II. (50 points) Let $x_{i} \in \mathbf{R}_{+}(i=1,2,3,4,5)$. Find the minimum value of $f=\frac{x_{1}+x_{3}}{x_{5}+2 x_{2}+3 x_{4}}+\frac{x_{2}+x_{4}}{x_{1}+2 x_{3}+3 x_{5}}+$ $\frac{x_{3}+x_{5}}{x_{2}+2 x_{4}+3 x_{1}}+\frac{x_{4}+x_{1}}{x_{3}+2 x_{5}+3 x_{2}}+\frac{x_{5}+x_{2}}{x_{4}+2 x_{1}+3 x_{3}}$.
II. Solution: Let $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=1$, then $f=\frac{5}{3}$.
The following proves: $f \geqslant \frac{5}{3}$. By the Cauchy-Schwarz inequality, we have
$$
f=\sum \frac{x_{1}+x_{3}}{x_{5}+2 x_{2}+3 x_{4}} \geqslant \frac{\sum\left(x_{1}+x_{3}\right)}{\sum\left(x_{1}+x_{3}\right)\left(x_{5}+2 x_{2}+3 x_{4}\right)} .
$$

Therefore, it suffices to prove $\frac{\sum\left(x_{1}+x_{3}\right)}{\sum\left(x_{1}+x_{3}\right)\left(x_{5}+2 x_{2}+3 x_{4}\right)} \geqslant \frac{5}{3}$
$$
\Leftrightarrow 3 \sum x_{1}^{2}-4 \sum x_{1} x_{2} \geqslant-\sum x_{1} x_{3} \Leftrightarrow 4 \sum\left(x_{1}-x_{2}\right)^{2} \geqslant \sum\left(x_{1}-x_{3}\right)^{2} \text {. }
$$

By the Cauchy-Schwarz inequality, we always have $(1+1)\left[\left(x_{i}-x_{i+1}\right)^{2}+\left(x_{i+1}-x_{i+2}\right)^{2}\right] \geqslant\left(x_{i}-x_{i+2}\right)^{2}$. Taking $i=1,2,3,4,5$ and summing up, we get the desired result. Therefore, $f_{\text {min }}=\frac{5}{3}$.