4-123 Let $a, b, c$ be given positive real numbers, try to determine all positive real numbers $x, y, z$ that satisfy the system of equations
$$\left\{\begin{array}{l}
x+y+z=a+b+c \\
4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c .
\end{array}\right.$$
［Solution］ The second equation above is equivalent to
$$\frac{a^{2}}{y z}+\frac{b^{2}}{x z}+\frac{c^{2}}{x y}+\frac{a b c}{x y z}=4$$

Let $x_{1}=\frac{a}{\sqrt{y z}}, y_{1}=\frac{b}{\sqrt{z x}}, z_{1}=\frac{c}{\sqrt{x y}}$, then
$$x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1}=4$$

Obviously, $00$, thus
$$z_{1}=-2 \sin u \sin v+2 \cos u \cos v=2 \cos (u+v)$$

Therefore, we have
$$\left\{\begin{array}{l}
2 \sin u \sqrt{y z}=a \\
2 \sin v \sqrt{z x}=b \\
2 \cos (u+v) \sqrt{x y}=c
\end{array}\right.$$

Thus,
$$\begin{aligned}
& (\sqrt{x} \cos v-\sqrt{y} \cos u)^{2}+(\sqrt{x} \sin v+\sqrt{y} \sin u-\sqrt{z})^{2} \\
= & x+y+z-2 \sqrt{x y} \cos u \cos v+2 \sqrt{x y} \sin u \sin v-2 \sqrt{z} x \sin v \\
& -2 \sqrt{y z} \sin u \\
= & a+b+c-c-b-a \\
= & 0
\end{aligned}$$

Therefore,
$$\begin{aligned}
\sqrt{z} & =\sqrt{x} \sin v+\sqrt{y} \sin u \\
& =\sqrt{x} \cdot \frac{b}{2 \sqrt{z x}}+\sqrt{y} \cdot \frac{a}{2 \sqrt{y z}} \\
& =\frac{1}{\sqrt{z}} \cdot \frac{a+b}{2} \\
z & =\frac{a+b}{2}
\end{aligned}$$

By the symmetry of the original equation, we get
$$x=\frac{b+c}{2}, y=\frac{c+a}{2}$$

It is easy to verify that the triplet $\left(\frac{b+c}{2}, \frac{c+a}{2}, \frac{a+b}{2}\right)$ satisfies the given system of equations.

Therefore, the original system of equations has only one positive real solution:
$$x=\frac{b+c}{2}, \quad y=\frac{c+a}{2}, \quad z=\frac{a+b}{2} .$$