Example 7 Let positive numbers $a, b, c, x, y, z$ satisfy
$$
c y+b z=a, a z+c x=b, b x+a y=c .
$$

Find the minimum value of the function $f(x, y, z)=\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}$.
(2005, National High School Mathematics Competition)
Given the conditions:
$$
\begin{array}{l}
b(a z+c x)+c(b x+a y)-a(c y+b z) \\
=b^{2}+c^{2}-a^{2},
\end{array}
$$

we have \( x=\frac{b^{2}+c^{2}-a^{2}}{2 b c} \).
Similarly, \( y=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \), \( z=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \).
Using \(\sum\) to denote cyclic summation, we have:
$$
\begin{array}{l}
f(x, y, z)=\sum \frac{x^{2}}{1+x} \\
=\sum \frac{\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 b^{2} c^{2}+2 b c\left(b^{2}+c^{2}-a^{2}\right)} \\
\geqslant \sum \frac{\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 b^{2} c^{2}+\left(b^{2}+c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)} .
\end{array}
$$

Let \( u_{i}>0, v_{i}>0 \) for \( i=1,2,3 \).
By the Cauchy-Schwarz inequality, we have:
$$
\left(\sum_{i=1}^{3} u_{i}\right)^{2}=\left(\sum_{i=1}^{3} \frac{u_{i}}{\sqrt{v_{i}}} \cdot \sqrt{v_{i}}\right)^{2} \leqslant \sum_{i=1}^{3} \frac{u_{i}^{2}}{v_{i}} \cdot \sum_{i=1}^{3} v_{i},
$$

which implies:
$$
\sum_{i=1}^{3} \frac{u_{i}^{2}}{v_{i}} \geqslant \frac{\left(\sum_{i=1}^{3} u_{i}\right)^{2}}{\sum_{i=1}^{3} v_{i}}.
$$
Equality holds if and only if \(\frac{u_{1}}{v_{1}}=\frac{u_{2}}{v_{2}}=\frac{u_{3}}{v_{3}}\).

Thus,
$$
\sum \frac{\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 b^{2} c^{2}+\left(b^{2}+c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)}
$$
$$
\begin{array}{l}
\geqslant \frac{\left[\sum\left(b^{2}+c^{2}-a^{2}\right)\right]^{2}}{4 \sum b^{2} c^{2}+\sum\left(b^{2}+c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)}(2) \\
=\frac{\left(b^{2}+c^{2}+a^{2}\right)^{2}}{4 \sum b^{2} c^{2}+\sum\left(b^{2}+c^{2}\right)\left(b^{2}+c^{2}+c^{2}\right)-2 \sum\left(b^{2}+c^{2}\right) a^{2}} \\
=\frac{\left(b^{2}+c^{2}+a^{2}\right)^{2}}{4 \sum b^{2} c^{2}+2\left(b^{2}+c^{2}+a^{2}\right)^{2}-4 \sum b^{2} c^{2}} \\
=\frac{\left(b^{2}+c^{2}+a^{2}\right)^{2}}{2\left(b^{2}+c^{2}+a^{2}\right)^{2}}=\frac{1}{2} .
\end{array}
$$

Equality in (1) holds if and only if \( a=b=c \), and in this case, equality in (2) also holds. Given the conditions, we have \( x=y=z=\frac{1}{2} \).
Therefore, when and only when \( x=y=z=\frac{1}{2} \),
$$
f(x, y, z)_{\min }=\frac{1}{2} .
$$