```
Let $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, 5)$. Find
\[
\begin{array}{l} 
f= \frac{x_{1}+x_{3}}{x_{5}+2 x_{2}+3 x_{4}}+\frac{x_{2}+x_{4}}{x_{1}+2 x_{3}+3 x_{5}}+ \\
\frac{x_{3}+x_{5}}{x_{2}+2 x_{4}+3 x_{1}}+\frac{x_{4}+x_{1}}{x_{3}+2 x_{5}+3 x_{2}}+ \\
\frac{x_{5}+x_{2}}{x_{4}+2 x_{1}+3 x_{3}}
\end{array}
\]

the minimum value.
```
Let $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=1$, then $f(1,1,1,1,1)=\frac{5}{3}$.
The following proves: $f \geqslant \frac{5}{3}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
f=\sum_{i=1}^{5} \frac{x_{i}+x_{i+2}}{x_{i+4}+2 x_{i+6}+3 x_{i+8}} \\
\geqslant \frac{\left[\sum_{i=1}^{5}\left(x_{i}+x_{i+2}\right)\right]^{2}}{\sum_{i=1}^{5}\left(x_{i}+x_{i+2}\right)\left(x_{i+4}+2 x_{i+6}+3 x_{i+8}\right)}
\end{array}
$$
(understanding the subscripts modulo 5). Therefore, it suffices to prove
$$
\begin{array}{l}
\frac{\left[\sum_{i=1}^{5}\left(x_{i}+x_{i+2}\right)\right]^{2}}{\sum_{i=1}^{5}\left(x_{i}+x_{i+2}\right)\left(x_{i+4}+2 x_{i+6}+3 x_{i+8}\right)} \geqslant \frac{5}{3} \\
\Leftrightarrow 3 \sum_{i=1}^{5} x_{i}^{2}-4 \sum_{i=1}^{5} x_{i} x_{i+1} \geqslant-\sum_{i=1}^{5} x_{i} x_{i+2} \\
\Leftrightarrow 4 \sum_{i=1}^{5}\left(x_{i}-x_{i+1}\right)^{2} \geqslant \sum_{i=1}^{5}\left(x_{i}-x_{i+2}\right)^{2} . \text { (1) }
\end{array}
$$

Again by the Cauchy-Schwarz inequality, for $i=1,2, \cdots, 5$, we always have
$$
\begin{array}{l}
\left(1^{2}+1^{2}\right)\left[\left(x_{i}-x_{i+1}\right)^{2}+\left(x_{i+1}-x_{i+2}\right)^{2}\right] \\
\geqslant\left(x_{i}-x_{i+2}\right)^{2} .
\end{array}
$$

In the above inequality, taking $i=1,2, \cdots, 5$ and then summing up, we obtain inequality (1).
Therefore, $f_{\min }=\frac{5}{3}$.