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.

Given
$$
\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2) \ge K
$$
Question is: find $K$.
Write
$$
Q(x,y,z) = \tan^2(x) + \tan^2(y) + \tan^2(z),
$$
such that
$$
x+y+z = P,
$$
where $P$ is a constant. Then
$$
dQ = 2 \frac{\tan(x)}{\cos^2(x)} d x
+ 2 \frac{\tan(y)}{\cos^2(y)} dy
+ 2 \frac{\tan(z)}{\cos^2(z)} dz
$$
But as $x+y+z=P$, we also have
$$
dx+dy+dz = 0
$$
so we obtain
$$
dQ = 2 \left( \frac{\tan(x)}{\cos^2(x)} - \frac{\tan(z)}{\cos^2(z)} \right) d x
+ 2 \left( \frac{\tan(y)}{\cos^2(y)} - \frac{\tan(z)}{\cos^2(z)} \right) dy
$$
We find an extreme value for
$$
\frac{\tan(x)}{\cos^2(x)} - \frac{\tan(z)}{\cos^2(z)} = 0
$$
and
$$
\frac{\tan(y)}{\cos^2(y)} - \frac{\tan(z)}{\cos^2(z)}
$$
So
$$
x = y = z = P/3
$$
Given that
$$
A + B + C = 180^o,
$$
we have
$$
x+y+z = 90^o
$$
So minimum for
$$
x=y=z=30^o
$$
So we obtain
$$
K = 3 \tan^2(30^o) = 1,
$$
whence
$$
\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2) \ge 1
$$
