Problem: Let $ABCD$ be a quadrilateral circumscribed about a circle,
whose interior and exterior angles are at least $60$ degrees.
Prove that
\[ \frac{1}{3}|AB^3 - AD^3|
  \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.  \]
When does equality hold?

Solution: Clearly it suffices to show the left inequality.
Since $AB+CD = BC+AD \implies |AB-AD| = |BC-CD|$, it suffices to prove
\[ \frac13(AB^2 + AB \cdot AD + AD^2)
\le BC^2 + BC \cdot CD + CD^2. \]
This follows by noting that
\begin{align*}
  BC^2 + BC \cdot CD + CD^2
  &\ge BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD) \\
  &= BD^2 \\
  &= AB^2 + AD^2 - 2(AB)(AD)\cos(\angle BAD) \\
  &\ge AB^2 + AD^2 - AB \cdot AD \\
  &\ge \tfrac13(AB^2 + AD^2 + AB \cdot AD)
\end{align*}
the last line following by AM-GM.

The equality holds iff $ABCD$ is a kite with $AB=AD$, $CB=CD$.