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)
 A: 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 $A = B = C = 60^\circ$.
A: You probably switched the inequality sign. If we have triangle with angles $\alpha, \beta, \gamma$, then inequality $$
\sin\left(\frac\alpha 2\right)\sin\left(\frac\beta 2\right)\sin\left(\frac\gamma 2 \right)\geq \frac 1 8
$$
does not need hold. Product on the LHS can be made arbitrarily small, as we can pick a right triangle with one very pointy corner (i.e. one of the angles can be arbitrarily small). We have $$
\sin\left(\frac\alpha 2\right)\sin\left(\frac\beta 2\right)\sin\left(\frac\gamma 2 \right)\leq \sin\left(\frac\alpha 2\right)
$$
(because all values are positive, and not grater than 1).

Let's prove that opposite equality holds. Define $f(x, y) = \sin(x/2)\sin(y/2)\sin((\pi - x - y)/2)$, and choose domain of $f$ to be
$$
D = \left\{ (x, y) \in \left[0, \pi  \right]^2 : x+ y \leq  \pi\right\}. $$ In that way, $D$ is compact, and every point on its interior describes a triangle with angles $x, y, \pi - x - y$. Since $f$ is continuous, we know that exists maximum on $D$. On the boundary, $f$ is equal to $0$, and on interior it is positive, so we know that maximum lies somewhere in the middle. Calculating gradient gives us $$
 \nabla f(x, y) = \left(\frac {\sin(\frac y 2) \cos(x + \frac y 2)} 2, \frac { \sin(\frac x 2) \cos(\frac x 2 + y)} 2\right)
$$ after we use angle sum formula for sine. The only point on the interior of $D$ on which the gradient vanishes is $(\pi / 3, \pi / 3)$, so there lies the maximum. We have
$$ f(x, y) \leq f\left(\frac \pi  3, \frac \pi  3\right) = \sin^3\left(\frac \pi  6\right) = \frac 1 8.$$
A: 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 $A = B = C = 60^\circ$.
