How to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz \geq 0$

Question

I know how to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz -2xz \geq 0$ since I can split it up in the three terms $(x-y)^2, (y-z)^2, (z-x)^2$ which are all greater than or equal to zero, but how do I prove this without the last term: $2x^2 + 2y^2 + 2z^2 -2xy -2yz \geq 0$?

Answer 1

It is $$x^2+y^2-2xy+y^2+z^2-2yz+z^2+x^2\geq 0$$ so $$(x-y)^2+(y-z)^2+z^2+x^2\geq0$$

Answer 2

$2x^2 + 2y^2 + 2z^2 -2xy -2yz = (y-x-z)^2+(x-z)^2+y^2 \ge 0$

Answer 3

Slightly differently:

Let us write a quadratic for $x$ as $$2x^2-2yx+(2y^2+2z^2) \ge 0, \forall x \in R$$ this means that $D=B^2- 4AC\le 0$ for all real values of $y,z$ and this true as $$D=4y^2-4.2.(2y^2+2z^2)=-12y^2-16z^2 \le 0$$