leq\frac{\pi}{4}$ our inequality is obviously true.
Let $\frac{\pi}{4}<x<\frac{\pi}{2}$ and $t=\tan{x}$. 
Hence, $t>1$ and we need to prove that
$$(\pi-2x)x\tan{x}>4x-\pi$$ or
$$2x^2\tan{x}+(4-\pi\tan{x})x-\pi>0$$ or
$$x<\frac{-4+\pi\tan{x}+\sqrt{16+\pi^2\tan^2x}}{4\tan{x}}$$
or $f(x)>0$, where
$$f(x)=\frac{-4+\pi\tan{x}+\sqrt{16+\pi^2\tan^2x}}{4\tan{x}}-x.$$
We see that
$$f'(x)=\frac{\cos^2x\sqrt{16+\pi^2\tan^2x}-4}{\sin^2x\sqrt{16+\pi^2\tan^2x}}.$$
We'll prove that $f'(x)<0$ for all $\frac{\pi}{4}<x<\frac{\pi}{2}$.
Indeed, it's enough to prove that
$$\cos^2x\sqrt{16+\pi^2\tan^2x}<4$$ or
$$\sqrt{16+\pi^2t^2}<4(1+t^2)$$ or
$$16t^4+(32-\pi^2)t^2>0,$$
which is obvious.
Id est, $$f(x)>\lim\limits_{x\rightarrow\frac{\pi}{2}^+}f(x)=\frac{\pi}{4}+\frac{\pi}{4}-\frac{\pi}{2}=0$$
and we are done!
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