13. (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 $f(x, y, z) = \frac{x^2}{1 + x} + \frac{y^2}{1 + y} + \frac{z^2}{1 + z}$.
13. Based on the conditional inequality and the structure of the required fraction, construct the distribution of the random variable $\xi$ as follows:
\begin{tabular}{|c|c|c|c|}
\hline$\xi$ & $\frac{x}{1+x}$ & $\frac{y}{1+y}$ & $\frac{z}{1+z}$ \\
\hline$P$ & $\frac{1+x}{3(x+y+z)}$ & $\frac{1+y}{3(x+y+z)}$ & $\frac{1+z}{3(x+y+z)}$ \\
\hline
\end{tabular}

By $E \xi=\left(\frac{x}{1+x}\right)\left[\frac{1+x}{3(x+y+z)}\right]+\left(\frac{y}{1+y}\right)\left[\frac{1+y}{3(x+y+z)}\right]+\left(\frac{z}{1+z}\right)\left[\frac{1+z}{3(x+y+z)}\right]=\frac{1}{3}$,
By $E \xi^{2}=\left(\frac{x}{1+x}\right)^{2}\left[\frac{1+x}{3(x+y+z)}\right]+\left(\frac{y}{1+y}\right)^{2}\left[\frac{1+y}{3(x+y+z)}\right]+\left(\frac{z}{1+z}\right)^{2}\left[\frac{1+z}{3(x+y+z)}\right]$ $=\frac{1}{3(x+y+z)}\left(\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}\right)$,

Also, by the variance formula
$$
\begin{aligned}
D \xi=( & \left.\frac{x}{1+x}-E \xi\right)^{2} \cdot\left[\frac{1+x}{3(x+y+z)}\right]+\left(\frac{y}{1+y}-E \xi\right)^{2} \cdot\left[\frac{1+y}{3(x+y+z)}\right] \\
& +\left(\frac{z}{1+z}-E \xi\right)^{2} \cdot\left[\frac{1+z}{3(x+y+z)}\right] \geqslant 0 .
\end{aligned}
$$

We get $D \xi=E \xi^{2}-(E \xi)^{2} \geqslant 0$, which means $E \xi^{2} \geqslant(E \xi)^{2}$. The equality holds if and only if $\frac{x}{1+x}=\frac{y}{1+y}=\frac{z}{1+z}$. Therefore, $\frac{1}{3(x+y+z)}\left(\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}\right) \geqslant \frac{1}{9}$.

Thus, $f(x, y, z)=\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z} \geqslant \frac{(x+y+z)}{3}$, so when and only when $x=y=z=\frac{1}{2}$, the minimum value of $f(x, y, z)$ is $\frac{1}{2}$.