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.
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
I believe something similar was posted before:
http://www.mathhelpforum.com/math-he...ht=equilateral
$\displaystyle a^2+b^2+c^2-ab-ac-bc=0\Rightarrow\ 2a^2+2b^2+2c^2-2ab-2ac-2bc=0$
$\displaystyle \Rightarrow\ a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0$
$\displaystyle (a-b)^2+(b-c)^2+(a-c)^2=0$
but all three terms on the left are non-negative and sum to zero.
As there are no negative terms, they are all zero.
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?
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.