<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] 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 $$-x^4+3\sin(x)(\sin(x)-x\cos(x))<0$$.
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