TITLE: Prove that $\sin x \cdot \sin (2x) \cdot \sin(3x) < \tfrac{9}{16}$ for all $x$
QUESTION [13 upvotes]: Prove that
  $$ \sin (x) \cdot \sin (2x) \cdot \sin(3x) < \dfrac{9}{16} \quad \forall \ x \in \mathbb{R}$$   

I thought about using derivatives, but it would be too lengthy.
Any help will be appreciated.
Thanks.

REPLY [11 votes]: One has
$$2\sin x\sin(3x)=\cos(2x)-\cos(4x)\ ,$$
and therefore
$$2\sin x\sin(2x)\sin(3x)=(1+u-2u^2)\sqrt{1-u^2}=:f(u)\qquad(-1\leq u\leq1)\ ,$$
where we have put $\cos(2x)=:u$. One computes
$$\sqrt{1-u^2}f'(u)=6u^3-2u^2-5u+1$$
with zeros at $$u\in\left\{1, -{\sqrt{10}+2\over6},{\sqrt{10}-2\over6}\right\}\ .$$
The third of these leads to the maximal value of $f(u)$, which we then have to divide by $2$. The result can be simplified to
$$\sin x\sin(2x)\sin(3x)\leq{68+5\sqrt{10}\over108\sqrt{2}}\doteq0.548737<{9\over16}\ .$$