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 $A=B=C=\dfrac{\pi}{3}$.
$\blacksquare$
HINT: Use Lagrange multipliers to find the minimum of the function $f(A,B,C) = \frac{1}{\sin^2(A)} + \frac{1}{\sin^2(B)} +\frac{1}{\sin^2(C)}$ with the constraint $A+B+C = \pi$
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 $4$.
