Question

Please help!
Ive been stucked in this question for quite a while.
Heres the question:

Answer 1

Given a^2+b^2+c^2=ab+bc+ca
Prove a=b=c

Answer 2

we know that the following is true:
(a-b)^2 + (b-c)^2 + (a-c)^2 >= 0
because each of the added terms is greater than or equal to zero.

Multiplying out you get:

a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + a^2 - 2ac + c^2 >= 0

Collecting like terms and rearranging you get:

2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac >= 0

Then dividing each side of the equation by 2 and rearranging you get:

a^2 + b^2 + c^2 >= ab + ac + bc

Answer 3

does that help?

Answer 4

okeeeeey....bye i guess? give medal please :)

Answer 5

@mathmale does that make sense? i feel like it made a little but not a lot of sense...

Answer 6

we know that the following is true: \[ (a-b)^2 + (b-c)^2 + (a-c)^2 >= 0 \]because each of the added terms is greater than or equal to zero.

Multiplying out you get:
\[ a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + a^2 - 2ac + c^2 >= 0 \]

Collecting like terms and rearranging you get:
\[2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac >= 0 \]

Then dividing each side of the equation by 2 and rearranging you get:
\[a^2 + b^2 + c^2 >= ab + ac + bc\]

Answer 7

Yayyy okay thanks
Umm how do i give out a medal?

Answer 8

But I need to prove a=b=c
So wouldnt i need to prove a-b=0?

Answer 9

Oh and wouldn't i know a^2+b^2+c^2>= 0 since x^2>=0 must be true if x is a real number?
I wonder if i can get anything out of this:
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
Let a^2+b^2+c^2=x
this:
(a+b+c)^2=3x

Answer 10

Ohhh ive looked it up. I know how to give out medals now lol. Thanks anyway

Answer 11

wow i got five medals :) sorry i was asleep so i didnt see you replies...do you still need help?