Prove that$(\frac{\sin x}{x})^3\ge \cos x$, for all $0\lt x\le \frac{\pi}{2}$

Question

I would really appreciate it if anyone would mind helping me solve the following question:

I need to prove that:

$$\left(\frac{\sin x}{x}\right)^3\ge \cos x,\qquad \forall 0\lt x\le \frac{\pi}{2}\tag{*}$$

Can anyone help me to prove $(*)$, or can anyone provide a hint to get me started, so I can solve it?

Answer 1

For $x=\frac{\pi}{2}$ it's obvious.

Let $x\in\left(0,\frac{\pi}{2}\right)$.

Thus, we need to prove that $f(x)\geq0$, where $f(x)=\frac{\sin x}{\sqrt[3]{\cos x}}-x$,

which is true because by AM-GM

$f'(x)=\frac{1+2\cos^2x}{3\sqrt[3]{\cos^4x}}-1\geq0$ and we are done!