TITLE: Showing that $\left (\frac{\sin x}{x} \right )^3\geq \cos^{2}x$
QUESTION [6 upvotes]: Show that $$\left (\frac{\sin x}{x}  \right )^3\geq \cos^{2}x,\forall x\in \left ( 0;\frac\pi2 \right )$$ 


Firstly, I had use the differentiation of $f(x)=\left (\frac{\sin x}{x}  \right )^3- \cos^{2}x$,but I think it very complex: $$f'(x)=3\left ( \frac{\sin x}{x} \right )^2\frac{x\cos x-\sin x}{x^2}+\sin2x$$ Or I can't prove it be positive in $ \left ( 0;\frac\pi2 \right )$
Secondly, I had use the convex function, but function $f(x)=\left (\frac{\sin x}{x}  \right )^3-\cos^{2}x$ isn't convex or concave in all $ \left ( 0;\frac\pi2 \right )$

REPLY [2 votes]: We can have a stronger one which is easier:
Let $f(x)=\tan{x}\sin{x} -x^2,x\in[0,\pi/2),$
$$f'(x)>0\implies \tan{x}\sin{x}>x^2,x\in(0,\pi/2).$$
$$f'(x)=\sin{x}\dfrac{\cos^2{x}+1}{\cos^2{x}}-2x>0\iff \dfrac{\tan{x}}{x}> \dfrac{2}{\cos{x}+\dfrac{1}{\cos{x}}} $$
which is true, because:
$$\dfrac{\tan{x}}{x}>1,\cos{x}+\dfrac{1}{\cos{x}}>2 \implies \dfrac{2}{\cos{x}+\dfrac{1}{\cos{x}}}<1,\forall x\in(0,\pi/2).$$
it is trivial that
$$\dfrac{\tan{x}}{x} \cdot\tan{x}\sin{x}>x^2,x\in(0,\pi/2).$$