<s>[INST] What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \sin x \sin {2x}$$
I have done my work here 
$$f (x)=\sin x \sin 2x =\frac{\cos x - \cos3x}2 $$
$$f'(x)= \frac{- \sin x+3 \sin 3x}2 =4\sin x (2-3\sin^2 x)=0$$
$$x=0,\pi; \sin x= \sqrt{\frac {2}{3}}$$
$$f (0)=f (\pi)=0$$
$$f \left(\arcsin \sqrt{\frac{2}{3}}\right) =\frac{4}{3 \sqrt{3}}$$
If my work is not much right then please rectify it
 [/INST] Perhaps a more unorthodox approach: From the AM-GM inequality,
$$\sqrt[3]{\frac{1}{2}\sin^2x\cdot\frac{1}{2}\sin^2x\cdot\cos^2x} \le \frac{\frac{1}{2}\sin^2x\cdot + \frac{1}{2}\sin^2x\cdot + \cos^2x}{3}$$
$$\sqrt[3]{\frac{1}{4}\sin^4x\cos^2x} \le \frac{\sin^2x + \cos^2x}{3}$$
But it is well known that $\sin^2x + \cos^2x = 1$. Hence,
$$\sqrt[3]{\frac{1}{4}\sin^4x\cos^2x} \le \frac{1}{3}$$
$$\sin^4x\cos^2x\le\frac{4}{27}$$
Taking square roots of both sides,
$$-\frac{2}{\sqrt{27}} \le \sin^2x\cos x \le \frac{2}{\sqrt{27}}$$
Multiplying throughout by $2$,
$$-\frac{4}{\sqrt{27}} \le 2\sin^2x\cos x \le \frac{4}{\sqrt{27}}$$
But from the double angle formula, $2\sin^2x\cos x = \sin x \cdot2\sin x\cos x = \sin x \sin 2x$. Hence,
$$-\frac{4}{\sqrt{27}} \le \sin x \sin 2x \le \frac{4}{\sqrt{27}}$$
Equality for each bound occurs when $\frac{1}{2}\sin^2x = \cos^2x$, i.e. $\tan x = \pm \sqrt{2}$.
</s>