Three, (50 points) Find the smallest real number $\lambda$, such that for any positive integer $n$, we have $(n,[n \sqrt{5}])<\lambda \sqrt{n}$, where $(a, b)$ denotes the greatest common divisor of the positive integers $a$ and $b$, and $[x]$ denotes the greatest integer not exceeding the real number $x$.

将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Three, first prove: for any positive integer $n$, we have
$$
(n,[n \sqrt{5}])\sqrt[4]{20-\varepsilon} \sqrt{n} \text {. }
$$

Notice that, the second type of Pell's equation $x^{2}-5 y^{2}=-1$ has a positive integer solution $(x, y)=(2,1)$. Therefore, this equation has infinitely many positive integer solutions. Take any one of its positive integer solutions $(x, y)=(p, q)$, then
$$
\begin{array}{l}
p^{2}-5 q^{2}=-1 \Rightarrow p^{2}=5 q^{2}-1 \\
\Rightarrow(p q)^{2}=\left(5 q^{2}-1\right) q^{2} \\
\Rightarrow 5(2 p q)^{2}=\left(10 q^{2}-1\right)^{2}-1 . \\
\text { By }\left(10 q^{2}-2\right)^{2}(20-\varepsilon) 4\left(5 q^{2}-1\right) q^{2} \\
\Leftrightarrow q>\frac{2}{\sqrt{\varepsilon}} .
\end{array}
$$

At this point, let $n=2 p q$, then we have
$$
(n,[n \sqrt{5}])>\sqrt[4]{20-\varepsilon} \sqrt{n} \text {. }
$$

In summary, the smallest value of $\lambda$ that satisfies the condition is $\sqrt[4]{20}$.