Example 7 Given $0<a, b, c<1$, and
$$
\begin{array}{l}
\sqrt{a(1-b)(1-c)}+\sqrt{b(1-c)(1-a)} \\
+\sqrt{c(1-a)(1-b)}=\sqrt{a b c} .
\end{array}
$$

Try to solve the system of equations
$$
\left\{\begin{array}{l}
\sqrt{y-b c}+\sqrt{z-b c}=\sqrt{a}, \\
\sqrt{z-c a}+\sqrt{x-c a}=\sqrt{b}, \\
\sqrt{x-a b}+\sqrt{y-a b}=\sqrt{c} .
\end{array}\right.
$$
Solution 1: (Algebraic Solution) Equation (1) can be transformed as follows:
$$
\begin{array}{l}
\sqrt{1-a}[\sqrt{b(1-c)}+\sqrt{c(1-b)}] \\
=\sqrt{a}[\sqrt{b c}-\sqrt{(1-b)(1-c)}] . \\
\because a>0, \\
\therefore \sqrt{\frac{1-a}{a}}=\frac{\sqrt{b c}-\sqrt{(1-b)(1-c)}}{\sqrt{b(1-c)+\sqrt{c(1-b)}}} . \\
\text { Hence } \frac{1-a}{a}=\frac{[\sqrt{b c}-\sqrt{(1-b)(1-c)}]^{2}}{[\sqrt{b(1-c)}+\sqrt{c(1-b)}]^{2}} \\
\frac{1}{a}=\frac{b c+(1-b)(1-c)+b(1-c)+c(1-b)}{[\sqrt{b(1-c)}+\sqrt{c(1-b)}]^{2}}, \\
\therefore \sqrt{a}=\sqrt{b(1-c)}+\sqrt{c(1-b)} .
\end{array}
$$

Therefore, $(x, y, z)=(a, b, c)$ satisfies equation (2), and similarly, it satisfies equations (3) and (4), so it is one solution to the original system of equations.

If $x>a$, then $yc, x<a$, leading to a contradiction. Similarly, $x<a$ also leads to a contradiction, hence $x=a$. Similarly, $y=b, z=c$. Therefore, $(x, y, z)=(a, b, c)$ is the unique solution to the system of equations.

Solution 2: (Geometric Solution) Let $\sqrt{a}=\sin A, \sqrt{b}=\sin B, \sqrt{c}=\sin C$, where $A, B, C$ are acute angles. Then equation (1) can be transformed as follows:
$\sin A \cdot \cos B \cdot \cos C + \sin B \cdot \cos C \cdot \cos A + \sin C \cdot \cos A \cdot \cos B = \sin A \cdot \sin B \cdot \sin C$,
$$
\begin{array}{l}
\therefore \sin (A+B+C)=0, \\
A+B+C=\pi .
\end{array}
$$

Construct an acute triangle $\triangle ABC$ with angles $A, B, C$, as shown in Figure 2. Let the diameter of the circumcircle of $\triangle ABC$ be 1, then $BC=\sin A$
$$
\begin{array}{l}
=\sqrt{a}, CA=\sin B=\sqrt{b}, AB=\sin C=\sqrt{c}, \text { and } \\
\angle B H F=\angle C H E=A, \angle C H D=\angle A H F=B, \\
\angle A H E=\angle B H D=C, A H=\cos A, B H=\cos B, \\
C H=\cos C .
\end{array}
$$
$$
\begin{array}{l} 
\text { From } \sqrt{a}=B C=B D+C D=B H \cdot \sin \angle B H D \\
+C H \cdot \sin \angle C H D=\cos B \cdot \sin C+\sin B \cdot \cos C \\
= \sqrt{b(1-c)}+\sqrt{c(1-b)} \text {, it follows that }(x, y, z) \\
=(a, b, c) \text { is a solution to the original system of equations }(\text { uniqueness as }
\end{array}
$$
before, omitted).