Question

$ABCD$ is a quadrilateral, then prove that $AB+BC+CD+DA<2(AC+BD)$.

Answer 1

Sum of two sides of a triangle should be greater than the third side of the triangle

so

$in△AOB,AB<OA+OB....\left(1\right)in△BOC,BC<OB+OC....\left(2\right)in△COD,CD<OC+OD.....\left(3\right)in△DOA,DA<OD+OA....\left(4\right)\Rightarrow AB+BC+CD+DA<2OA+2OB+2OC+2OD\Rightarrow AB+BC+CD+DA<2(OA+OB+OC+OD)\Rightarrow AB+BC+CD+DA<2(OA+OC)+2(OB+OD)\Rightarrow AB+BC+CD+DA<2AC+2BD\Rightarrow AB+BC+CD+DA<2(AC+BD)$