Problem
Let $ABCD$ be any quadrilateral, and let there be an interior point $O$ which is not the intersection of the diagonals; then the sum of $OA, \ OB, \ OC$ and $OD$ is greater than the sum of both diagonals $AC$ and $BD$:
$$OA+OB+OC+OD＞AC+BD.$$
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Solution
For $△OAC$,
$$AC<OA+OC. \qquad [1]$$
For $△OBD$,
$$BD<OB+OD. \qquad [2]$$
From $[1]$ and $[2]$,
$$AC+BD<OA+OB+OC+OD.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.42.