Three. (50 points) Find the largest positive real number $\lambda$ such that the inequality
$$
1+\sum_{k=1}^{n} \frac{1}{a_{k}} \geqslant \lambda \sum_{k=1}^{n}\left(1+\sum_{i=1}^{k} a_{i}^{2}\right)^{-\frac{1}{2}}
$$

holds for all $n \in \mathbf{N}_{+}$ and $a_{i} \in \mathbf{R}$.
Three, first prove a lemma.
Lemma $\frac{1}{a_{k}}+\frac{1}{\sqrt{1+\sum_{i=1}^{k-1} a_{i}^{2}}} \geqslant \frac{2 \sqrt{2}}{\sqrt{1+\sum_{i=1}^{k} a_{i}^{2}}}$.
Proof Let $1+\sum_{i=1}^{k-1} a_{i}^{2}=a_{k}^{2} t_{k}\left(t_{k}>0\right)$.
Then (1) $\Leftrightarrow 1+\frac{1}{\sqrt{t_{k}}} \geqslant \frac{2 \sqrt{2}}{\sqrt{1+t_{k}}}$.
Let $t_{k}=\tan ^{2} \theta_{k}\left(\theta_{k} \in\left(0, \frac{\pi}{2}\right)\right)$.
Then (2) $\Leftrightarrow \frac{1+\tan \theta_{k}}{\tan \theta_{k}} \geqslant 2 \sqrt{2} \cos \theta_{k}$
$\Leftrightarrow \sin \theta_{k}+\cos \theta_{k} \geqslant 2 \sqrt{2} \sin \theta_{k} \cdot \cos \theta_{k}$
$\Leftrightarrow 1+2 \sin \theta_{k} \cdot \cos \theta_{k} \geqslant 8\left(\sin \theta_{k} \cdot \cos \theta_{k}\right)^{2}$
$\Leftrightarrow\left(1-2 \sin \theta_{k} \cdot \cos \theta_{k}\right)\left(1+4 \sin \theta_{k} \cdot \cos \theta_{k}\right) \geqslant 0$.
Since $2 \sin \theta_{k} \cdot \cos \theta_{k}=\sin 2 \theta_{k} \leqslant 1$,
$1+4 \sin \theta_{k} \cdot \cos \theta_{k}>0$,
Therefore, (1) holds.
Back to the original problem.
By the lemma,
$1+\sum_{k=1}^{n} \frac{1}{a_{k}}$
$\geqslant(2 \sqrt{2}-1) \sum_{k=1}^{n} \frac{1}{\sqrt{1+\sum_{i=1}^{k} a_{i}^{2}}}+\frac{1}{\sqrt{1+\sum_{i=1}^{n} a_{i}^{2}}}$
$\geqslant(2 \sqrt{2}-1) \sum_{k=1}^{n}\left(1+\sum_{i=1}^{k} a_{i}^{2}\right)^{-\frac{1}{2}}$.

Let $a_{k}=2^{\frac{k-1}{2}}(k=1,2, \cdots, n)$.
When $n \rightarrow \infty$, we get $\lambda \leqslant 2 \sqrt{2}-1$.
In summary, $\lambda_{\max }=2 \sqrt{2}-1$.