Prove that if $\alpha, \beta, \gamma$ are angles in triangle, then $(tan(\frac{\alpha}{2}))^2+(tan(\frac{\beta}{2}))^2+(tan(\frac{\gamma}{2}))^2\geq1$ We have an acute triangle with angles $\alpha, \beta, \gamma$ and we need to tell whether the following inequality is true for all such triangles:
$$(\tan(\frac{\alpha}{2}))^2+(\tan(\frac{\beta}{2}))^2+(\tan(\frac{\gamma}{2}))^2\geq1$$
We are also asked if $1%$ is the greatest number, for which it is true.
I'm having huge problems when trying to solve such optimization calculus problems. I think we must somehow find triangle for which we can prove, that the upper sum is minimum, than evaluate it and prove that it is equal to $1$. As with most optimization problems, it turns out that if $\alpha=\beta=\gamma=\frac{\pi}{3}$, then the sum evaluates to $1$. But how to prove that this is infimum of this sum?
 A: $\tan^2(x/2)$ is a convex function on $(-\pi, \pi)$ so by Jensen's inequality $$\frac{1}{3}\left(\tan^2(\alpha/2)+\tan^2(\beta/2)+\tan^2(\gamma/2)\right) \geq \tan^2\left(\frac{\alpha+\beta+\gamma}{6}\right)=\tan^2(\pi/6)=\frac{1}{3}.$$
A: @WimC has given a very simple proof (+1). Here is a slightly longer proof.
Using $$\tan^2 \frac{\alpha}{2} = \frac{1 - \cos \alpha}{1 + \cos \alpha}$$
We can reduce the given inequality to,
$$1 \geq \cos \alpha \cos \beta + \cos \alpha \cos \gamma + \cos \beta \cos \gamma + 2 \cos \alpha \cos \beta \cos \gamma $$

Now, given $\alpha + \beta + \gamma = \pi$, we can show that,
$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma = 1$$

This further reduces the inequality to,
$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma \geq \cos \alpha \cos \beta + \cos \alpha \cos \gamma + \cos \beta \cos \gamma$$
Multiply by 2 and rearrange to get,
$$(\cos \alpha - \cos \beta)^2 + (\cos \beta - \cos \gamma)^2 + (\cos \alpha - \cos \gamma)^2 \geq 0$$
Which is trivially true. Hence proved.
A: Possible plan: (Which I don't know if it solves it!)
Draw the angle bisectors. Let $r$ be the in-radius and $d_1,d_2,d_3$ the distances from the vertices to the points there the sides touch the in-circle. 
The inequality you want is $$\frac{r^2}{d_1^2}+\frac{r^2}{d_2^2}+\frac{r^2}{d_3^2}\geq1$$
Or what is the same 
$$3r^2\geq\frac{3}{\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}}$$
I would try to use the inequalities between the means 
$$\frac{d_1+d_2+d_3}{3}\geq\left(\frac{3}{\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2}}\right)^{1/2}$$
That $d_1+d_2+d_3=p$, the semiperimeter, that $A=pr$, the area of the triangle, and that the incircle is inside the triangle and therefore has smaller area.
