Example 4—(2005 National High School Mathematics League Additional Question 2) 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 $f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}$.
Given the conditions, we have $b(a z+c x) + c(b x+a y) - a(c y+b z) = b^{2} + c^{2} - a^{2}$, which simplifies to $2 b c x = b^{2} + c^{2} - a^{2}$. Therefore, $x = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$,

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

The right side of inequality (5) is derived from the equivalent form of the Cauchy-Schwarz inequality (2), and equality in (5) holds if and only if $a = b = c$, i.e., $x = y = z = \frac{1}{2}$. At this point, equality in (6) also holds, so when $x = y = z = \frac{1}{2}$, $f(x, y, z)$ achieves its minimum value of $\frac{1}{2}$.