Is my proof that $x^2+y^2+z^2 ≥ xy+yz+xz$ correct? The question:
Prove that $x^2+y^2+z^2 ≥ xy+yz+xz$ for all real numbers $x$, $y$ and $z$.
This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different.
If $x$, $y$ and $z$ are real numbers then $(x-y-z)^2 \geqslant 0$. That is $x^2 + y^2 +z^2 -2xy-2xz-2yz \geqslant 0$. But this implies that $\frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$, so because $x^2 + y^2 +z^2 \geqslant \frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$ we get the desired result. 

Your concerns were already addressed in the comments, so there is no need to keep underlying your actual sign mistake. But on a meta-mathematical point of view, something should have struck you, since your method would have derived a symmetric inequality from a non-symmetric one, kind of strange.
A working and fast proof:
$$\color{red}{0\leq} (x-y)^2+(x-z)^2+(y-z)^2 = 2\color{red}{\left(x^2+y^2+z^2-xy-xz-yz\right)}. $$
