Question

$PQRS$ is a quadrilateral in which the diagonals intersect each other at O. Prove that $PQ+QR+RS+SP<2(PR+QS)$

Answer 1

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

In $△PQO,PQ+QO>PQ\to \left(1\right)$

In $△SOP,SO+PO>PS\to \left(2\right)$

In $△SOR,SO+OR>RS\to \left(3\right)$

In $△QOR,QO+OR>QR\to \left(4\right)$

Adding equations (1) , (2) , (3) & (4) , we get

$2(PO+OQ+SO+OR)>PQ+QR+RS+SP$

$PQ+QR+RS+SP<2(PQ+OR+SO+OQ)$

$PQ+QR+RS+SP<2(PR+SQ)$