Question

$O$ is any interior point of triangle $ABC$. Prove that $OA+OB+OC>1/2(AB+BC+AC)$

Answer 1

'O' is any interior point of triangle $\left(\Delta ABC\right)$

$OA+OB>AB$ …..$\left(1\right)$

($\because$ Sum of two sides in a triangle always greater than third side).

Similarly,

$OB+OC>BC$ …..$\left(2\right)$

$OA+OC>AC$ …..$\left(3\right)$

Add, $\left(1\right),\left(2\right),\left(3\right)$

$2(OA+OB+OC)>AB+BC+AC$

$OA+OB+OC>\frac{1}{2}(AB+BC+AC)$.