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.