<s>[INST] Trigonometric Triangle Equality $A, B, C$ are the angles of a  triangle  then  $tan^2(A/2)+tan^2(B/2)+tan^2(C/2)$ is
 always greater than what integral value.
 [/INST] Since $A,B,C$ are angles of a triangle, we have $0<\dfrac{A}{2},\dfrac{B}{2},\dfrac{C}{2}<\dfrac{\pi}{2}$. For this range of values, $\tan^2 x$ is a convex function. Hence, from Jensen's inequlaity,
$$\tan^2\left(\frac{1}{3}\frac{A}{2}+\frac{1}{3}\frac{B}{2}+\frac{1}{3}\frac{C}{2}\right) \leq \frac{1}{3}\tan^2\frac{A}{2}+\frac{1}{3}\tan^2\frac{B}{2}+\frac{1}{3}\tan^2\frac{C}{2}$$
$$\tan^2\left(\frac{A+B+C}{6}\right) \leq \frac{1}{3}\left(\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\right)$$
Since $A+B+C=\pi$, hence
$$\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2} \geq 1$$
$\blacksquare$
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