Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ 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?

$$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}=\frac{r}{4R}\leq\frac{1}{8}.$$
Another way.
Since $f(x)=\ln\sin\frac{x}{2}$ is a concave function, by Jensen and AM-GM we obtain:
$$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\leq\sin^2\frac{\frac{\alpha}{2}+\frac{\beta}{2}}{2}\sin\frac{\gamma}{2}=$$
$$=\frac{1}{2}\left(1-\sin\frac{\gamma}{2}\right)\sin\frac{\gamma}{2}\leq\frac{1}{2}\left(\frac{1-\sin\frac{\gamma}{2}+\sin\frac{\gamma}{2}}{2}\right)^2=\frac{1}{8}.$$
Done!
