Question

In a triangle ABC if a point O is inside the triangle the please prove that AB + BC + CA > AO + BO + CO.

Answer 1

In ∆ABC,
AB +AC >BC ….(1)
And in ∆OBC,
OB + OC > BC …(2)
Subtracting 1 from 2 we get,
(AB + AC) – (OB + OC ) > (BC – BC )
Ie AB + AC > OB + OC
From ׀, AB + AC > OB + OC
Similarly, AB + BC > OA + OC
And AC + BC > OA + OB
Adding both sides of these three inequalities, we get,
(AB + AC ) + (AB + BC) + (AC + BC) > (OB + OC) + (OA + OC) + (OA + OB)
Ie. 2(AB + BC + AC ) > 2(OA + OB + OC)
∴ AB + BC + AC > OA + OB + OC