Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$

Question

Let $ABC$ be a triangle. Prove that: $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) \ge \sqrt{3} $$ My attempt: $$P=\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) = \frac{1}{2}\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)$$ $$\alpha = \cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$$ $$\Leftrightarrow \alpha \ge3.\sqrt[3]{\alpha} \Leftrightarrow \alpha^2\ge27\Leftrightarrow \alpha\ge3\sqrt{3}$$ $$\Rightarrow P=\frac{1}{2}\alpha\ge\frac{3\sqrt{3}}{2}$$ Hmm where I was wrong ?

Answer 1

You are not wrong because $$ \frac32\sqrt{3}\ge \sqrt3. $$

You just have proved a stronger inequality than required.

Answer 2

Let $p$ is the semi-perimeter, $R$ is circumradius and $r$ is radius.

We have $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C} = \frac{2R}{a} + \frac{2R}{b} + \frac{2R}{c} = \frac{4R+r}{2p}+\frac{p}{2r},$$ and $$\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2} = \sum \frac{r}{b+c-a} = \frac{4R+r}{p}.$$ The inequality become $$p \geqslant 2\sqrt3 r.$$ Which is true by know inequality $$p \geqslant 3\sqrt 3 r.$$