Find the minimum of $\displaystyle \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$ Is it possible to find the minimum value of $E$ where
$$E = \frac{1}{\sin^2(\angle A)} + \frac{1}{\sin^2(\angle B)} + \frac{1}{\sin^2(\angle C)}$$for any $\triangle ABC$.
I've got the feeling that $\min(E) = 4$ and that the critical value occurs when $ABC$ is equilateral.

$$E=\csc^2A+\csc^2B+\csc^2C$$
Since $f(x)=\csc^2x$ is a convex function in $(0,\pi)$, we have from Jensen's inequality:
$$\csc^2\left(\frac{A+B+C}{3}\right) \leq \frac{1}{3}\csc^2A+\frac{1}{3}\csc^2B+\frac{1}{3}\csc^2C$$
Since $A+B+C=\pi$
$$\Rightarrow \frac{\csc^2A+\csc^2B+\csc^2C}{3}\geq \frac{4}{3}$$
$$\Rightarrow \csc^2A+\csc^2B+\csc^2C \geq 4$$
And equality occurs when $A=B=C=\dfrac{\pi}{3}$.
$\blacksquare$
