ACCCGS8 02.03.2013 12:00 $A,B,C,D$ are points on a circle in that order. Prove that $|AB-CD|+|AD-BC| \ge 2|AC-BD|$.
oneplusone 17.03.2013 07:04 We just show $|AB-CD|\geq |AC-BD|$. Let $AC$ intersect $BD$ at $E$. Let $EA=a,EB=b,AB=c$ and $ED=ka,EC=kb,CD=kc$. Then we want to show $|(k-1)c|\geq |ka+b-kb-a|=|(k-1)(a-b)|$, which is just $c\geq |a-b|$, true by triangle inequality.