<s>[INST] If $xyz = x+y+z$ for $x,y,z > 0$, then $\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$ 
If $x,y,z \in \Bbb R_{>0}$ satisfy $xyz = x+y+z$, prove that $$\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$$

We can express $1+x^2$ as
$$1+x^2 = (1-xy)(1-xz) = \frac{(y-xyz)(z-xyz)}{yz} = \frac{(x+z)(x+y)}{yz}$$
since $x + y + z = xyz$ implies $1+ x^2 + xy + xz = 1 + x^2yz$. Thus, our desired inequality boils down to
$$\sqrt{\frac{(x+z)(x+y)}{yz}} + \sqrt{\frac{(y+z)(y+x)}{xz}} + \sqrt{\frac{(z+x)(z+y)}{xy}} \ge 6$$
which looks more complicated. Perhaps I am not headed in the right direction? I'd appreciate any hints or solutions. Thank you!
 [/INST] Your idea gives a simple solution.
Indeed, by AM-GM twice we obtain: $$\sum_{cyc}\sqrt{\frac{(x+y)(x+z)}{yz}}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(x+y)}{xyz}}\geq3\sqrt[3]{\frac{8xyz}{xyz}}=6.$$
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