Problem

Source: Iran Third Round MO 1998, Exam 4, P2

Tags: inequalities, geometry unsolved, geometry



Let $ABCDEF$ be a convex hexagon such that $AB = BC, CD = DE$ and $EF = FA$. Prove that \[\frac{AB}{BE}+\frac{CD}{AD}+\frac{EF}{CF} \geq \frac{3}{2}.\]