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.
 A: 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
$$
\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2) \ge 1
$$
A: 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
$$\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2} \geq 1$$
$\blacksquare$
A: 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
$$\displaystyle \tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{B}{2}\right)+\tan^2\left(\frac{C}{2}\right)\geq 1$$
A: Minimum value is achieved when  symmetric in $ A,B,C,a,b,c $ equilateral triangle case when they are equal, i.e., $ A = B = C = \pi/3$, which gives to the expression a minimum value
$$ 1/3 + 1/3 + 1/3 = 1.$$
