Given a positive integer $n(n \geqslant 3)$. Try to find the largest constant $\lambda(n)$, such that for any $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}_{+}$, we have
$$
\prod_{i=1}^{n}\left(a_{i}^{2}+n-1\right) \geqslant \lambda(n)\left(\sum_{i=1}^{n} a_{i}\right)^{2} .
$$
Prove that if $a_{1}=a_{2}=\cdots=a_{n}=1$, then $\lambda(n) \leqslant n^{n-2}$.
It is easy to prove the following lemma by mathematical induction.
Lemma Let $n \in \mathbf{Z}_{+}, b_{1}, b_{2}, \cdots, b_{n} \in(-1$, $+\infty)$, and these $n$ numbers have the same sign. Then
$$
\prod_{i=1}^{n}\left(1+b_{i}\right) \geqslant 1+\sum_{i=1}^{n} b_{i} \text {. }
$$

Returning to the original problem.
Without loss of generality, assume
$$
\begin{array}{l}
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{k} \leqslant 1<a_{k+1} \leqslant \cdots \leqslant a_{n} . \\
\text { Then } \frac{a_{1}^{2}-1}{n}, \frac{a_{2}^{2}-1}{n}, \cdots, \frac{a_{k}^{2}-1}{n} \in(-1,0], \\
\frac{a_{k+1}^{2}-1}{n}, \frac{a_{k+2}^{2}-1}{n}, \cdots, \frac{a_{n}^{2}-1}{n} \in(0,+\infty)
\end{array}
$$

By the lemma, we have
$$
\begin{array}{l}
\prod_{i=1}^{k} \frac{n-1+a_{i}^{2}}{n}=\prod_{i=1}^{k}\left(1+\frac{a_{i}^{2}-1}{n}\right) \\
\geqslant 1+\sum_{i=1}^{k} \frac{a_{i}^{2}-1}{n}=\frac{1}{n}\left(\sum_{i=1}^{k} a_{i}^{2}+n-k\right), \\
\prod_{i=k+1}^{n} \frac{n-1+a_{i}^{2}}{n}=\prod_{i=k+1}^{n}\left(1+\frac{a_{i}^{2}-1}{n}\right) \\
\geqslant 1+\sum_{i=k+1}^{n} \frac{a_{i}^{2}-1}{n}=\frac{1}{n}\left(k+\sum_{i=k+1}^{n} a_{i}^{2}\right) .
\end{array}
$$

Multiplying the above two inequalities, and using the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
\prod_{i=1}^{n} \frac{n-1+a_{i}^{2}}{n} \\
\geqslant \frac{1}{n^{2}}\left(\sum_{i=1}^{k} a_{i}^{2}+n-k\right)\left(k+\sum_{i=k+1}^{n} a_{i}^{2}\right) \\
\geqslant \frac{1}{n^{2}}\left(\sum_{i=1}^{n} a_{i}\right)^{2} \\
\Rightarrow \prod_{i=1}^{n}\left(a_{i}^{2}+n-1\right) \geqslant n^{n-2}\left(\sum_{i=1}^{n} a_{i}\right)^{2} .
\end{array}
$$

Therefore, the largest constant $\lambda(n)$ is $n^{n-2}$.