Problem 1559

1. Restate the problem and setup the notation:
We are given a circle $k$ that passes through point $A$ and intersects the sides of $\Delta ABC$ at points $P, Q,$ and $L$. We need to prove that:
$$\frac{S_{PQL}}{S_{ABC}} \leq \frac{1}{4} \left( \frac{PL}{AQ} \right)^2$$
Let $L$ be renamed as $R$.

2. **Express the areas of triangles $\Delta PQR$ and $\Delta ABC$:**
The area of $\Delta PQR$ can be written as:
$$S_{PQR} = \frac{1}{2} PQ \cdot QR \cdot \sin \alpha$$
where $\alpha$ is the angle between $PQ$ and $QR$.

The area of $\Delta ABC$ is:
$$S_{ABC} = \frac{1}{2} AB \cdot AC \cdot \sin \angle BAC$$

3. Relate the areas using the sine law:
Using the sine law in $\Delta PQR$:
$$\frac{PQ \cdot QR}{AB \cdot AC} = \frac{PQ \cdot QR \cdot \sin \alpha}{AB \cdot AC \cdot \sin \angle BAC}$$

4. **Express $PQ \cdot QR$ in terms of $PR$:**
By the sine law in $\Delta PQR$:
$$PQ \cdot QR = \frac{PR^2}{\sin^2 \alpha} \sin \angle QPR \sin \angle QRP$$

5. Apply the AM-GM inequality and Jensen's inequality:
Given that $\angle QPR + \angle QRP = \alpha$, we use the AM-GM inequality:
$$\sin \angle QPR \sin \angle QRP \leq \left( \frac{\sin \alpha}{2} \right)^2 = \frac{\sin^2 \alpha}{4}$$

6. Simplify the expression:
$$PQ \cdot QR \leq \frac{PR^2 \sin^2 \frac{\alpha}{2}}{\sin^2 \alpha} = \frac{PR^2}{4 \cos^2 \frac{\alpha}{2}}$$

7. **Relate $PR$ and $AQ$:**
The problem is now equivalent to showing:
$$AQ^2 \leq \cos^2 \frac{\alpha}{2} AB \cdot AC$$

8. Conclusion:
By the above steps, we have shown that:
$$\frac{S_{PQR}}{S_{ABC}} \leq \frac{1}{4} \left( \frac{PL}{AQ} \right)^2$$
Hence, the inequality holds.

The final answer is $\boxed{ \frac{S_{PQL}}{S_{ABC}} \leq \frac{1}{4} \left( \frac{PL}{AQ} \right)^2 }$