Example 6 Let $a, b, c, x, y, z \in \mathbf{R}_{+}$, satisfying:
$$
c y + b z = a, \quad a z + c x = b, \quad b x + a y = c.
$$

Try to find the minimum value of the function
$$
f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}.
$$

[Analysis] On the surface, this problem is about an algebraic function, but it is actually a problem of finding the maximum value of a trigonometric function. The three equations given are the projection theorems of a triangle, and $x, y, z$ are used to hide $\cos A, \cos B, \cos C$ respectively.
From the conditions, we have
$$
\begin{array}{l}
b(a z+c x-b)+c(b x+a y-c)-a(c y+b z-a)=0 \\
\quad \Rightarrow 2 b c x+a^{2}-b^{2}-c^{2}=0 \\
\quad \Rightarrow x=\frac{b^{2}+c^{2}-a^{2}}{2 b c} .
\end{array}
$$

Similarly, \( y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}, z=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \).
Since \( a, b, c, x, y, z \in \mathbf{R}_{+} \), 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, a triangle \( \triangle ABC \) with side lengths \( a, b, c \) can be formed, and it is an acute triangle.
Thus, \( x=\cos A, y=\cos B, z=\cos C \).
The problem is reduced to: In an acute triangle \( \triangle ABC \), find the minimum value of the function
$$
f(\cos A, \cos B, \cos C)=\sum \frac{\cos ^{2} A}{1+\cos A}
$$

Let \( u=\cot A, v=\cot B, w=\cot C \).
Then \( u, v, w \in \mathbf{R}_{+}, u v+v w+w u=1 \), and
$$
\begin{array}{l}
u^{2}+1=(u+v)(u+w) \\
v^{2}+1=(u+v)(v+w) \\
w^{2}+1=(u+w)(v+w)
\end{array}
$$

Thus, \( \frac{\cos ^{2} A}{1+\cos A}=\frac{\frac{u^{2}}{u^{2}+1}}{1+\frac{u}{\sqrt{u^{2}+1}}} \)
$$
\begin{array}{l}
=\frac{u^{2}}{\sqrt{u^{2}+1}\left(\sqrt{u^{2}+1}+u\right)} \\
=\frac{u^{2}\left(\sqrt{u^{2}+1}-u\right)}{\sqrt{u^{2}+1}}=u^{2}-\frac{u^{3}}{\sqrt{u^{2}+1}} \\
=u^{2}-\frac{u^{3}}{\sqrt{(u+v)(u+w)}} \\
\geqslant u^{2}-\frac{u^{3}}{2}\left(\frac{1}{u+v}+\frac{1}{u+w}\right) .
\end{array}
$$

Then \( f \geqslant \sum u^{2}-\frac{1}{2} \sum \frac{u^{3}+v^{3}}{u+v} \)
$$
\begin{aligned}
= & u^{2}+v^{2}+w^{2}-\frac{1}{2}\left(\left(u^{2}-u v+v^{2}\right)+\right. \\
& \left.\left(v^{2}-v w+w^{2}\right)+\left(u^{2}-u w+w^{2}\right)\right) \\
= & \frac{1}{2}(u v+v w+u w)=\frac{1}{2},
\end{aligned}
$$

The equality holds if and only if \( u=v=w \), i.e., \( a=b=c, x=y=z=\frac{1}{2} \).