The key Lemma is: $$\sqrt{a-1}+\sqrt{b-1} \le \sqrt{ab}$$ for all $a,b \ge 1$ . Equality holds when $(a-1)(b-1)=1$ .
This is proven easily. $$\sqrt{a-1}+\sqrt{b-1} = \sqrt{a-1}\sqrt{1}+\sqrt{1}\sqrt{b-1} \le \sqrt{(a-1+1)(b-1+1)} = \sqrt{ab}$$ by Cauchy.
Equality then holds when $a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1$ .
Now assume that $x = \min(x,y,z)$ . Now note that, by the Lemma,
$$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}$$ . So equality must hold in order for the condition in the problem statement to be met.
So $(y-1)(z-1) = 1$ and $(x-1)(yz) = 1$ . If we let $z = c$ , then we can easily compute that $y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}$ .
Now it remains to check that $x \le y, z$ .
But by easy computations, $x = \frac{c^2+c-1}{c^2} \le c = z \Longleftrightarrow (c^2-1)(c-1) \ge 0$ , which is obvious.
Also $x = \frac{c^2+c-1}{c^2} \le \frac{c}{c-1} = y \Longleftrightarrow 2c \ge 1$ , which is obvious, since $c \ge 1$ .
So all solutions are of the form $\boxed{\left(\frac{c^2+c-1}{c^2}, \frac{c}{c-1}, c\right)}$ , and all permutations for $c > 1$ .
Remark: An alternative proof of the key Lemma is the following:
By AM-GM, $$(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}$$ $$ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}$$ . Now taking the square root of both sides gives the desired. Equality holds when $(a-1)(b-1) = 1$ .