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?

Like my answers in
Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $
Trigonometric Inequality $\sin{A}+\sin{B}-\cos{C}\le\frac32$
In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$
let $S=\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$
$2S=\left(\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right)\sin\dfrac C2$
As $\dfrac{A+B}2=\dfrac\pi2-\dfrac C2,$
$2S=\left(\cos\dfrac{A-B}2-\sin\dfrac C2\right)\sin\dfrac C2$
$\iff\sin^2\dfrac C2-\sin\dfrac C2\cos\dfrac{A-B}2+2S=0$
As $\sin\dfrac C2$ is real, the discriminant must be $\ge0$
i.e., $$\cos^2\dfrac{A-B}2-8S\ge0\iff8S\le\cos^2\dfrac{A-B}2\le1$$
So, $8S\le1\implies S\le\frac{1}{8}$
the equality occurs if $\cos\dfrac{A-B}2=1\iff\cos(A-B)=1\iff A=B$ as $0<A,B<\pi$
and hence $\sin\dfrac C2=\dfrac12\implies C=\dfrac\pi3$ as $0<C<\pi$
$\implies A=B=\dfrac{A+B}2=\dfrac\pi3=C$
