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. 
 A: 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:
$$\color{red}{0\leq} (x-y)^2+(x-z)^2+(y-z)^2 = 2\color{red}{\left(x^2+y^2+z^2-xy-xz-yz\right)}. $$
A: another way $$xy+yz+zx\le \sqrt { \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right)  } =x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\\ \\ \\ $$
A: Based on hagnatural's comment: No, your proof is incorrect since one of your signs is wrong:
$$(x-y-z)^2 = x^2 + y^2 + z^2 - 2xy - 2xz \color{red}{+} 2yz.$$
And unfortunately there is no direct term $(\pm x \pm y \pm z)^2$ which gives you opposite signs on the cross terms, compared to the squares. (But as noted in the other answers, it is possible to use similar ideas by combining the three terms $(x - y)^2$, $(x - z)^2$, and $(y - z)^2$.)
