Here is a proof for the more general inequality $ \sum_{i=1}^n \frac{a_i}{1 + \sum_{j=1}^i a_j^2} \leq \sqrt{n}. $ where $a_1, \ldots, a_n$ range over $\mathbb{R}$. The method is a little cumbersome so I would be interested in a more direct proof.
I assume this problem comes from a math competition and would be interested to know its source.
Let $S_n$ denote the sum. Clearly we can assume that each $a_i \geq 0$, as otherwise we could increase $S_n$ by swapping the sign of one of the variables. Now make the change of variables $ a_i = t_i \sqrt{1 + a_1^2 + \cdots + a_{i-1}^2} $ for each $i$ so that $ \frac{a_i}{1 + \sum_{j=1}^{i-1} a_j^2 + a_i^2} = \frac{1}{\sqrt{1 + \sum_{j=1}^{i-1} a_j^2}} \frac{t_i}{1 + t_i^2} $ and furthermore $ \frac{1}{\sqrt{1 + \sum_{j=1}^{i-1} a_j^2}} = \prod_{j=1}^{i-1} \frac{1}{\sqrt{1 + t_j^2}}. $ Now we have $ S_n = \sum_{i=1}^n \frac{t_i}{1 + t_i^2} \prod_{j=1}^{i-1} \frac{1}{\sqrt{1 + t_j^2}}. $ Grouping terms, we get $ S_n = \frac{t_1}{1 + t_1^2} + \frac{1}{\sqrt{1 + t_1^2}} \left( \frac{t_2}{1 + t_2^2} + \frac{1}{\sqrt{1 + t_2^2}} \left( \cdots \right)\right) $ We therefore define $f : [n] \to \mathbb{R}^+$ inductively by $ f(n) = \max_{t_n \geq 0} \frac{t_n}{1 + t_n^2} $ and $ f(k) = \max_{t_k \geq 0} \frac{t_k}{1 + t_k^2} + f(k+1) \frac{1}{\sqrt{1 + t_k^2}}. $ Then we have $S_n \leq f(1)$ (and in fact, $f(1) = \max_{t_1, \ldots, t_n} S_n$).
Note that if $M$ is an upper bound for $f(k+1)$, then $ \max_{t_k \geq 0} \frac{t_k}{1 + t_k^2} + M \frac{1}{\sqrt{1 + t_k^2}} $ is an upper bound for $f(k)$. It therefore suffices to show that $ f(k) \leq \sqrt{n+1-k} $ inductively in $k$, starting at $n$.
We have the base case $ f(n) = \max_{t_n \geq 0} \frac{t_n}{1 + t_n^2} \leq \frac{1}{2}. $ It therefore suffices for us to show $ \frac{t}{1 + t^2} + \sqrt{m} \frac{1}{\sqrt{1 + t^2}} \leq \sqrt{m+1} $ for every $m \geq 1$ uniformly in $t \geq 0$. With the trivial bound $ \frac{t}{1 + t^2} \leq \frac{t}{\sqrt{1 + t^2}} $ it suffices to show $ \frac{t + \sqrt{m}}{\sqrt{1 + t^2}} \leq \sqrt{m+1}. $ Squaring, it suffices to show $ \frac{t^2 + 2 t \sqrt{m} + m}{1 + t^2} \leq m + 1 $ Or, equivalently, $ t^2 + 2 t \sqrt{m} + m \leq m + 1 + (m + 1) t^2 $ or $ 2 t \sqrt{m} \leq 1 + m t^2. $ But this is the AM-GM inequality for the pair $(1, mt^2)$, and the proof is complete.
Note that the method of proof showed that the inequality is not sharp for any $n$.