<s>[INST] 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:


*

*$>1$

*$<1$

*$\ge1$

*$\le1$

 [/INST] 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 $[1,+\infty)$.
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