Example 3 Solve the system of equations
$$
\left\{\begin{array}{l}
\sqrt{y-a}+\sqrt{z-a}=1, \\
\sqrt{z-b}+\sqrt{x-b}=1, \\
\sqrt{x-c}+\sqrt{y-c}=1 .
\end{array}\right.
$$

where $a, b, c$ are positive numbers, and
$$
\sqrt{a}+\sqrt{b}+\sqrt{c}=\frac{\sqrt{3}}{2} \text{. }
$$
(2nd National Mathematical Olympiad Training Examination)
Solution 1: Construct an equilateral triangle $PQR$ with side length 1, then the sum of the distances from any point $O$ inside the triangle to the three sides is always equal to the height $\frac{\sqrt{3}}{2}$ (proof omitted). Due to equation (4), we can choose point $O$ such that the distances from $O$ to the three sides are $OA=\sqrt{a}, OB=\sqrt{b}, OC=\sqrt{c}$, as shown in Figure 1, the construction method is omitted. Then
$$
(x, y, z)=\left(OP^{2}, OQ^{2}, OR^{2}\right)
$$

satisfies the given system of equations, which is a set of solutions we seek. We can obtain the specific values of equation (5) as
$$
(x, y, z)=\left(x_{0}, y_{0}, z_{0}\right),
$$

where $x_{0}=\frac{4}{3}(b+\sqrt{bc}+c), y_{0}=\frac{4}{3}(c+\sqrt{ca}+\sqrt{a}), z_{0}=\frac{4}{3}(a+\sqrt{ab}+b)$. The method is: let $\angle OPB=\alpha_{1}, \angle OPC=\alpha_{2}$, then
$$
\begin{array}{l}
\frac{1}{2}=\cos 60^{\circ}=\cos \left(\alpha_{1}+\alpha_{2}\right) \\
=\cos \alpha_{1} \cdot \cos \alpha_{2}-\sin \alpha_{1} \cdot \sin \alpha_{2} \\
=\frac{\sqrt{x-b}}{\sqrt{x}} \cdot \frac{\sqrt{x-c}}{\sqrt{x}}-\frac{\sqrt{b}}{\sqrt{x}} \cdot \frac{\sqrt{c}}{\sqrt{x}} \\
=\frac{1}{x}(\sqrt{(x-b)(x-c)}-\sqrt{bc}), \\
x+2 \sqrt{bc}=2 \sqrt{(x-b)(x-c)}, \\
x^{2}+4 \sqrt{bc} x+4 bc \\
=4 x^{2}-4(b+c) x+4 bc, \\
3 x^{2}-4(b+\sqrt{bc}+c) x=0 . \\
\because x \neq 0, \\
\therefore x=\frac{4}{3}(b+\sqrt{bc}+c)=x_{0} .
\end{array}
$$
The methods for finding $y$ and $z$ are similar.
If the original system of equations has a solution $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$, and $x^{\prime}>x_{0}$, then by equation (3) we get $y^{\prime}z_{0}$, and by equation (2) we get $x^{\prime}<x_{0}$, leading to a contradiction.
Similarly, $x^{\prime}<x_{0}$ also leads to a contradiction.
Therefore, $x^{\prime}=x_{0}$. Similarly, $y^{\prime}=y_{0}, z^{\prime}=z_{0}$.
Thus, equation (6) is the unique solution to the original system of equations.
The ingenious part of Solution 1 lies in its full utilization of the information carried by the condition (4) that parameters $a, b, c$ satisfy. Inspired by this, we can also provide an algebraic solution to Example 3.
Solution 2: From equation (4), we get
$$
\begin{array}{c}
\frac{1}{\sqrt{3}}(2 \sqrt{c}+\sqrt{a})+\frac{1}{\sqrt{3}}(\sqrt{a}+2 \sqrt{b})=1 . \\
\therefore \sqrt{\frac{1}{3}(4 c+4 \sqrt{aa}+a)}+\sqrt{\frac{1}{3}(a+4 \sqrt{ab}+4 b)}=1, \\
\sqrt{\frac{4}{3}(c+\sqrt{aa}+a)-a}+\sqrt{\frac{4}{3}(a+\sqrt{ab}+b)-a}=1 .
\end{array}
$$

Therefore, equation (6) satisfies equation (1), and similarly, it also satisfies equations (2) and (3), hence it is a set of solutions to the original system of equations (uniqueness is the same as before, omitted).