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. $>1$
  2. $<1$
  3. $\ge1$
  4. $\le1$

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 $\pi$.



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 $$\left(\tan\dfrac A2-\tan\dfrac B2\right)^2+\left(\tan\dfrac B2-\tan\dfrac C2\right)^2+\left(\tan\dfrac C2-\tan\dfrac A2\right)^2\ge0$$


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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