# [Math] Prove that $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$

a.m.-g.m.-inequalitygeometric-inequalitiesjensen-inequalitytrigonometry

Prove that $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.

I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a solution, but hit and trial. Can this conclusion be derived using maximum/mininum concept?

Note that $\sin x$ is concave function when $0\leq x\leq \dfrac{\pi}{2}$. Then by Jensen's inequality on concave function, we have:
$$\frac{\sin \dfrac{A}{2}+\sin \dfrac{B}{2}+\sin \dfrac{C}{2}}{3}\leq \sin \dfrac{A+B+C}{2\cdot 3}=\sin\dfrac{\pi}{6}=\frac{1}{2}$$
Now by AM-GM inequality $$\frac{\sin \dfrac{A}{2}+\sin \dfrac{B}{2}+\sin \dfrac{C}{2}}{3}\geq \Big(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\Big)^{\dfrac{1}{3}}$$
Hence $$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$$
Equality holds when $A=B=C=\dfrac{\pi}{3}$