<s>[INST] 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.
 [/INST] fix $A$, and now we have:
$$\frac{1}{\sin^2(B)}+\frac{1}{\sin^2(C)}\geq \frac{2}{\sin^2\left(\frac{B+C}{2}\right)}$$
becaues $\dfrac{1}{\sin^2(x)}$ is a convex function over $[0,\pi]$. thus when we fix an angle like $A$ it is better that two other angles be equal. Now the function $\dfrac{1}{\sin^2(A)}+\dfrac{1}{\sin^2(B)}+\dfrac{1}{\sin^2(C)}$ is defined over 
$$X=\{(A,B,C)| A,B,C \geq 0 , A+B+C=\pi\}$$
that is a compact subest of $\mathbb{R}^3$. and this continous function will get its minimum over its domain. now if minimum takes in a place like $(A,B,C)$ then if we have $B\neq C$ then by the above we can find another minimum that is less than this minimum. thus $B=C$ and $A=B$ too. thus $A=B=C=\dfrac{\pi}{3}$ and minimum is $4$.
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