Consider a cyclic quadrilateral such that the midpoints of its sides form another cyclic quadrilateral. Prove that the area of the smaller circle is less than or equal to half the area of the bigger circle
If $ABCD$ is the given cyclic quadrilateral and $P, Q, R, S$ its side midpoints, then because $PQRS$ is a parallelogram, $PQRS$ is cyclic $\iff PQRS$ is rectangle $\iff AC\perp BD$. Thus $PR$ is the diameter of the circumcircle of $PQRS$. Let $AE$ be diameter in the big circle. We need to prove that $\sqrt{2}PR\le AE$. Note that
$$PR=\left|\frac{\overrightarrow{AD}+\overrightarrow{BC}}{2}\right|\le \frac{AD+BC}{2}\le \sqrt{\frac{AD^2+BC^2}{2}}=\sqrt{\frac{AD^2+DE^2}{2}}=\frac{AE}{\sqrt{2}}$$as desired.