2017/018) ABC is a triangle and O is a point in it. Prove that $(AB + BC + AC ) > (OA + OB + OC)$

Let BO meet AC in D

Then $AB+AD > BD = OB+OD$

And $OD+DC > OC$

Sum these for $AB+(AD+DC)+OD > OC+OB+OD$

or $AB+AC > OC+OB$

Similarly for BA+BC and CB+CA and sum to get

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

or $AB+BC+CA > OA+OB+OC$