<s>[INST] 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?
 [/INST] Let us consider the function $f(\theta)=\log\sin\theta$ on the interval $\theta\in\left(0,\frac{\pi}{2}\right)$. It is a concave function
since $f''(\theta)=-\frac{1}{\sin^2\theta}<0$, hence by Jensen's inequality
$$ f(a)+f(b)+f(c) \leq 3\,f\left(\frac{a+b+c}{3}\right) $$
holds for any $a,b,c\in\left(0,\frac{\pi}{2}\right)$. By considering $a=\frac{A}{2},b=\frac{B}{2},c=\frac{C}{2}$ and exponentiating both sides of the previous inequality turns into
$$ \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\leq \sin^3\left(\frac{A+B+C}{6}\right) = \frac{1}{8}.$$
</s>