If P denotes the perimeter and S denotes the sum of the distances of a point within a triangle from its angular points, then

BO is extended to D

In $$\triangle$$ABD

AB + AD > BD

=> AB + AD > OB + OD ------------Eqn(1)

In $$\triangle$$ODC

OD + DC > OC -------------Eqn(2)

Adding eqn (1) & (2)

AB + AD + OD + DC > OB + OD + OC

=> AB + AC > OB + OC ----------Eqn(3)

Similarly,

BC + BA > OA + OC -----------Eqn(4)

and, CA + CB > OA + OB ----------Eqn(5)

Adding eqns (3),(4) & (5), we get :

2 (AB + BC + CA) > 2 (OA + OB + OC)

=> AB + BC + CA > OA + OB + OC

$$\therefore$$ P > S