Three. (50 points) Let $x_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n)$, and $x_{1} x_{2} \cdots x_{n}=1, n$ be a given positive integer. Try to find the smallest positive number $\lambda$, such that the inequality $\frac{1}{\sqrt{1+2 x_{1}}}+\frac{1}{\sqrt{1+2 x_{2}}}+\cdots+\frac{1}{\sqrt{1+2 x_{n}}} \leqslant \lambda$ always holds.
Let $2 x_{1}=\tan ^{2} \theta_{1}, 2 x_{2}=\tan ^{2} \theta_{2}, \cdots, 2 x_{n}=\tan ^{2} \theta_{n}$, where $\theta_{i} \in(0, \pi / 2)$, $(i=1,2, \cdots, n)$, then multiplying these $n$ equations gives $\tan ^{2} \theta_{1} \tan ^{2} \theta_{2} \cdots \tan ^{2} \theta_{n}=2^{n}$.
The original inequality $\Leftrightarrow \cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n} \leqslant \lambda \Leftrightarrow \lambda \geqslant \max \left\{\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n}\right\}$.
When $n=1$, from $\tan ^{2} \theta_{1}=2$ we get $\cos \theta_{1}=\frac{\sqrt{3}}{3}$; when $n=2$, $\tan \theta_{1} \tan \theta_{2}=2 \Leftrightarrow$
$$
\begin{array}{l}
\sin ^{2} \theta_{1} \sin ^{2} \theta_{2}=4 \cos ^{2} \theta_{1} \cos ^{2} \theta_{2} \Leftrightarrow \cos ^{2} \theta_{1}+\cos ^{2} \theta_{2}=1-3 \cos ^{2} \theta_{1} \cos ^{2} \theta_{2} \Leftrightarrow \\
\left(\cos \theta_{1}+\cos \theta_{2}\right)^{2}=-3 \cos ^{2} \theta_{1} \cos ^{2} \theta_{2}+2 \cos \theta_{1} \cos \theta_{2}+1,
\end{array}
$$

When $\cos \theta_{1} \cos \theta_{2}=\frac{1}{3}$, $\left(\cos \theta_{1}+\cos \theta_{2}\right)_{\text {max }}=\frac{2 \sqrt{3}}{3}$.
When $n \geqslant 3$, we now prove $\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n}7$, then $\tan ^{2} \theta_{1} \geqslant \tan ^{2} \theta_{2}>$ $\frac{7}{2}, \therefore \cos \theta_{1} \leqslant \cos \theta_{2}$ $\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n-1}=r$, contradiction! $\therefore n-1$ is the smallest upper bound of the left side of inequality (1).
When $n \geqslant 3$, the smallest positive number $\lambda=n-1$, in summary, the smallest positive number $\lambda=\left\{\begin{array}{l}\frac{\sqrt{3} n}{3}(n=1,2) \\ n-1(n \geqslant 3)\end{array}\right.$.