<s>[INST] If $x+y+z \geq xyz$ Prove that $x^2 + y^2 + z^2 \geq xyz$. $x,y,z \in \mathbb R$
My attempt:
Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So :
$$x^2 + y^2 + z^2 \geq x+y+z $$
But this is actually not always true, take the case when $ 0<x,y,z <1$ And you will see that this is not working.
Maybe I’m wrong.
 [/INST] I guess you guys are too fast and / or assume too much prior knowledge. Let's digest it.
We need to be careful since all numbers are reals. Hence AM-GM and square root  methods don't work.
First, note that if $xyz < 0$, then there is nothing to prove, since in this case $x^2 + y^2 + z^2 \geq 0 \geq xyz$ holds in all cases. So we need to consider  $xyz \ge  0$ only.
We want to establish the following train of inequalities:
$$
(x^2+y^2+z^2)^2 \ge (xy+yz+zx)^2 \ge {3xyz(x+y+z)} \ge  {3}  (xyz)^2 = 3 |xyz|^2
$$
The first one follows from
$$
 x^2+y^2+z^2 - ( xy+yz+zx)  = \frac12 \Big[ (x-y)^2+(y-z)^2+ (z-x)^2 \Big] \ge 0
$$
The second one follows from
$$
(xy+yz+zx)^2 -  3xyz(x+y+z) =  \frac12 \Big[(xy-yz)^2+(zx-xy)^2+(yz-zx)^2 \Big] \ge 0 
$$
For the third one, for $xyz > 0$, then this follows from the question's condition: $x + y+ z \ge xyz$.
So indeed it holds for all reals:
$$
x^2+y^2+z^2 \ge \sqrt{3} |xyz| \ge \sqrt{3} \cdot xyz 
$$
This is also the tightest inequality, since we observe that equality holds for the question's condition and for this solution at $x=y=z = \sqrt{3}$. (River Li asked for that constant $k = \sqrt{3}$).
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