1. O is known to be a point in △ABC. Verification: 1/2 (BC+CA+AB) < OA+OB+OC.

Prove that connecting OA, OB, OC,

In δ△OAB, ab < OA+ob.

At △OAC, AC < OA+OC.

In △OBC, BC < OB+OC.

Add up the above three inequalities,

Get: ab+AC+BC < 2 (OA+ob+oc)

Divide both sides by 2, and you get

(AB+AC+BC)