O is any point inside the triangle ABC. Let’s prove that

AB+BC+AC>OA+OB+OC

From the above proof, we have

AB + AC > OB + OC …….. (I)

Let us modify the above figure a bit by connecting OA

Now from Δ ABC

AC + BC > AB ……….. (i)

Similarly from Δ OAB

OA + OB > AB ………. (ii)

Now let us subtract equation (ii) from equation (i)

AC + BC – (OA + OB) > AB – AB

AC + BC – (OA + OB) > 0

AC + BC > OA + OB ………….. (II)

Now let us again consider Δ ABC,

AB + BC > AC ………….. (iii)

Similarly from Δ OAC

OA + OC > AC …………. (iv)

Now let us subtract equation (iv) from equation (iii)

AB + BC – (OA + OC) > AC – AC

AB + BC – (OA + OC) > 0

AB + BC > OA + OC ……. (III)

Now let us add equations (I), (II) and (III)

AB + BC + AC + BC + AB + AC > OB + OA + OA + OC + OC + OB

2AB + 2BC + 2AC > 2OB + 2OC + 2OA

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

(AB + BC + AC) > (OB + OC + OA)

Hence Proved.