<s>[INST] How prove $\bigl(\frac{\sin x}{ x}\bigr)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$ How prove $\left(\frac{\sin x}{ x}\right)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$.
Can this be proved with simple way?
 [/INST] Set $y = 2\log\left(\frac{x}{\sin x}\right)$ and $z=\log\left(\frac{\tan x}{x}\right)$. 
Exploting Weierstrass products, for every $x\in I=\left(0,\frac{\pi}{2}\right)$ we have:
$$ z-y = \sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\frac{x^{2m}}{m\cdot \pi^{2m}}\left(\frac{4^m}{(2n-1)^{2m}}-\frac{3}{n^{2m}}\right), $$
so, given that $ \zeta(2m) = \sum_{n=1}^{+\infty} \frac{1}{n^{2m}},$
$$ z-y = \sum_{m=1}^{+\infty}\frac{x^{2m}\cdot\zeta(2m)}{m\cdot \pi^{2m}}(4^m-4).\tag{1} $$
Since every term in the RHS of $(1)$ is non-negative, it follows that $z > y$ for every $x\in I$. 
So we have:
$$ \left(\frac{\sin x}{x}\right)^2+\frac{\tan x}{x} = e^z + e^{-y} > e^{y}+e^{-y} \geq 2. \tag{2}$$
</s>