Circle $k$ passes through $A$ and intersects the sides of $\Delta ABC$ in $P,Q$, and $L$. Prove that:
$\frac{S_{PQL}}{S_{ABC}}\leq \frac{1}{4} (\frac{PL}{AQ})^2$.
Let L be named R.
$$ \frac{[PQR]}{[ABC]} = \frac{1/2 \cdot PQ \cdot QR \cdot \sin \alpha}{1/2 \cdot AB \cdot AC \cdot \sin 180^{\circ}-\alpha} = \frac{PQ\cdot QR}{AB\cdot AC}$$By the Sine law in $PQR$
$$PQ\cdot QR =\frac{PR^2}{\sin^2 \alpha} \sin \angle QPR \sin \angle QRP$$And by GM-AM and Jensen, given that $\angle QPR + \angle QRP = \alpha$
$$PQ \cdot QR \leq \frac{PR^2 \sin^2 \frac{\alpha}{2}}{\sin^2 \alpha} = \frac{PR^2}{4\cos^2 \frac{\alpha}{2}}$$So the problem is equivalent to
$$AQ^2 \leq \cos^2 \frac{\alpha}{2} AB \cdot AC$$
