Let $n$ be a positive integer. Find the largest real number $\lambda$ such that for all positive real numbers $x_1,x_2,\cdots,x_{2n}$ satisfying the inequality
\[\frac{1}{2n}\sum_{i=1}^{2n}(x_i+2)^n\geq \prod_{i=1}^{2n} x_i,\]
the following inequality also holds
\[\frac{1}{2n}\sum_{i=1}^{2n}(x_i+1)^n\geq \lambda\prod_{i=1}^{2n} x_i.\]
1. **Identify the problem and the goal:**
   We need to find the largest real number \(\lambda\) such that for all positive real numbers \(x_1, x_2, \ldots, x_{2n}\) satisfying the inequality
   \[
   \frac{1}{2n}\sum_{i=1}^{2n}(x_i+2)^n \geq \prod_{i=1}^{2n} x_i,
   \]
   the following inequality also holds:
   \[
   \frac{1}{2n}\sum_{i=1}^{2n}(x_i+1)^n \geq \lambda \prod_{i=1}^{2n} x_i.
   \]

2. **Assume a candidate for \(\lambda\):**
   We will prove that \(\lambda = \left(\frac{3}{4}\right)^n\).

3. **Test the candidate with specific values:**
   Taking \(x_i = 2\) for all \(i\), we have:
   \[
   \frac{1}{2n}\sum_{i=1}^{2n}(2+2)^n = \frac{1}{2n}\sum_{i=1}^{2n}4^n = 4^n \geq 2^{2n} = \prod_{i=1}^{2n}2.
   \]
   This implies \(\lambda \leq \left(\frac{3}{4}\right)^n\).

4. **Define new variables for simplification:**
   Let
   \[
   S = \left(\frac{1}{2n}\sum_{i=1}^{2n}(x_i+2)^n\right)^{1/n}, \quad S' = \left(\frac{1}{2n}\sum_{i=1}^{2n}(x_i+1)^n\right)^{1/n}, \quad P = \left(\prod_{i=1}^{2n}x_i\right)^{1/n}.
   \]
   We need to show that \(S \geq P \implies S' \geq \frac{3}{4}P\).

5. **Case 1: \(P \geq 4\):**
   By Minkowski's inequality,
   \[
   S' + 1 = \left(\frac{1}{2n}\sum_{i=1}^{2n}(x_i+1)^n\right)^{1/n} + \left(\frac{1}{2n}\sum_{i=1}^{2n}1^n\right)^{1/n} \geq \left(\frac{1}{2n}\sum_{i=1}^{2n}(x_i+2)^n\right)^{1/n} = S \geq P.
   \]
   Since \(P \geq 4\), we have:
   \[
   P \geq \frac{3}{4}P + 1 \implies S' + 1 \geq \frac{3}{4}P + 1 \implies S' \geq \frac{3}{4}P.
   \]

6. **Case 2: \(0 \leq P \leq 4\):**
   By AM-GM and Hölder's inequality,
   \[
   S' \geq \left(\prod_{i=1}^{2n}(x_i+1)\right)^{1/2n} \geq \left(\prod_{i=1}^{2n}x_i\right)^{1/2n} + \left(\prod_{i=1}^{2n}1\right)^{1/2n} = \sqrt{P} + 1.
   \]
   Since \(0 \leq \sqrt{P} \leq 2\), we have:
   \[
   1 + \sqrt{P} - \frac{3}{4}P = \frac{1}{4}(2 - \sqrt{P})(2 + 3\sqrt{P}) \geq 0.
   \]
   Therefore,
   \[
   S' \geq \frac{3}{4}P.
   \]

Thus, we have shown that \(\lambda = \left(\frac{3}{4}\right)^n\) is the largest real number satisfying the given conditions.

The final answer is \( \boxed{ \left(\frac{3}{4}\right)^n } \).