18. (2005 National High School Mathematics League Additional Question) 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 $f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}$.
18. Using $\sum$ to denote cyclic sums, from the condition we get $b(a z+c x)+c(b x+a y)-a(c y+b z)=b^{2}+c^{2}-a^{2}$, which simplifies to $2 b x \cdot x = b^{2}+c^{2}-a^{2}$, so $x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$,
Similarly, $y=\frac{c^{2}+a^{2}-b^{2}}{2 a c}, z=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$,
Thus,
$$
\begin{aligned}
f(x, y, z) & =\sum \frac{x^{2}}{1+x}=\sum \frac{\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 b^{2} c^{2}+2 b c\left(b^{2}+c^{2}-a^{2}\right)} \\
& \geqslant \sum \frac{\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 b^{2} c^{2}+\left(b^{2}+c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)} \\
& \geqslant \frac{\left[\sum\left(b^{2}+c^{2}-a^{2}\right)\right]^{2}}{4 \sum b^{2} c^{2}+\sum\left(b^{2}+c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)} \\
& =\frac{\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 \sum b^{2} c^{2}+\sum\left(b^{2}+c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)-2 \sum\left(b^{2}+c^{2}\right) a^{2}} \\
& =\frac{\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 \sum b^{2} c^{2}+2\left(b^{2}+c^{2}-a^{2}\right)^{2}-4 \sum b^{2} c^{2}}=\frac{1}{2}
\end{aligned}
$$

The right-hand side of inequality (1) is obtained from the equivalent form (I) of the Cauchy-Schwarz inequality, and equality in (1) holds if and only if $a=b=c$, i.e., $x=y=z=\frac{1}{2}$, at which point equality in (2) also holds. Therefore, when $x=y=z=\frac{1}{2}$, $f(x, y, z)$ attains its minimum value of $\frac{1}{2}$.