<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] First, we prove for all positive $x,y,z$.
If all of $x,y,z\geq1$ we have
$$x^2\geq x ,\\ y^2\geq y \\ z^2\geq z$$ $$⇒x^2+y^2+z^2\geq x+y+z \geq xyz\\$$
If at least one of $x,y,z$ is less than $1$ then take (WLOG) $x\geq y\geq z>0$ & $1>z>0$
$$\frac{x}{y}\geq1 $$
$$⇒\frac{x}{yz}\geq1 $$
$$⇒\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\geq 1$$
$$⇒x^2+y^2+z^2\geq xyz$$
If two of $x,y,z$ is negative (say $x$ and $y$)
we have $$-x-y+z ≥ x+y+z ≥xyz=(-x)(-y)z$$
$$⇒-x-y+z ≥(-x)(-y)z$$ Here $-x,-y,z$ are positive numbers. So from the earlier result we can say,
$$(-x)^2+(-y)^2+z^2\geq (-x)(-y)z$$ $$⇒x^2+y^2+z^2\geq xyz$$
The inequality is obvious when one of $x,y,z$ is $0$ or if one of $x,y,z$ is negative or if all of $x,y,z$ is negative.
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