Answer :

From the above figure,

In Δ PAB,


PA + PB > AB ……… (i)


Similarly in Δ PBC


PB + PC > BC ……… (ii)


Also in Δ PAC,


PA + PC > AC …….. (iii)


Now let us add the equations (i), (ii) and (iii)


AP + BP + BP + CP + AP + CP > AB + BC + AC


2AP+ 2BP + 2CP > AB + BC + AC


2 (AP + BP + CP) > AB + BC + AC.


It can be written in other way as follows:


AB + BC + AC < 2 (AP + BP + CP)


Hence Proved


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