Let be distinct points on a plane, four of which form a quadrangle, and three of which are in the interior or boundary of this quadrangle. Show that the diagonals of this quadrangle are longer than the double of the minimum of the distances between any two of these seven points. Paul Erdős

HIDE: Side note If the quadrangle is convex, the constant from the inequality can be improved from $ 2 $ to $ \sqrt{\frac{3\pi}{2}}. $