Prove that $\sin x \cdot \sin (2x) \cdot \sin(3x) < \tfrac{9}{16}$ for all $x$ 
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.
 A: We have the following equality:
$$ \sin(x) \sin(3x) = \frac{\cos(3x - x) - \cos(3x + x)}{2} = \frac{\cos(2x) - \cos(4x)}{2} = \frac{c - 2c^2 + 1}{2} $$
where $ c = \cos(2x) $. The function $ f(x) = -2x^2 + x + 1 $ attains its global maximum at $ x = 1/4 $, which means that we have
$$ \sin(x)\sin(2x)\sin(3x) \leq \sin(x)\sin(3x) = \frac{f(\cos(2x))}{2} \leq \frac{f(1/4)}{2} = \frac{9}{16} $$
if at most one of $ \sin(x), \sin(2x), \sin(3x) $ are negative.
Note: To show that the inequality is strict, it suffices to observe that $ \sin(2x) < 1 $ when $ \cos(2x) = 1/4 $.
There should be some way to fix this solution so it applies more generally. I will revisit this answer sometime in the future if I find such an argument. If you find a way, feel free to suggest an edit.
A: 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
$$\sin x\sin(2x)\sin(3x)\leq{68+5\sqrt{10}\over108\sqrt{2}}\doteq0.548737<{9\over16}\ .$$
