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

$$(a^2+b^2+c^2)-(ab+ac+bc) = \frac{1}{2}\left[(a-b)^2+(a-c)^2+(b-c)^2\right]\geq 0 $$
and the middle term equals zero iff $a=b=c$.
Alternative approach: by the Cauchy-Schwarz inequality
$$\left|ab+ac+bc\right|\leq \sqrt{a^2+b^2+c^2}\sqrt{b^2+c^2+a^2} = a^2+b^2+c^2$$
and the equality holds iff the vectors $(a,b,c)$ and $(b,c,a)$ are linearly dependent, i.e. iff $a=b=c$.
