Prove that a triangle with angles $a, b, c$ the inequality $\sin \frac{a}{2} \cos\frac{b}{2} \sin\frac{c}{2}\leq\frac{1}{8}$ holds

Question

Prove, using Lagrange Multipliers (or so it seems) that a triangle with angles $a, b, c$ holds: $$\sin \frac{a}{2} \cos\frac{b}{2} \sin\frac{c}{2}\leq\frac{1}{8}$$.

Thank you!

Answer 1

Why not use a (slightly) less overpowered technique?

By AM GM,

$LHS \le \frac{(\sin \frac{a}2 + \sin \frac{c}2 + \cos \frac{b}2)^3}{27}$

Then $\cos \frac{b}2 = \sin \frac{b}2$ since $a+b+c = \pi$.

Then by Jensen's Inequality, since $\sin x$ is concave on the interval $[0, \pi]$, we have

$\sin \frac{a}2 + \sin \frac{c}2 + \cos \frac{b}2 \le 3\sin \left( \frac{a+b+c}6 \right) = \frac{3}2$. Then the desired inequality easily follows.