<s>[INST] Trigonometric Functions And Identities Question . If $A$ , $B$ and $C$ are the angles of a triangle , I have to show that : 
$ \tan^2 \cfrac{A}{2} + \tan^2 \cfrac{B}{2} + \tan^2 \cfrac{C}{2} \ge 1 $ . 
I had only arrived at $A+B+C = \pi $ , thus $\cfrac{A}{2} + \cfrac{B}{2} + \cfrac{C}{2} = \cfrac{\pi}{2} $ 
What to do next? 
 [/INST] First note that $\dfrac{A}2, \dfrac{B}2, \dfrac{C}2 \in (0, \pi/2)$. Now consider the function $f(x) = \tan^2(x)$. For $x \in (0,\pi/2)$, $f(x)$ is convex. Hence, we have
$$\dfrac{f(x_1) + f(x_2) + f(x_3)}3 \geq f \left(\dfrac{x_1 + x_2 + x_3}3 \right)$$
Take $x_1 = \dfrac{A}2$, $x_2 = \dfrac{B}2$ and $x_3 = \dfrac{C}2$. We have $x_1 + x_2 + x_3 = \dfrac{A+B+C}2 = \dfrac{\pi}2$.
Hence, we get
$$\dfrac{f(A/2) + f(B/2) + f(C/2)}3 \geq f \left(\dfrac{\pi}6 \right)$$
This gives us
$$\dfrac{\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2)}3 \geq \tan^2(\pi/6) = \dfrac13$$
Hence,
$$\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2) \geq 1$$

If $f(x)$ is convex, a $n$-gon with vertices on the curve will always lie on the upper side of the curve (In fact, this is the definition of convexity. A quick way to check if a function is convex is to check if its second derivative (if it exists) is non-negative.) and hence the center of mass of the $n$-gon will also lie on the upper side of the curve. In our case, $n=3$.
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