This is somewhat of a trick question . We can prove that $|AB-CD|\ge|AC-BD|\ (*)$: Let $P=AC\cap BD$, and assume WLOG that $DC>AB$ (when we have equality, it's clear). $PAB$ and $PDC$ are similar, so let $k=\frac{DC}{AB}>1$. We have $|AB-CD|=(k-1)AB$, and $|AC-BD|=|PA+PC-PB-PD|=|(PA-PD)-(PB-PC)|$, which is equal to $|(k-1)PA-(k-1)PB|=(k-1)|PA-PB|$, and $(*)$ is now reduced to $AB\ge|PA-PB|$ (true by the triangle inequality).

Suppose without loss of generality that $AB \ge CD$ and $AD \ge BC$. We now consider two cases: Case I: $AC \ge BD$ Since we have $AC \ge BD$, $AB \ge CD$, $AD \ge BC$, $AC(AB+AD) \ge BD(BC+CD)$ $AB \cdot AC + AD \cdot AC + 2BD \cdot AC \ge BC \cdot BD + CD \cdot BD + 2AC \cdot BD$ $AB + AD + 2BD \ge BC + CD + 2AC$ $AB - CD + AD - BC \ge 2AC - 2BD$ Case II: $BD \ge AC$ We have $BD \ge AC$, $AB \ge CD$, $AD \ge BC$ and thus $BD(AB+AD) \ge AC(BC+CD)$. Using similar logic to Case I, we get a similar result: $AB - CD + AD - BC \ge 2BD - 2AC$ So therefore $|AB - CD| + |AD - BC| \ge 2|BD - AC|$ QED.

MithsApprentice wrote: Let $ABCD$ be a cyclic quadrilateral. Prove that \[ |AB - CD| + |AD - BC| \geq 2|AC - BD|. \] This is an old, old post, anyway, I solved this problem recently and strangely enough my solution wasn't posted so here it goes. 1) Assume $AB \geq DC$ and $AD \geq BC$. Let $D'$· be the point in the circle with $BD' \parallel CD$ and $B'$ the point with $DB' \parallel BC$ 2) Note $CD' = DB = CB'$, $DD' = BC$, $BB' = CD$ 3) We have the two triangle inequalities $AD + D'C \geq DD' + AC$ $AB + CB' \geq BB' + AC$ The result follows from summing the inequalities and using 2) Daniel

sorry to bring this topic again to life, but i found an interesting "reference" previous to this olympiad... http://www.imo.org.yu/othercomp/Rom/RomMO98.pdf check out the 9th-form problem 2... it's exactly what grobber said in his post... however both of them are not too difficult.

Let angles $\alpha,\beta,\gamma,\delta$ have sum $180^\circ$ so we want to show $|\sin\alpha-\sin\gamma|+|\sin\beta-\sin(\alpha+\beta+\gamma)|\ge 2|\sin(\alpha+\beta)-\sin(\beta+\gamma)|$. Equivalently, we want to show \[|\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|+|\cos(\tfrac{\alpha+2\beta+\gamma}{2})\sin(\tfrac{-\alpha-\gamma}{2})|\ge 2|\cos(\tfrac{\alpha+2\beta+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|.\]With respect to $\beta$, note the difference between the two sides is minimized at the extrema of $|\cos(\tfrac{\alpha+2\beta+\gamma}{2})|$. Hence it suffices to argue \[|\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|\ge 0\]and \[|\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|+|\cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{-\alpha-\gamma}{2})|\ge 2| \cos(\tfrac{\alpha+\gamma}{2})\sin(\tfrac{\alpha-\gamma}{2})|.\]The former is trivial. For the latter, it is equivalent to demonstrate \[\sin(\tfrac{\alpha+\gamma}{2})\ge |\sin(\tfrac{\alpha-\gamma}{2})|.\]This is trivial: WLOG $\alpha>\gamma$, then it follows from noting $0<\tfrac{\alpha-\gamma}{2}<\tfrac{\alpha+\gamma}{2}<\pi/2$.