2) (This question is worth 50 points) 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
$$
f(x, y, z)=\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}
$$
Given the condition, we have:
\[ b(a z + c x - b) + c(b x + a y - c) - a(c y + b z - a) = 0, \]
which simplifies to:
\[ 2 b c x + a^2 - b^2 - c^2 = 0, \]
so
\[ x = \frac{b^2 + c^2 - a^2}{2 b c}. \]

Similarly, we get:
\[ y = \frac{a^2 + c^2 - b^2}{2 a c}, \quad z = \frac{a^2 + b^2 - c^2}{2 a b}. \]

Since \(a, b, c, x, y, z\) are positive, from the above three equations, we know that \(b^2 + c^2 > a^2\), \(a^2 + c^2 > b^2\), and \(a^2 + b^2 > c^2\). Therefore, with \(a, b, c\) as side lengths, a right-angled triangle \(ABC\) can be formed, so \(x = \cos A\), \(y = \cos B\), \(z = \cos C\). Thus, the problem is transformed into finding the minimum value of the function \(f(\cos A, \cos B, \cos C)\) in the right-angled triangle \(ABC\):
\[ f = \frac{\cos^2 A}{1 + \cos A} + \frac{\cos^2 B}{1 + \cos B} + \frac{\cos^2 C}{1 + \cos C}. \]

Let \(u = \cot A\), \(v = \cot B\), \(w = \cot C\). Then:
\[ u, v, w \in \mathbf{R}^+, \]
\[ u v + v w + w u = 1, \]
\[ w^2 + 1 = (u + w)(v + w). \]

We have:
\[ \frac{\cos^2 A}{1 + \cos A} = \frac{\frac{u^2}{u^2 + 1}}{1 + \frac{u}{\sqrt{u^2 + 1}}} = \frac{u^2}{\sqrt{u^2 + 1}(\sqrt{u^2 + 1} + u)} = \frac{u^2 (\sqrt{u^2 + 1} - u)}{\sqrt{u^2 + 1}} = u^2 - \frac{u^3}{\sqrt{u^2 + 1}} = u^2 - \frac{u^3}{\sqrt{(u + v)(u + w)}} \geq u^2 - \frac{u^3}{2} \left( \frac{1}{u + v} + \frac{1}{u + w} \right). \]

Similarly:
\[ \frac{\cos^2 B}{1 + \cos B} \geq v^2 - \frac{v^3}{2} \left( \frac{1}{u + v} + \frac{1}{v + w} \right), \]
\[ \frac{\cos^2 C}{1 + \cos C} \geq w^2 - \frac{w^3}{2} \left( \frac{1}{u + w} + \frac{1}{v + w} \right). \quad \cdots \cdots \cdot 40 \text{ points} \]

Thus:
\[ f \geq u^2 + v^2 + w^2 - \frac{1}{2} \left( \frac{u^3 + v^3}{u + v} + \frac{v^3 + w^3}{v + w} + \frac{u^3 + w^3}{u + w} \right) = u^2 + v^2 + w^2 - \frac{1}{2} \left[ (u^2 - u v + v^2) + (v^2 - v w + w^2) + (u^2 - u w + w^2) \right] = \frac{1}{2} (u v + v w + u w) = \frac{1}{2}. \]

The equality holds when \(u = v = w\), i.e.,
\[ a = b = c, \quad x = y = z = \frac{1}{2}. \]

Thus, the minimum value of \(f(x, y, z)\) is:
\[ [f(x, y, z)]_{\min} = \frac{1}{2}. \]

Note: If a candidate directly sets \(x = y = z\) and gets the answer \(f(x, y, z)_{\min} = \frac{1}{2}\), only 10 points are given.

Solution 2: First, as in Solution 1, we can find:
\[ x = \frac{b^2 + c^2 - a^2}{2 b c}, \quad y = \frac{a^2 + c^2 - b^2}{2 a c}, \quad z = \frac{a^2 + b^2 - c^2}{2 a b}. \]

Let \(\sum\) denote the cyclic summation symbol, such as:
\[ \sum a = a + b + c, \]
\[ \sum a b = a b + a c + b c, \]
\[ \sum a^2 (b + c) = a^2 (b + c) + b^2 (a + c) + c^2 (a + b). \]

Then:
\[ f(x, y, z) = \sum \frac{\left( \frac{b^2 + c^2 - a^2}{2 b c} \right)^2}{1 + \frac{b^2 + c^2 - a^2}{2 b c}} = \sum \frac{(b^2 + c^2 - a^2)^2}{2 b c \left[ (b + c)^2 - a^2 \right]} \geq \frac{\left[ \sum (b^2 + c^2 - a^2) \right]^2}{2 (a + b + c) \sum b c (b + c - a)}. \]

(Cauchy-Schwarz inequality, because \(x > 0 \Rightarrow b^2 + c^2 - a^2 > 0\), similarly:
\[ a^2 + b^2 - c^2 > 0, \quad a^2 + c^2 - b^2 > 0). \]

Thus:
\[ = \frac{\left( \sum a^2 \right)^2}{2 (a + b + c) \left( \sum b c (b + c) - 3 a b c \right)} = \frac{\sum a^4 + 2 \sum a^2 b^2}{2 \left[ 2 \sum a^2 b c + 2 \sum a^2 b^2 + \sum a^3 (b + c) - 3 \sum a^2 b c \right]}. \]

Next, we prove \(f(x, y, z) \geq \frac{1}{2}\):
\[ \Leftrightarrow \frac{\sum a^4 + 2 \sum a^2 b^2}{2 \sum a^2 b^2 + \sum a^3 (b + c) - \sum a^2 b c} \geq 1, \]
\[ \Leftrightarrow \sum a^4 + \sum a^2 b c \geq \sum a^3 (b + c), \]
\[ \Leftrightarrow \sum a^2 (a - b)(a - c) \geq 0. \]

Assume \(a \geq b \geq c > 0\), then \(c^2 (c - a)(c - b) \geq 0\), so:
\[ a^2 (a - b)(a - c) + b^2 (b - a)(b - c) = (a - b)(a^3 - a^2 c - b^3 + b^2 c) = (a - b)^2 (a^2 + a b + b^2 - a c - b c) \geq 0 \quad \text{(since } a^2 \geq a c, \quad b^2 \geq b c). \]

Thus, (1) holds, i.e., \(f(x, y, z) \geq \frac{1}{2}\). The equality holds when \(a = b = c\), i.e., \(x = y = z = \frac{1}{2}\). Therefore, the minimum value of \(f(x, y, z)\) is \(\frac{1}{2}\).

Note: In fact, (1) is the famous Schur's inequality.