P is any point inside ABC let’s prove that,

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

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