Question 136
Prove that
$$
\frac{AO^2}{BC} + \frac{BO^2}{CA} + \frac{CO^2}{AB} \ge \frac{AO + BO + CO}{\sqrt{3}}
$$
where $O$ is the point inside triangle $ABC$ such that $\angle AOB = \angle BOC = \angle COA = 120^\circ$.
$$
\frac{AO^2}{BC} + \frac{BO^2}{CA} + \frac{CO^2}{AB} \ge \frac{AO + BO + CO}{\sqrt{3}}
$$
where $O$ is the point inside triangle $ABC$ such that $\angle AOB = \angle BOC = \angle COA = 120^\circ$.