$4 \cdot 146$ Solve the system of equations
$$
\left\{\begin{array}{l}
\frac{y z}{y+z}=a, \\
\frac{x z}{x+z}=b, \\
\frac{x y}{x+y}=c .
\end{array}\right.
$$
[Solution] Let $a \neq 0, b \neq 0, c \neq 0$, note that $x \neq 0, y \neq 0, z \neq 0$. Therefore, the original system of equations can be rewritten as
$$
\left\{\begin{array}{l}
\frac{1}{y}+\frac{1}{z}=\frac{1}{a} \\
\frac{1}{x}+\frac{1}{z}=\frac{1}{b} \\
\frac{1}{x}+\frac{1}{z}=\frac{1}{c}
\end{array} .\right.
$$

Adding the three equations, we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
By comparing the last equation with each equation in the system $\circledast$, the solution to the original system of equations is
$$
\left\{\begin{array}{l}
x=\frac{2 a b c}{a c+a b-b c}, \\
y=\frac{2 a b c}{a b+b c-a c}, \\
z=\frac{2 a b c}{a c+b c-a b} .
\end{array}\right.
$$