$x=y=1$ and $z<0$ will give $-\infty$ when $z$ is close to $0$.

okay, but what if $x,y,z$ are positive reals?

orl wrote: Let $x, y, z$ be real numbers such that $x + y + z \geq xyz$. Find the smallest possible value of $\frac {x^2 + y^2 + z^2}{xyz}$. See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17142

hello, if the variables are positive the searched minimum is $\sqrt{3}$. Sonnhard.

Let $x + y + z = kxyz$ Then $\sqrt {(\frac {x^2 + y^2 + z^2}{xyz})^2} = \sqrt {\frac {x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2}{x^2y^2z^2}} = \sqrt {\frac {x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2}{(xyz)(\frac {x + y + z}{k})}}$ Hence, we need to minimize $\frac {x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2}{(xyz)(\frac {x + y + z}{k})}$ I claim that $x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2 \ge 3xyz(x + y + z)$ Which is true by 2 trivial applications of muirheads inequality. Hence, since $3xyz(x + y + z) \ge 3xyz(\frac {x + y + z}{k})$, we have: $\frac {x^4 + y^4 + z^4 + 2x^2y^2 + 2y^2z^2 + 2z^2x^2}{(xyz)(\frac {x + y + z}{k})}\ge 3$ equality will hold when $x = y = z = \sqrt {3}$. The final minimum is the square root of this answer, which is $\sqrt {3}$.

Actually to show that $x^2+y^2+z^2\ge\sqrt{3}xyz$ appeared on the Balkan 2001 paper http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=19&year=2001&sid=5c22a3c467b0fb6290e11cdb1f19768c (this problem is Chinese Olympiad 2001).