To prove AB + BC + CA > OA +OB+ OC?

Question

ABC is any Triangle and O is any point inside the triangle

Answer 1

I am sorry to say that the previous proof is totally wrong.

Draw the parallel to each side passing through O. You obtain three parallelograms say ODAE, OFBG, OHCI. You have OA< OD+DA, OB< OF+FC and OC< OH+HC. You add up everything and notice that OD=AE, OF=BG and OH=IC and that AD+IC

Answer 2

Given that ABC is any Triangle and O is any point inside the triangle:

Any segnment from a point O inside a triangle to any vertex of the triangle cannot be longer than the two sides adjasent to that vertex, i.e. OA cannot be longer than both AB and CA. This is because if this is false, point O would be outside of the triangle.

(an attempt at a picure)
B_\ .o
___\
____\/ C
____A

Segnment OA is longer than AC, but shorter than AB.

AB or CA (or both) > OA,
BC or AB (or both) > OB, and
CA or BC (or both) > OC;

(AB or CA) + (BC or AB) + (CA or BC) > OA +OB+ OC;

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

Comment:
I've read the answer below, and still can't figure out what it is implying. The answer does not provide any reason for why this proof is wrong or any less suficient that the one provided. It also does not reach any resolution.

Answer 3

1) we knw that side opposite to greatest angle (out of 3 angles of triangle) is greatest in length.
2) Now any point inside a triangle will b greatest one if we join the points i.e in AOB or AOC or BOC,etc. Try it...
So the first 3 greatest line segments according to size in triangle wil be AB,BC and CD.
3) Hope u r satisfied :)

Answer 4

i m not sure but all i no is that that forms tree more triangles inside the big 1

Answer 5

Nice proof gianlino!

Answer 6

i agree with northstar.a nice and innovative proof presented very well by gianlino

Answer 7

i know it is right but dont know how to explain