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
$$
f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}.
$$

$$
\begin{array}{l}
\Rightarrow 2 b c x + a^{2} - b^{2} - c^{2} = 0 \\
\Rightarrow x = \frac{b^{2} + c^{2} - a^{2}}{2 b c}.
\end{array}
$$

Similarly, $y = \frac{a^{2} + c^{2} - b^{2}}{2 a c}, \quad z = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$.
Since $a, b, c, x, y, z \in \mathbf{R}_{+}$, from the above three equations, we know
$$
b^{2} + c^{2} > a^{2}, \quad a^{2} + c^{2} > b^{2}, \quad a^{2} + b^{2} > c^{2}.
$$

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

Let $u = \cot A, \quad v = \cot B, \quad w = \cot C$.
Then $u, v, w \in \mathbf{R}_{+}, \quad u v + v w + w u = 1$, 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, $\frac{\cos^{2} A}{1 + \cos A} = \frac{\frac{u^{2}}{u^{2} + 1}}{1 + \frac{u}{\sqrt{u^{2} + 1}}}$
$$
= \frac{u^{2}}{\sqrt{u^{2} + 1} \left( \sqrt{u^{2} + 1} + u \right)}
$$
$= \frac{u^{2} \left( \sqrt{u^{2} + 1} - u \right)}{\sqrt{u^{2} + 1}} = u^{2} - \frac{u^{3}}{\sqrt{u^{2} + 1}}$
$= u^{2} - \frac{u^{3}}{\sqrt{(u + v)(u + w)}}$
$\geqslant u^{2} - \frac{u^{3}}{2} \left( \frac{1}{u + v} + \frac{1}{u + w} \right)$.
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., $a = b = c, \quad x = y = z = \frac{1}{2}$.
Prove starting from a constant, to achieve the structural transformation of the relational expression.
Notice,
$$
\begin{array}{l}
2 \leqslant \frac{\sin A}{\sin B}+\frac{\sin B}{\sin A}, \\
\sin A \cdot \cos A+\sin B \cdot \cos B \\
=\frac{\sin 2 A+\sin 2 B}{2}=\sin (A+B) \cdot \cos (A-B) \\
=\sin C \cdot \cos (A-B) \leqslant \sin C .
\end{array}
$$

Then \(2 \sum \cos A \cdot \cos B\)
$$
\begin{array}{l}
\leqslant \sum\left(\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A}\right) \cos A \cdot \cos B \\
=\sum \frac{\cos A}{\sin A}(\sin B \cdot \cos B+\sin C \cdot \cos C) \\
\leqslant \sum \cos A,
\end{array}
$$

The equality holds if and only if \(\angle A=\angle B=\angle C\). Therefore, the conclusion is proved.