AB < AP + PB (Shortest distance between two points is a straight line.)
Similarly
BC < BP + PC and CA < CP + PA
Add all three inequalities
AB + BC +CA < AP + PB + BP + PC + CP + PA
So
AB + BC + CA < 2(AP + BP + PC) (AP same as PA etc)
Also
AB + AC > PB + PC (Sum of 2 lines drawn from a point to the extremities of a straight line)
Similarly
BA + BC > PA +PC and CB + CA > PB + PC
Add all three inequalities
AB + AC + BA + BC + CB + CA > PB + PC + PA + PC +PB + PA
2(AB +BC +CA) > 2(PA + PB + PC) (AB same as BA etc)
So
AB + BC +CA > PA + PB + PC
Rewrite as
PA + PB + PC < AB + BC +CA