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. 

another way $$xy+yz+zx\le \sqrt { \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right)  } =x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\\ \\ \\ $$
