Show that $\csc^n\frac{A}{2}+\csc^n\frac{B}{2}+\csc^n\frac{C}{2}$ has the minimum value $3.2^n$ Show that in a $\Delta ABC$, $\sin\frac{A}{2}\leq\frac{a}{b+c}$

Hence or otherwise show that $\csc^n\frac{A}{2}+\csc^n\frac{B}{2}+\csc^n\frac{C}{2}$ has the minimum value $3.2^n$ for all $n\geq1$.

In this problem, I have proved the first part that $\sin\frac{A}{2}\leq\frac{a}{b+c}$ but second part I could not prove. I tried using the AM-GM inequality but could not succeed. Please help.
 A: $\frac{1}{\sin\frac{x}{2}}$ is a convex function over $I=(0,\pi)$ as well as $\left(\frac{1}{\sin\frac{x}{2}}\right)^n$ since it is positive and log-convex: 
$$\frac{d^2}{dx^2}(-\log\sin(x/2))=\frac{1}{4}\,\csc^2\frac{x}{2}.$$
The claim hence follows from Jensen's inequality.
A: As a different approach, Lagrange multipliers work here...the constraint function $$A+B+C=\pi$$ has gradient $(1,1,1)$.  Hence at an "internal" critical point the three partials of your function must be equal.  Note that the boundary doesn't matter...the boundary will have one or more angles equal to $0$ and the function is clearly not minimal there.  Set the partial in $A$ equal to the partial in $B$. After some simple algebra we get $$\frac {cos(\frac A2)}{sin^n(\frac A2)}\;=\;\frac {cos(\frac B2)}{sin^n(\frac B2)}$$  But, for positive integers $n$ the function $\frac {cos(x)}{sin^n(x)}$ is strictly decreasing in $x$ (for $x\in [0,\frac {\pi}{2}]$) so we must have $A=B$.  Similarly, all three angles are equal, and the minimum follows by taking all three equal to $\frac {\pi}{3}$.
