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 $b c=1$ and $1 / a=\sqrt{b}+\sqrt{c}$. Letting $c=t^{2}$ for $t \geq 1$, we recover $b=t^{-2} \leq t^{2}$ and $a=\frac{1}{t+1 / t} \leq t^{2}$.
Hence the solution set is
$$
(x, y, z)=\left(1+\left(\frac{t}{t^{2}+1}\right)^{2}, 1+\frac{1}{t^{2}}, 1+t^{2}\right)
$$
and permutations, for any $t>0$.