Angles inequality in acute triangle

Question

Let $\alpha$, $\beta$, $\gamma$ be angles of acute triangle. How to prove that $(\tan(\frac{\alpha}{2}))^2 + (\tan(\frac{\beta}{2}))^2 + (\tan(\frac{\gamma}{2}))^2 \ge 1$? Does left side of equation have bigger than 1 bound?

Answer 1

Note in $\Delta ABC$ $$\sum_{cyc}\tan{\dfrac{A}{2}}\tan{\dfrac{B}{2}}=1$$ because $$\tan{\dfrac{C}{2}}=\tan{(\dfrac{\pi}{2}-\dfrac{A+B}{2})}=\dfrac{1}{\tan{(\dfrac{A+B}{2})}}=\dfrac{1-\tan{\dfrac{A}{2}\tan{\dfrac{B}{2}}}}{\tan{\dfrac{A}{2}}+\tan{\dfrac{B}{2}}}$$ and use $$x^2+y^2+z^2\ge xy+yz+xz$$ so $$\sum_{cyc}\tan^2{\dfrac{A}{2}}\ge 1$$