Without loss of generality we can suppose $a = \max\{a, b, c\}$. From the first equation of the given system we obtain
$$c(\sqrt{b} - 1) \le a(\sqrt{b} - 1) = c, \quad \text{i.e.} \quad b \le 4.$$
Similarly, from the second equation of the given system we have
$$b(\sqrt{c} - 1) = a \ge b, \quad \text{thus} \quad c \ge 4.$$
Using these results together with the third equation we get
$$4 \le c \le c(\sqrt{c} - 1) \le c(\sqrt{a} - 1) = b \le 4$$
which yields $a = b = c = 4$.

Conclusion. The unique solution of the given system of equations is therefore $(a, b, c) = (4, 4, 4)$.