How to prove $x^2 + y^2 + z^2\geq xy + xz + yz$ Result: Let $, ,  ∈ ℝ$. Then we have  $^2 + ^2 + ^2 ≥  +  + $
Need some help proving this, just a few steps with work.
Was thinking you start with $x^2+y^2+z^2−xy−xz−yz$
then factor?
can anyone show me how to solve this?
$x^2+y^2+z^2≥xy+yz+zx⇔ $
$⇔2(x^2+y^2+z^2)≥2(xy+yz+zx)⇔$
$⇔x^2−2xy+y^2+y^2−2yz+z^2+z^2−2xz+x^2≥0⇔$
$⇔(x−y)^2+(y−z)^2+(z−x)^2≥0$
Should I do anything else? Or is this right?

Another way.
Let $y=x+u$ and $z=x+v$.
Thus, $$x^2+y^2+z^2-xy-xz-yz=$$
$$=x^2+(x+u)^2+(x+v)^2-x(x+u)-x(x+v)-(x+u)(x+v)=$$
$$=x^2+x^2+2xu+u^2+x^2+2xv+v^2-x^2-xu-x^2-xv-x^2-xv-xu-uv=$$
$$=u^2-uv+v^2=\left(u-\frac{v}{2}\right)^2+\frac{3v^2}{4}\geq0.$$
