Prove that $ (\tan\frac{A}{2})^2 + (\tan \frac{B}{2})^2 + (\tan \frac{C}{2})^2 \ge 1$ when $A,B,C$ are the angles in a triangle.
I am trying to solve it using Jensen's Equality but not getting any desired result. Is there any other methodology of solving this problem.
From Jensen's inequality the finite form you have: $\frac 1 3((\tan\frac{A}{2})^2 + (\tan \frac{B}{2})^2 + (\tan \frac{C}{2})^2) \ge (\tan(\frac {A + B + C}{6}))^2 = \frac 1 3 $
You'll have to justify Jensen's inequality by proving $f(x) = \tan^2(x), x \in (0, \frac {\pi}{2})$ is a convex function, which is quite obvious since the composition of two convex nondecreasing functions is convex.
Let $a$, $b$ and $c$ be sides-lengths of the triangle.
Thus, by law of cosines we need to prove that: $$\sum_{cyc}\frac{1-\frac{b^2+c^2-a^2}{2bc}}{1+\frac{b^2+c^2-a^2}{2bc}}\geq1$$ or $$\sum_{cyc}\frac{(a+b-c)(a+c-b)}{a+b+c)(b+c-a)}\geq1$$ or $$\sum_{cyc}(a+b-c)^2(a+c-b)^2\geq(a+b+c)\prod_{cyc}(a+b-c)$$ or $$\sum_{cyc}(a^2-b^2-c^2+2bc)^2\geq\sum_{cyc}(2a^2b^2-a^4)$$ or $$\sum_{cyc}(3a^4+4a^2b^2-2a^2b^2-2a^2b^2+4a^2bc+2a^2b^2-4a^3b-4a^3c)\geq \sum_{cyc}(2a^2b^2-a^4)$$ or $$\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)\geq0,$$ which is Schur.
Done!
Also we can use the following. $$\sum_{cyc}\tan^2\frac{\alpha}{2}\geq\sum_{cyc}\tan\frac{\alpha}{2}\tan\frac{\beta}{2}=1$$ for which we need to prove that $$\sum_{cyc}\tan\frac{\alpha}{2}\tan\frac{\beta}{2}=1,$$ which is $$\tan\frac{\alpha}{2}\left(\tan\frac{\beta}{2}+\tan\frac{\gamma}{2}\right)=1-\tan\frac{\beta}{2}\tan\frac{\gamma}{2}$$ or $$\tan\frac{\alpha}{2}\tan\frac{\beta+\gamma}{2}=1$$ or $$\tan\frac{\alpha}{2}\cot\frac{\alpha}{2}=1,$$ which is obvious.
