Amir Hossein 31.10.2010 22:46 Let $ABCDEF$ be a convex hexagon such that $AB = BC, CD = DE$ and $EF = FA$. Prove that \[\frac{AB}{BE}+\frac{CD}{AD}+\frac{EF}{CF} \geq \frac{3}{2}.\]
newsun 25.11.2011 13:37 Setting $AC=e, CE=a, AE=c$ Using Ptolemy's inequality for the convex quadrilateral $ACEF$ gives $AF.CE+AC.EF \ge AE.CF \Leftrightarrow EF.(AC+CE) \ge CF.AE \Leftrightarrow \frac{EF}{CF} \ge \frac{c}{a+e} $ Similarly, \[\frac{AB}{BE}+\frac{CD}{AD}+\frac{EF}{CF} \geq \frac{a}{c+e}+\frac{e}{a+c}+\frac{c}{a+e} \ge \frac 32 \]