If $A+B+C=\pi$ then what is the minimal value of $$\tan^{2}A/2+\tan^{2}B/2+\tan^{2}C/2$$
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http://mathworld.wolfram.com/JensensInequality.html. It can be made arbitrarily large by making $A\to\pi$ – lab bhattacharjee Sep 25 '14 at 06:42
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As $x \mapsto \tan^2(x/2)$ is convex, by Jensen's inequality we get $$\tan^2(A/2)+\tan^2(B/2)+\tan^2(C/2) \ge 3\tan^2\frac{\pi}6=1$$
Macavity
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Here $A,B,C$ are the angles of a $\triangle.$ So Here $$\displaystyle 0<\frac{A}{2},\frac{B}{2},\frac{C}{2}<\frac{\pi}{2}$$.
Now Using $\bf{A.M\geq G.M}\;,$ We Get
So we get $$\displaystyle \frac{\tan^2 \left(\frac{A}{2}\right)+\tan^2 \left(\frac{B}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{A}{2}\right)\cdot \tan^2 \left(\frac{B}{2}\right)} = \tan\left(\frac{A}{2}\right)\cdot \tan \left(\frac{B}{2}\right)\color{red}\checkmark$$
Similarly $$\displaystyle \frac{\tan^2 \left(\frac{B}{2}\right)+\tan^2 \left(\frac{C}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{B}{2}\right)\cdot \tan^2 \left(\frac{C}{2}\right)} = \tan \left(\frac{B}{2}\right)\cdot \tan \left(\frac{C}{2}\right)\color{red}\checkmark$$
Similarly $$\displaystyle \frac{\tan^2 \left(\frac{C}{2}\right)+\tan^2 \left(\frac{A}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{C}{2}\right)\cdot \tan^2 \left(\frac{A}{2}\right)} = \tan \left(\frac{C}{2}\right)\cdot \tan \left(\frac{A}{2}\right)\color{red}\checkmark$$
Now Add all Three, We Get
$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq \tan\left(\frac{A}{2}\right)\cdot \tan\left(\frac{B}{2}\right)+\tan\left(\frac{B}{2}\right)\cdot \tan\left(\frac{C}{2}\right)+\tan\left(\frac{C}{2}\right)\cdot \tan\left(\frac{A}{2}\right)\color{blue}\checkmark\color{blue}\checkmark$$
Now Here $$\displaystyle \frac{A}{2}+\frac{B}{2}=\frac{C}{2}\Rightarrow \tan\left(\frac{A}{2}+\frac{B}{2}\right)=\tan \frac{C}{2}$$
So We Get $$\displaystyle \tan\left(\frac{A}{2}\right)\cdot \tan\left(\frac{B}{2}\right)+\tan\left(\frac{B}{2}\right)\cdot \tan\left(\frac{C}{2}\right)+\tan\left(\frac{C}{2}\right)\cdot \tan\left(\frac{A}{2}\right)=1$$
Put into $\color{blue}\checkmark\color{blue}\checkmark\;,$ We Get
$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq 1$$
juantheron
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That the exact same answer provides a solution to two different questions is an indication that one should probably be closed as a duplicate of the other. – user642796 Apr 24 '15 at 06:21