Second 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 $f(x, y, z) = \frac{x^{2}}{1+x} + \frac{y^{2}}{1+y} + \frac{z^{2}}{1+z}$.
Given the conditions, we have
$$
\begin{array}{l}
b(a z+c x-b)+c(b x+a y-c)- \\
a(c y+b z-a)=0,
\end{array}
$$

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, $y=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$ and $z=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$.
Thus, $f(x, y, z)$
$$
\begin{array}{c}
=\frac{\left(b^{2}+c^{2}-a^{2}\right)^{2}}{2 b c\left(b^{2}+c^{2}+2 b c-a^{2}\right)}+ \\
\frac{\left(a^{2}+c^{2}-b^{2}\right)^{2}}{2 a c\left(a^{2}+c^{2}+2 a c-b^{2}\right)}+ \\
\frac{\left(a^{2}+b^{2}-c^{2}\right)^{2}}{2 a b\left(a^{2}+b^{2}+2 a b-c^{2}\right)} .
\end{array}
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
f(x, y, z) \cdot\left[2 b c\left(b^{2}+c^{2}+2 b c-a^{2}\right)+\right. \\
2 a c\left(a^{2}+c^{2}+2 a c-b^{2}\right)+ \\
\left.2 a b\left(a^{2}+b^{2}+2 a b-c^{2}\right)\right] \\
\geqslant\left[\left(b^{2}+c^{2}-a^{2}\right)+\left(a^{2}+c^{2}-b^{2}\right)+\right. \\
\left.\quad\left(a^{2}+b^{2}-c^{2}\right)\right]^{2} \\
=\left(a^{2}+b^{2}+c^{2}\right)^{2} .
\end{array}
$$

Thus, $f(x, y, z) \geqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{4\left(b^{2} c^{2}+a^{2} c^{2}+b^{2} a^{2}\right)-2 M}$,

where $M=a^{2} b c+b^{2} a c+c^{2} a b-b^{3} c-b c^{3}-$ $a^{3} c-a c^{3}-a^{3} b-a b^{3}$.
Next, we prove
$$
\begin{array}{l}
a^{2}(a-b)(a-c)+b^{2}(b-a)(b-c)+ \\
c^{2}(c-a)(c-b) \geqslant 0\left(a, b, c \in \mathbf{R}_{+}\right) . \text {(2) }
\end{array}
$$

Since $a, b, c \in \mathbf{R}_{+}$ and the expression is cyclic, we can assume without loss of generality that $a \geqslant b \geqslant c>0$, then
$$
\begin{array}{l}
c^{2}(c-a)(c-b) \geqslant 0, \\
a^{2}(a-b)(a-c) \geqslant 0 .
\end{array}
$$

Thus, we have
$$
\begin{array}{l}
a^{2}(a-b)(a-c)+b^{2}(b-a)(b-c)+ \\
c^{2}(c-a)(c-b) \\
\geqslant b^{2}(a-b)(a-c)+b^{2}(b-a)(b-c) \\
=b^{2}(a-b)^{2} \geqslant 0 .
\end{array}
$$

Expanding equation (2), we get
$$
\begin{array}{l}
a^{4}+b^{4}+c^{4}+a^{2} b c+a b^{2} c+a b c^{2} \\
\geqslant a^{3} b+a^{3} c+b^{3} c+b^{3} a+c^{3} a+c^{3} b .
\end{array}
$$

From equations (1) and (3), we have
$$
\begin{array}{l}
f(x, y, z) \geqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{4\left(b^{2} c^{2}+a^{2} c^{2}+b^{2} a^{2}\right)-2 M} \\
\geqslant \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{4\left(b^{2} c^{2}+a^{2} c^{2}+b^{2} a^{2}\right)-2\left(-a^{4}-b^{4}-c^{4}\right)} \\
=\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{2\left(a^{4}+b^{4}+c^{4}+2 b^{2} c^{2}+2 a^{2} c^{2}+2 b^{2} a^{2}\right)} \\
=\frac{1}{2} .
\end{array}
$$

The equality holds if and only if $a=b=c$, i.e., $x=y=z=\frac{1}{2}$.