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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


marked as duplicate by davidlowryduda Jun 19 '16 at 18:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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[\ $.


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