Famous Inequality

By AM-GM inequality , we have:

$\begin{aligned} \frac 1{\sin^n \frac A2} + \frac 1{\sin^n \frac B2} + \frac 1{\sin^n \frac C2} & \ge \frac 3{\sqrt[3]{\sin^n \frac A2 \sin^n \frac B2 \sin^n \frac C2}} & \small \blue{\text{Equality occurs when }A=B=C = \frac \pi 3} \\ & \ge 3\cdot 2^n \end{aligned}$

Therefore $x+y = 3+2 = \boxed 5$ .

In any triangle $\triangle ABC$ and $n\ge 0$ , the following inequality is true: $\frac{1}{\sin^n \frac{A}{2}} + \frac{1}{\sin^n \frac{B}{2}} + \frac{1}{\sin^n \frac{C}{2}} \ge 3\times2^n$

$\implies x + y = 5$ .

Proof:

Let $\triangle ABC$ be any acute angled triangle , then:

$f(x) = \csc^n \frac{x}{2}\text{ for all } x \in (0,\frac{\pi}{2})$

$\implies f'(x) = \frac{-n}{2} \csc^n \frac{x}{2} \cot \frac{x}{2}$

$\implies f"(x) = \frac{n^2}{4} \csc^n \frac{x}{2} \cot^2\frac{x}{2} + \frac{n}{4} \csc^{n+2} \frac{x}{2} > 0$

By applying Jensen's Inequality:

$\sum_{cyc} \frac{\csc^n \frac{A}{2}}{3} \ge \csc^n (\frac{\pi}{6}) = 2^n$

$\implies \sum_{cyc} \csc^n \frac{A}{2} \ge 3\times 2^n$

If $\triangle ABC$ was an obtuse triangle then:

$\frac{a}{b+c} = \frac{\sin A}{\sin B +\sin C} = \frac {\sin A}{\sin (\frac{B-C}{2})} \ge \sin \frac{A}{2}$

$\implies \csc\frac{A}{2} \ge \frac{b+c}{a}$

$\sum_{cyc} \csc^n\frac{A}{2} \ge \sum \Big(\frac{b+c}{a}\Big)^n \ge 3 \sqrt[3]{\prod \Big(\frac{b+c}{a}\Big)^n} \ge 3 \times 2^n$