I've been trying to prove (maybe even disprove) the following inequality:
$$ 
\sum\_{n=1}^{N} \frac{a\_n}{\sqrt{\sum\_{i=1}^{n}a\_i}} \leq C \sqrt{\sum\_{n=1}^{N}a\_n}
$$
Where $ a\_1,...,a\_N\geq 0 $ are some non-negative numbers, and $C$ is an absolute constant.
Help will be much appreciated.



An inequality involving square roots and sums
For every $n\in\{1,\dotsc,N\}$, we have
$$2\sqrt{\sum\_{i\leq n} a\_i}-2\sqrt{\sum\_{i\leq n-1} a\_i}=\frac{2a\_n}{\sqrt{\sum\_{i\leq n} a\_i}+\sqrt{\sum\_{i\leq n-1} a\_i}}>\frac{a\_n}{\sqrt{\sum\_{i\leq n} a\_i}}.$$
Summing these up, we obtain the inequality with $C=2$. It is also straightforward to see that for $C<2$ the inequality fails, hence $C=2$ is the optimal constant.


