Range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ with $A,B,C$ angles in a triangle 
What is the range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ if $A,B,C$ are angles in a triangle?

For $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$, I know we have to apply the AM–GM inequality. So
$$\begin{align}\frac12\left(\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2\right)&\ge\sqrt{\tan^2\frac A2\tan^2\frac B2\tan^2\frac C2}\\
&\ge\tan\frac A2\tan\frac B2\tan\frac C2\\
\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2&\ge2\tan\frac A2\tan\frac B2\tan\frac C2
\end{align}$$
After this, the LHS is of the required form, but the RHS is having half-angle terms. Then again, if $A+B+C=\pi$, $\tan\frac{A+B+C}2=\tan90^\circ=\infty$. Now I am stuck; how to solve further?
The given options are:


*

*$>1$

*$<1$

*$\ge1$

*$\le1$

 A: Note $\tan^2(x)$ is convex, so by Jensen's inequality
$$\sum_{cyc} \tan^2 \frac{A}2 \geqslant 3\tan^2 \frac{A+B+C}{2\cdot3} = 1$$
Further as $\tan(x)$ is unbounded, clearly we can have the sum unbounded by allowing one of the angles to approach $\pi$.
A: HINT:
As $\tan\left(\dfrac{A+B}2\right)=\tan\left(\dfrac{\pi- C}2\right)$
$$\implies\dfrac{\tan\dfrac A2+\tan\dfrac B2}{1-\tan\dfrac A2\tan\dfrac B2}=\dfrac1{\tan\dfrac C2}$$
$$\implies\tan\dfrac A2\tan\dfrac B2+\tan\dfrac B2\tan\dfrac C2+\tan\dfrac C2\tan\dfrac A2=1$$
Now $$\left(\tan\dfrac A2-\tan\dfrac B2\right)^2+\left(\tan\dfrac B2-\tan\dfrac C2\right)^2+\left(\tan\dfrac C2-\tan\dfrac A2\right)^2\ge0$$
A: Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Hence, by C-S $\sum\limits_{cyc}\tan^2\frac{\alpha}{2}=\sum\limits_{cyc}\frac{yz}{x(x+y+z)}=\sum\limits_{cyc}\frac{y^2z^2}{xyz(x+y+z)}\geq\frac{(xy+xz+yz)^2}{3xyz(x+y+z)}\geq1$.
The equality occurs for $x=y=z$, id est, for $a=b=c$.
In another hand, our sum $\rightarrow+\infty$ for $\alpha\rightarrow180^{\circ}$.
Id est, the answer is $[1,+\infty)$.
