Given, ABCD is a quadrilateral and AC and BD are the diagonals intersect at the point O.

We know that the sum of the two sides of a triangle is greater than the third side.

So, considering the triangle ABC, BCD, CAD and BAD, we get

AB + BC > AC

CD + AD > AC

AB + AD > BD

BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)

=> 2(AB + BC + CA + AD) > 2(AC + BD)

=> (AB + BC + CA + AD) > (AC + BD)