<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 $a, b, c$ be the sides of a triangle with given angles $A, B, C$.
Recall for any $3$ positive numbers $a, b, c$, it can form the sides of a non-degenerate triangle if and only if we can find $3$ positive numbers
$u,v,w$ such that
$$a = v + w,\quad b = u + w,\quad c = u + v$$
This is known as Ravi substitution for a triangle.
In terms of them, we have
$$
\sin^2\frac{A}{2} 
= \frac12( 1 - \cos A) = \frac12\left(1 - \frac{b^2+c^2-a^2}{2bc}\right)
= \frac{a^2 - (b-c)^2}{4bc} = \frac{vw}{bc}
$$
By a similar argument, we have
$$\sin^2\frac{B}{2}  = \frac{uw}{ac}\quad\text{ and }\quad\sin^2\frac{C}{2} = \frac{uv}{ab}$$
Multiply these 3 relations together, taking square root and apply AM $\ge$ GM to the factors in denominator, we obtain:
$$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}
= \frac{uvw}{abc}
= \frac{uvw}{(v+w)(u+w)(u+v)}
\le \frac{uvw}{(2\sqrt{vw})(2\sqrt{uw})(2\sqrt{uv})} = \frac18$$
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