Question

O is any point in the interior of triangle ABC. Prove that AB+ BC+AC > OA + OB+OC

Pls provide the full explanation for EACH step.

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 + OA > OA + OB + OC
In ∆OAB,
OA + OB > AB …(1)
In ∆OBC,
OB + OC > BC …(2)
In ∆OCA
OC + OA > CA …(3)
Adding 1,2 and 3,
(OA + OB) + (OB + OC) + (OC+ OA) >AB + BC +CA
Ie. 2(OA + OB + OC) > AB + BC + CA
∴ OA + OB + OC > ( AB + BC + CA)

Answer 2

$\huge{\pink{\underline{\ulcorner{\star\: Solution\: \star}\urcorner}}}$

Statement follow :

The sum of any two sides of triangle is always greater than third


[A] In ∆ ABF

AB + AF > BF

AB + AF > BO + OF .....(a)

In ∆ COF

CF + FO > OC .....(b)

add both equation a & b

AB + AF + CF + FO > BO + OF + OC

Substitute "OF"

AB + AF +FC > BO + OC

AB + AC > BO + OC ......(1)



[B] In ∆ BFC

FC + BC > BF

FC + BC > BO + OF ......(c)

In ∆ AFO

OF + AF >AO .....(d)

Add equation c & d

FC + BC + OF + AF > BO + OF + AO

Substitute " OF "

FC + AF + BC > BO + AO

AC + BC > BO + AO ......(2)



[C] In ∆ EBC

EB + BC > EC

EB + BC > EO + OC....(e)

In ∆ AEO

EA + EO > AO....(f)

add both equation e & f

EB + BC + EA + EO > EO + OC + AO

Substitute " EO "

(EB + EA) + BC > AO +OC

AB + BC > OA + OC ......(3)



add equation 1 , 2 & 3


$AB + AC +AC +BC + AB + BC > BO + OC +BO + OA + OA + OC$

2AB + 2AC + 2BC > 2BO + 2OC +2OA

2(AB + AC + BC) > 2(BO + OC + OA)

$\boxed{AB + BC + AC > BO + OC + OA}$

Hence Proved