From the given question, we have

In ∆OAB, ∆OBC and ∆OCA we have:

OA + OB > AB

OB + OC > BC

And, OC + OA > AC

Adding all these, we get:

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

(OA + OB + OC >

∴ Option (C) is correct