Problem 1464

Partition the set of indices $\{1,2, \ldots, n\}$ into disjoint subsets $M_{1}, M_{2}, \ldots, M_{k}$ so that $a_{\ell} \in\left[2^{j-1}, 2^{j}\right]$ for $\ell \in M_{j}$. Then, if $\left|M_{j}\right|=: p_{j}$, we have
$$\sum_{\ell \in M_{j}} \frac{a_{\ell}}{\sqrt{a_{1}^{2}+\ldots+a_{\ell}^{2}}} \leqslant \sum_{i=1}^{p_{j}} \frac{2^{j}}{2^{j-1} \sqrt{i}}=2 \sum_{i=1}^{p_{j}} \frac{1}{\sqrt{i}}$$
where we used that $a_{\ell} \leqslant 2^{j}$ and in the denominator every index from $M_{j}$ contributes at least $\left(2^{j-1}\right)^{2}$. Now, using $\sqrt{i}-\sqrt{i-1}=\frac{1}{\sqrt{i}+\sqrt{i-1}} \geqslant \frac{1}{2 \sqrt{i}}$, we deduce that
$$\sum_{\ell \in M_{j}} \frac{a_{\ell}}{\sqrt{a_{1}^{2}+\ldots+a_{\ell}^{2}}} \leqslant 2 \sum_{i=1}^{p_{j}} \frac{1}{\sqrt{i}} \leqslant 2 \sum_{i=1}^{p_{j}} 2(\sqrt{i}-\sqrt{i-1})=4 \sqrt{p_{j}}$$
Therefore, summing over $j=1, \ldots, k$ and using the QM-AM inequality, we obtain
$$\sum_{\ell=1}^{n} \frac{a_{\ell}}{\sqrt{a_{1}^{2}+\ldots+a_{\ell}^{2}}} \leqslant 4 \sum_{j=1}^{k} \sqrt{\left|M_{j}\right|} \leqslant 4 \sqrt{k \sum_{j=1}^{k}\left|M_{j}\right|}=4 \sqrt{k n}$$
Comment. Consider the function $f\left(a_{1}, \ldots, a_{n}\right)=\sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{1}^{2}+\ldots+a_{i}^{2}}}$. One can see that rearranging the variables in increasing order can only increase the value of $f\left(a_{1}, \ldots, a_{n}\right)$. Indeed, if $a_{j}>a_{j+1}$ for some index $j$ then we have
$$f\left(a_{1}, \ldots, a_{j-1}, a_{j+1}, a_{j}, a_{j+2}, \ldots, a_{n}\right)-f\left(a_{1}, \ldots, a_{n}\right)=\frac{a}{S}+\frac{b}{\sqrt{S^{2}-a^{2}}}-\frac{b}{S}-\frac{a}{\sqrt{S^{2}-b^{2}}}$$
where $a=a_{j}, b=a_{j+1}$, and $S=\sqrt{a_{1}^{2}+\ldots+a_{j+1}^{2}}$. The positivity of the last expression above follows from
$$\frac{b}{\sqrt{S^{2}-a^{2}}}-\frac{b}{S}=\frac{a^{2} b}{S \sqrt{S^{2}-a^{2}} \cdot\left(S+\sqrt{S^{2}-a^{2}}\right)}>\frac{a b^{2}}{S \sqrt{S^{2}-b^{2}} \cdot\left(S+\sqrt{S^{2}-b^{2}}\right)}=\frac{a}{\sqrt{S^{2}-b^{2}}}-\frac{a}{S} .$$
Comment. If $k<n$, the example $a_{m}:=2^{k(m-1) / n}$ shows that the problem statement is sharp up to a multiplicative constant. For $k \geqslant n$ the trivial upper bound $n$ becomes sharp up to a multiplicative constant.