Example 4 Given a positive integer $n$. Find the largest constant $\lambda$, such that for all positive real numbers $x_{1}, x_{2}, \cdots, x_{2 n}$ satisfying
$$
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+2\right)^{n} \geqslant \prod_{i=1}^{2 n} x_{i}
$$

we have
$$
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n} \geqslant \lambda \prod_{i=1}^{2 n} x_{i} \cdot
$$
【Analysis】When $x_{i}=2(1 \leqslant i \leqslant 2 n)$, the equality in formula (1) holds, at this time,
$$
\lambda \leqslant \frac{\frac{1}{2 n} \sum_{i=1}^{2 n}(2+1)^{n}}{2^{2 n}}=\frac{3^{n}}{2^{2 n}} \text {. }
$$

Below, we prove by contradiction: For all positive real numbers $x_{1}, x_{2}, \cdots, x_{2 n}$ that satisfy formula (1), we have
$$
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n} \geqslant \frac{3^{n}}{2^{2 n}} \prod_{i=1}^{2 n} x_{i} \text {. }
$$

Assume there exists a set of $x_{1}, x_{2}, \cdots, x_{2 n}$ such that
$$
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n} < \frac{3^{n}}{2^{2 n}} \prod_{i=1}^{2 n} x_{i} \text {. }
$$
For each $k(0 \leqslant k \leqslant n)$, using the above inequality, we have
$$
\begin{array}{l}
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{k} \geqslant \prod_{i=1}^{2 n}\left(x_{i}+1\right)^{\frac{k}{2 n}} \\
\geqslant \prod_{i=1}^{2 n} 3^{\frac{k}{2 n}}\left(\frac{x_{i}}{2}\right)^{\frac{k}{3 n}} \geqslant 3^{k} .
\end{array}
$$

For each $k$, by Chebyshev's inequality and formula (4), we get
$$
\begin{array}{l}
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n} \\
\geqslant\left[\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n-k}\right]\left[\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{k}\right] \\
\geqslant 3^{n-k} \times \frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{k} \\
\Rightarrow \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{k} \leqslant 3^{k-n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n} . \\
\text { Hence } \sum_{i=1}^{2 n}\left(x_{i}+2\right)^{n}=\sum_{i=1}^{2 n}\left(x_{i}+1+1\right)^{n} \\
=\sum_{i=1}^{2 n} \sum_{k=0}^{n} \mathrm{C}_{n}^{k}\left(x_{i}+1\right)^{k} \\
=\sum_{k=0}^{n}\left[\mathrm{C}_{n}^{k} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{k}\right] \\
\leqslant\left(\sum_{k=0}^{n} \mathrm{C}_{n}^{k} 3^{k-n}\right)\left[\sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n}\right] \\
=\frac{4^{n}}{3^{n}} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n} .
\end{array}
$$

Combining with formula (1), we have
$$
\begin{array}{l}
\frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+1\right)^{n} \geqslant \frac{3^{n}}{2^{2 n}} \times \frac{1}{2 n} \sum_{i=1}^{2 n}\left(x_{i}+2\right)^{n} \\
\geqslant \frac{3^{n}}{2^{2 n}} \prod_{i=1}^{2 n} x_{i} .
\end{array}
$$

This contradicts formula (3). Therefore, the assumption is false.
Thus, for all $x_{1}, x_{2}, \cdots, x_{2 n}$, formula (2) holds.
In summary, the required $\lambda_{\text {max }}=\frac{3^{n}}{2^{2 n}}$.