In a $\triangle ABC,$ Evaluation of minimum value of $\cot^2 A+\cot^2 B+\cot^2 C$ 
In a $\triangle ABC,$ Evaluation of minimum value of $\cot^2 A+\cot^2 B+\cot^2 C$,
Given $A+B+C = \pi$

$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M}$
$$\frac{\cot^2 A+\cot^2 B}{2}\geq \cot A\cdot \cot B\Rightarrow \cot^2 A+\cot^2B \geq 2\cot A\cdot \cot B$$
Similarly $$\cot^2 B+\cot^2 C\geq 2\cot B\cdot \cot C$$
and $$\cot^2 C+\cot^2 A\geq 2\cot C\cdot \cot A$$
So $$\cot^2 A+\cot^2 B+\cot^2 C\geq \cot A\cdot \cot B+\cot B\cdot \cot C+\cot C\cdot \cot A=1$$
Using $$A+B+C = \pi\Rightarrow A+B=\pi-C$$
So $$\cot\left(A+B\right) = \cot\left(\pi-C\right)\Rightarrow \frac{\cot B\cdot \cot A-1}{\cot B+\cot A} = -\cot C$$
So we get $$\cot A\cdot \cot B+\cot B\cdot \cot C+\cot C\cdot \cot A=1$$
My question is Instead of using $\bf{A.M\geq G.M}$ Inequality, Can we use Jensen inequality directly
$$\frac{\cot^2A+\cot^2 B+\cot^2 C}{3}\geq \cot^2\left(\frac{A+B+C}{3}\right)$$
So we can Write it as $$\cot^2A+\cot^2 B+\cot^2 C \geq 1$$
If no, then what wrong with it, Thanks

You can use Jensen's inequality directly, if you first prove that the function
$$f(x):=\cot^2 x\qquad(0<x<\pi)$$
is convex. To this end compute $$f''(x)={2(1+2\cos^2 x)\over\sin^4 x}$$
and verify that $f''(x)$ is $>0$ on the interval $\ ]0,\pi[\ $.
