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.
Geometric Inequality
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: Geometric Inequality
Any hints? I'm stuck...