Saudi Arabian 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
$$\frac{AC}{AF} \cdot \frac{BD}{FD} \ge \frac{3}{4}. \quad (1)$$
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.