Problem: Find all real numbers $x,y,z \ge 1$ satisfying
\[ \min \left( \sqrt{x+xyz}, \sqrt{y+xyz}, \sqrt{z+xyz} \right)
  = \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}. \]

Solution: Set $x = 1+a$, $y = 1+b$, $z = 1+c$
which eliminates the $x,y,z \ge 1$ condition.
Assume without loss of generality that $a \leq b \leq c$.
Then the given equation rewrites as
\[ \sqrt{(1+a)\left( 1+(1+b)(1+c) \right)} = \sqrt a + \sqrt b + \sqrt c. \]

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{align*}
  (1+a)\left( 1 + (1+b)(1+c) \right)
  &= (a+1)\left( 1 + (b+1)(1+c) \right) \\
  &\ge (a+1) \left( 1 + \left( \sqrt b + \sqrt c \right)^2 \right) \\
  &\ge \left( \sqrt a + \left( \sqrt b + \sqrt c \right) \right)^2
\end{align*}
by two applications of Cauchy-Schwarz.

Equality holds if $bc = 1$ and $1/a = \sqrt b + \sqrt c$.
Letting $c = t^2$ for $t \ge 1$,
we recover $b = t^{-2} \le t^2$ and $a = \frac{1}{t+1/t} \le t^2$.

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 $t > 0$.