Example 4 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, we have
$$
b(a z+c x-b)+c(b x+a y-c)-a(c y+b z-a)=0,
$$

which simplifies to $2 b c x+a^{2}-b^{2}-c^{2}=0$.
Solving for $x$, we get $x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$.
Similarly, $y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}, z=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$.
Since $a, b, c, x, y, z$ are positive numbers, from the above three equations, we know
$$
b^{2}+c^{2}>a^{2}, a^{2}+c^{2}>b^{2}, a^{2}+b^{2}>c^{2}.
$$

Therefore, with $a, b, c$ as side lengths, a right-angled $\triangle ABC$ can be formed.
Thus, $x=\cos A, y=\cos B, z=\cos C$.
The problem is then transformed into: In the right-angled $\triangle ABC$, find the minimum value of the function
$f(\cos A, \cos B, \cos C)$
$$
=\frac{\cos ^{2} A}{1+\cos A}+\frac{\cos ^{2} B}{1+\cos B}+\frac{\cos ^{2} C}{1+\cos C}.
$$

Let $u=\cot A, v=\cot B, w=\cot C$, then $u, v, w \in \mathbf{R}_{+}, w v+v w+w u=1$, and
$u^{2}+1=(u+v)(u+w)$,
$v^{2}+1=(u+v)(v+w)$,
$w^{2}+1=(u+w)(v+w)$.
Thus, $\frac{\cos ^{2} A}{1+\cos A}=\frac{\frac{u^{2}}{u^{2}+1}}{1+\frac{u}{\sqrt{u^{2}+1}}}$
$=\frac{u^{2}}{\sqrt{u^{2}+1}\left(\sqrt{u^{2}+1}+u\right)}=\frac{u^{2}\left(\sqrt{u^{2}+1}-u\right)}{\sqrt{u^{2}+1}}$
$=u^{2}-\frac{u^{3}}{\sqrt{u^{2}+1}}=u^{2}-\frac{u^{3}}{\sqrt{(u+v)(u+w)}}$
$\geqslant u^{2}-\frac{u^{3}}{2}\left(\frac{1}{u+v}+\frac{1}{u+w}\right)$.
Similarly, $\frac{\cos ^{2} B}{1+\cos B} \geqslant v^{2}-\frac{v^{3}}{2}\left(\frac{1}{u+v}+\frac{1}{v+w}\right)$,
$\frac{\cos ^{2} C}{1+\cos C} \geqslant w^{2}-\frac{w^{3}}{2}\left(\frac{1}{u+w}+\frac{1}{v+w}\right)$.
Thus, $f \geqslant u^{2}+v^{2}+w^{2}-$
$$
\begin{aligned}
& \frac{1}{2}\left(\frac{u^{3}+v^{3}}{u+v}+\frac{v^{3}+w^{3}}{v+w}+\frac{u^{3}+w^{3}}{u+w}\right) \\
= & u^{2}+v^{2}+w^{2}-\frac{1}{2}\left[\left(u^{2}-u v+v^{2}\right)+\right. \\
& \left.\left(v^{2}-v w+w^{2}\right)+\left(u^{2}-u w+w^{2}\right)\right] \\
= & \frac{1}{2}(w v+v w+u w)=\frac{1}{2}.
\end{aligned}
$$

The equality holds if and only if $u=v=w$.
At this point, $a=b=c, x=y=z=\frac{1}{2}$.