<s>[INST] 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?
 [/INST] Your proof is right, but maybe it looks a bit of better by using a cyclic sum:
$$\sum_{cyc}(x^2-xy)=\frac{1}{2}\sum_{cyc}(2x^2-2xy)=\frac{1}{2}\sum_{cyc}(x^2-2xy+y^2)=\frac{1}{2}\sum_{cyc}(x-y)^2\geq0.$$
We can use also the following.
$$x^2+y^2+z^2-xy-xz-yz=x^2-(y+z)x+y^2-yz+z^2=$$
$$=\left(x-\frac{y+z}{2}\right)^2+y^2-yz+z^2-\frac{(y+z)^2}{4}=\left(x-\frac{y+z}{2}\right)^2+\frac{3(y-z)^2}{4}\geq0.$$
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