Found this problem in the national high school math competition today... I was completely dumbfounded.
Find the value of the smallest interior angle of triangle $ABC$ when $\sin 7A+\sin 7B+\sin 7C$ takes the maximum value.
Can anyone give me a hint on how to tackle this problem? This is where I'm stuck at: $$\sin 7A+\sin 7B+\sin 7C=-4\sin\frac{7A}{2}\sin\frac{7B}{2}\sin\frac{7C}{2}$$
Note,
$$I=\sin 7A+\sin 7B+\sin 7C$$ $$=\sin7A - 2\cos\frac{7A}2 \cos\frac{7(B-C)}2 \le \sin7A - 2\cos\frac{7A}2 $$
where the equality corresponds to $B=C$ and $\cos\frac{7A}2<0$ is assumed (to be verified below). Set the derivative of the RHS to zero,
$$\cos7A +\sin\frac{7A}2 = 1-2\sin^2\frac{7A}2 +\sin\frac{7A}2 =0$$
which yields $\sin\frac{7A}2 = -\frac12$. The valid solutions are $A=\frac\pi3,\>\frac{11\pi}{21},\>\frac{19\pi}{21} $, of which the first and the third, or $A=B=C=\frac\pi3$ and $A=\frac{19\pi}{21}, \>B=C=\frac{\pi}{21}$ respectively, produce the maximum value for the sum,
$$I \le \sin \frac{7\pi}3 +\sin \frac{7\pi}3+\sin \frac{7\pi}3= \sin \frac{19\pi}3 +\sin \frac{\pi}3+\sin \frac{\pi}3=\frac{3\sqrt3}2$$
Indeed, $\cos\frac{7A}2<0$ for $A = \frac\pi3,\>\frac{19\pi}{21}$ as assumed earlier.