<s>[INST] Solutions of Triangles inequalities - Duplicate A,B,C are angles of a triangle we are supposed to prove that $sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})$ $\leq$ $\frac{1}{8}$. I used trigonometric ratios of half angles which would give $\frac{(s-a)(s-b)(s-c)}{abc}$. How can I proceed after this step?
Any help would be appreciated ( I need the solution where angles are made in terms of sides and then the inequality is simplified)
 [/INST] As @Esgeriath's answer mentioned, the correct inequality should be
$$
\sin \frac{A}{2} \; \sin \frac{B}{2} \; \sin \frac{C}{2} \le \frac{1}{8}.
$$
That answer used multivariable calculus methods to calculate the extremum of the left-hand side.
Here is an elementary proof of this inequality.

You have shown that
$$
\sin \frac A2 \; \sin \frac B2 \; \sin \frac C2
= \frac{(s - a)(s - b)(s - c)}{abc}
$$
where $s = \frac{1}{2}(a + b + c)$.  Recall that the area of the triangle is
$$
T = \sqrt{s (s - a)(s - b)(s - c)}
$$
by Heron's formula, the inradius is
$$
r = \frac{T}{s},
$$
and the circumradius is
$$
R = \frac{abc}{4T},
$$
so
$$
\frac{(s - a)(s - b)(s - c)}{abc}
= \frac{T^2}{4 s R T} = \frac{r}{4R}.
$$
But $R \ge 2r$ by Euler's inequality, with equality if and only if the triangle is equilateral, so
$$
\sin \frac A2 \; \sin \frac B2 \; \sin \frac C2 \le \frac{1}{8}
$$
where equality holds if and only if $A = B = C = 60^\circ$.
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