Find all real numbers \( x, y, z \geq 1 \) satisfying 
\[ \min (\sqrt{x+x y z}, \sqrt{y+x y z}, \sqrt{z+x y z})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \]
Set $x=1+a, y=1+b, z=1+c$ which eliminates the $x, y, z \geq 1$ condition. Assume without loss of generality that $a \leq b \leq c$. Then the given equation rewrites as 
$$ \sqrt{(1+a)(1+(1+b)(1+c))}=\sqrt{a}+\sqrt{b}+\sqrt{c} $$
In fact, we are going to prove the left-hand side always exceeds the right-hand side, and then determine the equality cases. We have:
$$ \begin{aligned} (1+a)(1+(1+b)(1+c)) & =(a+1)(1+(b+1)(1+c)) \\ & \leq(a+1)\left(1+(\sqrt{b}+\sqrt{c})^{2}\right) \\ & \leq(\sqrt{a}+(\sqrt{b}+\sqrt{c}))^{2} \end{aligned} $$
by two applications of Cauchy-Schwarz. Equality holds if $bc=1$ and $1/a=\sqrt{b}+\sqrt{c}$. Letting $c=t^{2}$ for $t \geq 1$, we recover $b=t^{-2} \leq t^{2}$ and $a=\frac{1}{t+1/t} \leq t^{2}$. 
$$ (x, y, z)=\left(1+\left(\frac{t}{t^{2}+1}\right)^{2}, 1+\frac{1}{t^{2}}, 1+t^{2}\right) $$
and permutations, for any $t>0$.