Geometric Inequality
Posted: Wed May 06, 2015 5:30 pm
Let $P$ be a point in the interior of $\triangle ABC$ with circumradius $R$. Prove that \[\dfrac{AP}{BC^2}+\dfrac{BP}{CA^2}+\dfrac{CP}{AB^2}\ge \dfrac 1 R.\]
Source: CleverMath homepage.
Those who miss the old Brilliant.org, try CleverMath.
Source: CleverMath homepage.
Those who miss the old Brilliant.org, try CleverMath.
