Question

Let ‘O’ be any point in the interior of ΔACB. Prove that:

1. AB + AC > OB + ОС 
2. AB + BC + CA > (OA + OB + OC) 
3. AB + BC + CA < 2(OA + OB + ОС)

Answer 1

Given: O is any point in the interior of ΔACB.

To Prove:

1. AB + AC > OB + OC
2. AB + BC + CA > (OA + OB + OC)
3. AB + BC + CA < 2(OA + OB + OC)

Proof [Step-wise]:

a. AB + AC > OB + OC

1. Let the line BO meet AC at D.

Since O is interior, B, O and D are collinear with O between B and D.

2. In ΔABD, by the triangle inequality:

AB + AD > BD. 

But BD = BO + OD, 

So, AB + AD > BO + OD.   ...(1)

3. In ΔODC, by the triangle inequality:

OD + DC > OC.   ...(2)

4. Add (1) and (2):

AB + AD + OD + DC > BO + OD + OC.

Cancel AD + DC = AC and OD from both sides to get AB + AC > BO + OC.QED.

b. AB + BC + CA > OA + OB + OC

1. Apply part (a) and its cyclic versions by the same construction for the other vertices:

AB + AC > OB + OC,

BC + BA > OC + OA,

CA + CB > OA + OB.

2. Add the three inequalities:

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

Divide by 2 to obtain AB + BC + CA > OA + OB + OC.QED.

c. AB + BC + CA < 2(OA + OB + OC)

1. In ΔOAB: OA + OB > AB.

In ΔOBC: OB + OC > BC.

In ΔOCA: OC + OA > CA.

2. Add these three inequalities:

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

i.e. AB + BC + CA < 2(OA + OB + OC).QED.

All three inequalities are proved:

1. AB + AC > OB + OC,
2. AB + BC + CA > OA + OB + OC,
3. AB + BC + CA < 2(OA + OB + OC).