Let ABC be a triangle and P be an interior point. Prove that AB + BC + CA

Question

Let ABC be a triangle and P be an interior point. Prove that AB + BC + CA < 2 (PA + PB + PC).

Philoid · Accepted Answer

According to triangle inequality the sum of smaller two sides is always greater than the largest side

In ΔAPB

AP + PB>AB …(i)

In ΔAPC

AP + PC>AC …(ii)

In ΔCPB

CP + PB>CB …(iii)

Adding (i),(ii) and (iii) we get

2 (PA + PB + PC)> AB + BC + CA

or changing the sign of inequality we can say

AB + BC + CA < 2 (PA + PB + PC)