10、(20 points) Given $a_{1}, a_{2}, \cdots, a_{n} \in \mathbb{R}^{+}, a=\sum_{i=1}^{n} a_{i}(n \in \mathbb{N}, n \geq 2)$. Find the minimum value of $f\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\sum_{i=1}^{n} \frac{a_{i}}{3 a-a_{i}}$.
Notice
$$
\begin{aligned}
f\left(a_{1}, a_{2}, \cdots, a_{n}\right) & =\sum_{i=1}^{n} \frac{a_{i}}{3 a-a_{i}}=\sum_{i=1}^{n} \frac{a_{i}-3 a+3 a}{3 a-a_{i}}=\sum_{i=1}^{n}\left(-1+\frac{3 a}{3 a-a_{i}}\right) \\
& =-n+\sum_{i=1}^{n} \frac{3 a}{3 a-a_{i}}=-n+\sum_{i=1}^{n}\left(\frac{3 a-a_{i}}{3 a}\right)^{-1}=-n+\sum_{i=1}^{n}\left(1-\frac{a_{i}}{3 a}\right)^{-1}
\end{aligned}
$$

By Cauchy's inequality, we have $\left(\sum_{i=1}^{n}\left(1-\frac{a_{i}}{3 a}\right)\right) \sum_{i=1}^{n}\left(1-\frac{a_{i}}{3 a}\right)^{-1} \geq n^{2}$
(10 points)
Thus,
$$
f\left(a_{1}, a_{2}, \cdots, a_{n}\right) \geq-n+\frac{n^{2}}{\sum_{i=1}^{n}\left(1-\frac{a_{i}}{3 a}\right)}=-n+\frac{n^{2}}{n-\frac{\sum_{i=1}^{n} a_{i}}{3 a}}=-n+\frac{n^{2}}{n-\frac{1}{3}}=\frac{n}{3 n-1}
$$

Equality holds if and only if $a_{i}=\frac{n}{a}(i=1,2, \cdots, n)$.