Saudi Arabia Mathematical Competitions 2012

It is clear that $AB\parallel ED$ and $AE\parallel CD$. Let $F$ be the intersection point of lines $AB$ and $CD$. The inequality is equivalent to $$\frac{AC}{AF}\cdot \frac{BD}{FD}\ge \frac{3}{4}.(1)$$ Using the Law of Sines in triangles $ACF$ and $BDF$, we get that (1) is equivalent to $$\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{\sin \widehat{ACF}}\cdot \frac{\sqrt{3}}{\sin \widehat{DBF}}\ge \frac{3}{4},$$ so $\sin \widehat{ACF}\cdot \sin \widehat{DBF}\le 1$, which is clearly true. We have equality if and only if $\widehat{ACF}=\widehat{DBF}=90^{\circ }$, hence $BC=AB=CD=BF=CF$. This means $AEDF$ is a rhombus, and $B$ and $C$ are the midpoints of $AF$ and $DF$ respectively.