<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] $$^2 + ^2 + ^2 ≥  +  + $$
$$2x^2+2y^2+2z^2 ≥ 2xy+2xz+2yz$$
$$(x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2) ≥ 0$$
$$(x-y)^2+(x-z)^2+
(y-z)^2≥0$$ 
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