<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] Here is an alternative trigonometric proof of the inequality for fun.  Denote $$
x = \frac{A}{2}, \qquad y = \frac{B}{2}, \qquad \{x, y, x + y\} \subset [0, 90^\circ].
$$
Then, using the product-to-sum formula
$$
\sin x \sin y = \frac{\cos (x - y) - \cos (x + y)}{2}
$$
as well as the AM-GM inequality,
we have
$$ \begin{split}
\sin \frac{A}{2} \, \sin \frac{B}{2} \, \sin \frac{C}{2}
&= \sin x \sin y \sin (90^\circ - x - y) \\
&= \sin x \sin y \cos (x + y) \\
&= \frac{\cos (x - y) - \cos (x + y)}{2} \cos(x + y) \\
&\le \frac{1 - \cos(x + y)}{2} \cos(x + y) \\
&\le \frac{1}{2} \Bigl( \frac{1}{2} \Bigr)^2 = \frac{1}{8},
\end{split} $$
where equality holds for the first $\le$ sign if and only if $x = y$, and equality holds for the second $\le$ sign if and only if $\cos(x + y) = \frac{1}{2}$.  Therefore, we have
$$
\sin \frac{A}{2} \, \sin \frac{B}{2} \, \sin \frac{C}{2} \le \frac{1}{8}
$$
where equality holds if and only if $A = B = C = 60^\circ$.
</s>