Problem: Let $ABCD$ be a convex cyclic quadrilateral.
Prove that \[ |AB - CD| + |AD - BC| \geq 2|AC - BD|. \]

Solution: Let the diagonals meet at $P$,
and let $AP = pq$, $DP = pr$, $BP = qs$, $CP = rs$.
Then set $AB = qx$, $CD = rx$, $AD = py$, $BC = sy$.


In this way we compute
\[ \left\lvert AC-BD \right\rvert
  = \left\lvert (p-s)(q-r) \right\rvert  \]
and
\[ \left\lvert AB-CD \right\rvert
  = \left\lvert q-r \right\rvert x. \]
By triangle inequality on $\triangle AXB$,
we have $x \ge \left\lvert p-s \right\rvert$.
So $\left\lvert AB-CD \right\rvert \ge \left\lvert AC-BD \right\rvert$.