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?
 A: $$\left(\frac{\sin x}{x}\right)^2 + \left(\frac{\tan x}{x}\right) = \\ \left(\frac{x - \frac{x^3}{6} + \frac{x^5}{120} + \ldots}{x}\right)^2 + \left(\frac{x + \frac{x^3}{3} + \frac{2x^5}{15} + \ldots}{x}\right) = \\ \left(1 - \frac{x^2}{6} + \frac{x^4}{120} - \ldots\right)^2 + \left(1 + \frac{x^2}{3} + \frac{2x^4}{15} + \ldots\right) \simeq \\ \left(1 + 2 \left(-\frac{x^2}{6} + \frac{x^4}{120} - \ldots \right)\right) + 1 + \frac{x^2}{3} + \frac{2x^4}{15} + \ldots\\2 + x^2 \left(\frac{ - 2}{6} + \frac{1}{3}\right) + x^4 \left(\frac{2}{120} + \frac{2}{15} \right) + \ldots = \\2 + x^4 \left(\frac{9}{60} \right) + \ldots > 2.$$
A: 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:
$$ \left(\frac{\sin x}{x}\right)^2+\frac{\tan x}{x} = e^z + e^{-y} > e^{y}+e^{-y} \geq 2. \tag{2}$$
A: One could look at the function and see that it is monotonously rising in the given interval, one would have to look a the limits, if $x$ approaches $0$ or $\pi/2$.
This should work as well. Would you know, how to do that?
A: this is famous GENERALIZED WILKER inequality : see this interesting some reslut with two papers  1 and  2.
there have open problem:How to prove this general inequality $\displaystyle a\left(\frac{\sin{x}}{x}\right)^m+b\left(\frac{\tan{x}}{x}\right)^n>a+b$
A: since $\sin(x)>0$ and $x>0$ and $\tan(x)>0$ we have
$$\left(\frac{\sin(x)}{x}\right)^2+\frac{\tan(x)}{x}\geq 2\sqrt{\left(\frac{\sin(x)}{x}\right)^2\cdot \frac{\sin(x)}{x}\cdot\frac{1}{\cos(x)}}$$ Now we have to show that $$\frac{\sin(x)^3}{x^3}>\cos(x)$$ in the given interval, defining $$f(x)=\frac{\sin(x)^3}{x^3}-\cos(x)$$ and we get $$\lim_{x \to 0+}f(x)=0$$ and $$f'(x)=-{\frac {\sin \left( x \right)  \left( -{x}^{4}-3\,\sin \left( x
 \right) \cos \left( x \right) x+3\, \left( \sin \left( x \right) 
 \right) ^{2} \right) }{{x}^{4}}}
$$ this is positive since $-\sin(x)<0$ and $$-x^4+3\sin(x)(\sin(x)-x\cos(x))<0$$.
