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 $x \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$. 
 A: By performing the change of variable $x\mapsto\left(\frac{\pi}{2}-x\right)$ it is enough to show that
$$\left(\frac{\pi}{2}-x\right)\cot(x) > -2+\frac{\pi}{2x} $$
holds for any $x\in\left(0,\frac{\pi}{4}\right)$, or that over the same interval
$$\frac{\tan(x)}{x}<\frac{\frac{\pi}{2}-x}{\frac{\pi}{2}-2x} \tag{1} $$
holds. That can be done by comparing the coefficients in the Taylor series of the RHS and LHS at the origin. By considering the logarithmic derivative of the Weierstrass product for the cosine function,
$$ \frac{\tan(x)}{x}=\sum_{k\geq 1}\frac{2(4^k-1)\zeta(2k)}{\pi^{2k}}\,x^{2k-2} \tag{2}$$
while
$$\frac{\frac{\pi}{2}-x}{\frac{\pi}{2}-2x}=1+\sum_{n\geq 1}\frac{2^{2n-1}}{\pi^n}\,x^n \tag{3}$$
hence it is enough to show that
$$ \frac{2^{4k-3}}{\pi^{2k-1}}x^{2k-1}+\frac{2^{4k-1}}{\pi^{2k}}x^{2k}\geq \frac{2(4^{k+1}-1)\zeta(2k+2)}{\pi^{2k+2}}x^{2k}\tag{4} $$
holds for every $k\geq 1$ and every $x\in\left(0,\frac{\pi}{4}\right)$, but that is straightforward.
A: 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!
A: Substitute $x$ by $\frac \pi 2 - x$ (as in Jack D'Aurizio's answer), so that
we have to show that
$$
 \tan x  < \frac{x (\pi - 2x)}{\pi - 4x} \quad \text{for } 0 < x < \frac \pi 4 \, .
$$
Denote the right-hand side with $g(x)$. The idea is to use the fact
that $\tan$ is a solution of the differential equation (initial
value problem)
$$
  y' = 1 + y^2 \, , \quad y(0) = 0
$$
and to show that $g$ satisfies a corresponding differential inequality.
We have $g(0) = 0$, and a straight-forward calculation gives
$$
 g'(x) - g(x)^2 - 1 = \frac{x(\pi - 2x)(2x^2 - \pi x + 4)}{(\pi - 4x)^2} > 0
$$
for  $0 < x < \frac \pi 4$.  
It follows that
$$
 \arctan(g(x)) = \int_0^x  \frac{g'(t)}{1 + g(t)^2} \, dt > \int_0^x 1 \, dt = x
$$
and therefore $g(x) > \tan x$.
