II. (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
$$
f(x, y, z)=\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}
$$
From 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 $x$ gives $x=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$.
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, from the above three equations, we know
$$
b^{2}+c^{2}>a^{2}, a^{2}+c^{2}>b^{2}, a^{2}+b^{2}>c^{2} \text {. }
$$

Thus, a triangle $\triangle ABC$ with sides $a, b, c$ can be formed, and it is an acute triangle. Therefore, $x=\cos A, y=\cos B, z=\cos C$.
The problem then becomes: In an acute $\triangle ABC$, find the minimum value of the function
$$
\begin{array}{l}
f(\cos A, \cos B, \cos C) \\
=\frac{\cos ^{2} A}{1+\cos A}+\frac{\cos ^{2} B}{1+\cos B}+\frac{\cos ^{2} C}{1+\cos C}
\end{array}
$$

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

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

Similarly, $\frac{\cos ^{2} B}{1+\cos B} \geqslant v^{2}-\frac{v^{3}}{2}\left(\frac{1}{u+v}+\frac{1}{v+w}\right)$,
$$
\frac{\cos ^{2} C}{1+\cos C} \geqslant w^{2}-\frac{w^{3}}{2}\left(\frac{1}{u+w}+\frac{1}{v+w}\right) \text {. }
$$

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

Equality holds if and only if $u=v=w$. In this case,
$$
a=b=c, x=y=z=\frac{1}{2} \text {. }
$$

Therefore, $[f(x, y, z)]_{\min }=\frac{1}{2}$.