Verify $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0<x<\pi/2$.

Solution 1:

We need to prove that $f(x)>0$, where $f(x)=\frac{\sin{x}}{\sqrt[3]{\cos{x}}}-x$.

Indeed, by AM-GM $$f'(x)=\frac{\cos{x}\sqrt[3]{\cos{x}}+\frac{\sin^2x}{3\sqrt[3]{\cos^2x}}}{\sqrt[3]{\cos^2x}}-1=\frac{3\cos^2x+\sin^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=$$ $$=\frac{1+\cos^2x+\cos^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}\geq\frac{3\sqrt[3]{\cos^4x}-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=0.$$ Thus, $f(x)>f(0)=0$ and we are done!

Solution 2:

For $x > 0$ the well-known representation as alternating series give the estimates $$ \sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} > x - \frac{x^3}{6} \, ,\\ \cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} < 1 - \frac{x^2}{2} + \frac{x^4}{24} \, . $$

For $0 < x < \pi/2$ in particular $x^2 < 6$ holds and therefore $$ \left( \frac{\sin x}{x} \right )^3 - \cos x > \left( 1 - \frac{x^2}{6} \right )^3 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} \right ) \\ = \frac{x^4}{24}- \frac{x^6}{216} = \frac{x^4}{24} \left( 1 - \frac{x^2}{9} \right) > 0 \, . $$