, 2010

Note that $\angle B\Omega D = \angle AB\Omega + BA\Omega = B - \alpha + \alpha = B$, where $\alpha$ is the Brocard angle. Thus triangles $B\Omega D$ and $ABD$ are similar. It follows that $BD^2 = AD \cdot \Omega D$. Let $BD : DC = x : y$, $CE : EA = z : x$, $AF : FB = y : z$. Then $BD = a x/(x+y)$. We also have
$$\frac{\Omega D}{AD} = \frac{z}{x+y+z}, \quad \frac{A\Omega}{AD} = \frac{x+y}{x+y+z}.$$
Using this we get $x^2 a^2/(x+y)^2 = z AD^2/(x+y+z)$. This implies that
$$A\Omega^2 = \frac{(x+y)^2}{(x+y+z)^2} AD^2 = \frac{1}{x+y+z} \cdot \frac{x^2}{z} \cdot a^2.$$
Thus
$$\frac{A\Omega^2}{BC^2} = \frac{1}{x+y+z} \cdot \frac{x^2}{z}.$$

We obtain
$$\sum_{\text{cyclic}} \frac{A\Omega^2}{BC^2} = \frac{1}{x+y+z} \sum_{\text{cyclic}} \frac{x^2}{z} \geq 1,$$
as
$$(x + y + z)^2 \leq (x + y + z) \sum_{\text{cyclic}} \frac{x^2}{z},$$
by Cauchy-Schwarz inequality.