Let $ABCDEF$ be a convex hexagon such that $AB = BC, CD = DE, EF = FA$.
Prove that $\frac{BC}{BE} +\frac{DE}{DA} +\frac{FA}{FC} \ge \frac{3}{2}$ . When does equality occur?
By Ptolemy for \(ABCE\) we have \(\displaystyle\frac{BC}{BE}\ge \frac{AC}{AE+EC}\). Thus if \(CE=x, EA=y, AC=z\) it suffices to prove
\[\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge \frac 3 2\]which is true by C-B-S:
\[\sum \frac{x}{y+z}=\sum\frac{x^2}{xy+xz}\ge \frac{(x+y+z)^2}{2(xy+yz+zx)}\ge \frac 3 2.\]Equality holds iff \(ABCDEF\) is regular.