Showing that $\left (\frac{\sin x}{x} \right )^3\geq \cos^{2}x$ 
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 )$

Start with
$$
1-\cos(x)\ge0\tag{1}
$$
For $x\ge0$, successive integration yields
$$
x-\sin(x)\ge0\tag{2}
$$
and
$$
\frac{x^2}2-1+\cos(x)\ge0\tag{3}
$$
and
$$
\frac{x^3}6-x+\sin(x)\ge0\tag{4}
$$
and
$$
\frac{x^4}{24}-\frac{x^2}2+1-\cos(x)\ge0\tag{5}
$$
From $(4)$, we get
$$
\frac{\sin^3(x)}{x^3}\ge1-\frac{x^2}2+\frac{x^4}{12}-\frac{x^6}{216}\tag{6}
$$
From $(5)$, we get
$$
\cos(x)\le1-\frac{x^2}2+\frac{x^4}{24}\tag{7}
$$
Subtracting $(7)$ from $(6)$ yields
$$
\frac{\sin^3(x)}{x^3}-\cos(x)\ge\frac{x^4}{24}-\frac{x^6}{216}\tag{8}
$$
and for $|x|\le3$, $\frac{x^4}{24}-\frac{x^6}{216}\ge0$. Thus for, $x\in\left[0,\frac\pi2\right]$,
$$
\frac{\sin^3(x)}{x^3}\ge\cos(x)\ge\cos^2(x)\tag{9}
$$
