sqing wrote: In $\triangle ABC$ , prove that\[\frac{tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}}{\sqrt{3}}\geq\sqrt[6]{tan^2\frac{A}{2}+tan^2\frac{B}{2}+tan^2\frac{C}{2}}.\] Let $\tan\frac{\alpha}{2}=a$, $\tan\frac{\beta}{2}=b$ and $\tan\frac{\gamma}{2}=c$. Hence, $ab+ac+bc=1$ and after homogenization we need to prove that $(a+b+c)^6\geq27(a^2+b^2+c^2)(ab+ac+bc)^2$. Let $a^2+b^2+c^2=u(ab+ac+bc)$. Id est, we need to prove that $(u+2)^3\geq27u$, which is AM-GM: $(u+2)^3\geq\left(3\sqrt[3]u\right)^3=27u$.

sqing wrote: In $\triangle ABC$ , prove that\[\frac{tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}}{\sqrt{3}}\geq\sqrt[6]{tan^2\frac{A}{2}+tan^2\frac{B}{2}+tan^2\frac{C}{2}}.\] $\left(\frac{tan\frac{A}{2}+tan\frac{B}{2}+tan\frac{C}{2}}{\sqrt{3}}\right)^2=\frac{tan^2\frac{A}{2}+tan^2\frac{B}{2}+tan^2\frac{C}{2}+1+1}{3}$ $\geq\sqrt[3]{tan^2\frac{A}{2}+tan^2\frac{B}{2}+tan^2\frac{C}{2}}.$