1. ## Prove an identity

Hello,
show that if $\displaystyle a^2+b^2+c^2=ab+ac+bc$, then $\displaystyle a=b=c$.

I tried to roll stuff up using binomial expansions, eg. for (a+b+c)^2, bt it only yields some obvious things like a=a. Any ideas?

Thanks.

2. Try $\displaystyle (a-b)^2+(a-c)^2+(b-c)^2$

Edit: Oh never mind - I think that's just 0.

Edit 2: Lol. Oh yeah - that's why it works.

3. Assuming that a,b and c are positif real numbers .
Using the Arithmetic mean-Geometric mean inequality we get a˛+b˛>=2ab also a˛+c˛>=2ac and b˛+c˛>=2bc summing we get a˛+b˛+c˛>=ab+ac+bc and equality holds in this inequality when a=b=c !

You can also use the Cauchy-Schwartz inequality and in the same way you get equality holds when a=b=c

4. I believe something similar was posted before:
http://www.mathhelpforum.com/math-he...ht=equilateral

5. Originally Posted by atreyyu
Hello,
show that if $\displaystyle a^2+b^2+c^2=ab+ac+bc$, then $\displaystyle a=b=c$.

I tried to roll stuff up using binomial expansions, eg. for (a+b+c)^2, bt it only yields some obvious things like a=a. Any ideas?

Thanks.
$\displaystyle a^2+b^2+c^2-ab-ac-bc=0\Rightarrow\ 2a^2+2b^2+2c^2-2ab-2ac-2bc=0$

$\displaystyle \Rightarrow\ a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0$

$\displaystyle (a-b)^2+(b-c)^2+(a-c)^2=0$

but all three terms on the left are non-negative and sum to zero.
As there are no negative terms, they are all zero.

6. Originally Posted by snowtea
I believe something similar was posted before:
http://www.mathhelpforum.com/math-he...ht=equilateral
It's exactly the same problem with geometric interpertation (with some complex geometry from "Also sprach Zarathustra" )

7. Thanks for all, the replies, but all your solutions make use of the fact that a, b, c are positive. There is no word of it in the original wording I have got, and without that I don't think you can sum inequalities together or draw a concluion Archie Meade has done... maybe there is a problem with the phrasing of the question and a, b, c should indeed be positive?

8. We all haven't assumed that a,b,c are positive. We have used the fact that $\displaystyle x^2$ is nonnegative for all real $\displaystyle x$.

For example $\displaystyle (a-b)^2$ is always positive or zero regardless of what $\displaystyle a$ and $\displaystyle b$ are.

9. Yes,

by "all three terms", I meant the terms $\displaystyle (a-b)^2,\;\; (b-c)^2,\;\; (a-c)^2,$
not the terms a, b, c.

10. Well , i guess you have the solution now ! And all the solutions are with a,b and c real numbers ! (i made a mistake by assuming they were positive , but a˛+b˛>=2ab is true for all a and b real numbers )

11. That's correct. Thank you all a lot.