TITLE: Prove that if three real numbers $a, b, c$ satisfy $a^2+b^2+c^2=ab+bc+ac$, then $a=b=c$.
QUESTION [1 upvotes]: I don't have much to start the problem, as I have trouble approaching it. What method of solving should I use here?

REPLY [1 votes]: Suppose suppose loss of generality that $a\le b\le c$ then $\begin{cases}b=a+u&u\ge 0\\c=b+v&v\ge 0\end{cases}$

$a^2+b^2+c^2-(ab+bc+ca)=u^2+\underbrace{uv}_{\ge 0}+v^2=0\implies u=v=0\implies a=b=c$


Another interesting substitution is $c=ta+(1-t)b$

$a^2+b^2+c^2-(ab+bc+ca)=(a-b)^2\underbrace{(t^2-t+1)}_{\text{no real root}}=0\implies a=b\implies a=b=c$
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