We'll prove that $\lambda = (3/4)^n$. Taking $x_i=2$ for all $i$ we have $\lambda\le (3/4)^n$, so if \[S=\left(\frac{1}{2n}\sum_{i=1}^{2n}(x_i+2)^n\right)^{1/n},\; S'=\left(\frac{1}{2n}\sum_{i=1}^{2n}(x_i+1)^n\right)^{1/n},\; P=\left(\prod_{i=1}^{2n}x_i\right)^{1/n},\]it remains to show that $S\ge P\implies S'\ge 3P/4$. Case 1: $P\ge4$. Then by Minkowski's inequality, \begin{align*} 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} \\ &\ge \left(\frac{1}{2n}\sum_{i=1}^{2n}(x_i+2)^n\right)^{1/n}=S\ge P\ge \frac{3}{4}P+1, \end{align*}as desired. Case 2: $0\le P\le4$. Then by AM-GM and Holder's inequality, \[ S' \ge \left(\prod_{i=1}^{2n}(x_i+1)\right)^{1/2n} \ge \left(\prod_{i=1}^{2n}x_i\right)^{1/2n} + \left(\prod_{i=1}^{2n}1\right)^{1/2n} = \sqrt{P}+1. \]But \[0\le \sqrt{P}\le 2\implies 1+\sqrt{P}-\frac{3}{4}P=\frac{1}{4}(2-\sqrt{P})(2+3\sqrt{P})\ge0,\]so $S'\ge3P/4$ as desired.