Find minimum value of the trigonometrical expression If $A+B+C=\pi$
then find the minimum value of
$\sin 3A+\sin 3B+\sin 3C$ 
where $0\le A\le \pi,0\le B \le \pi,0\le C\le \pi$

The minimum value is $-2$.
Let be $A\leq B\leq C$. So we have $3A\leq \pi$ and so $\sin(3A)\geq 0$. So we get the minimum for $A=0, B=C=\frac{\pi}{2}$.
