Knowing the value of $S=\sum\_{k=1}^n s\_k$ with $s\_k\geq 0$,
is it possible to obtain an upper bound on $\sum\_{k=1}^n\sqrt{s\_k}$ better than $n \times \sqrt{\max\_{1\leq k\leq n} s\_k}$ ?



Bounding sum of square roots in function of the sum value
Yes: by the [generalized mean inequality](https://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality) (or, more specifically, by the [AM--QM inequality](https://en.wikipedia.org/wiki/HM-GM-AM-QM_inequalities)), $\sqrt{nS}$ is an upper bound on $\sum\_{k=1}^n\sqrt{s\_k}$, which is better than
$n\sqrt{\max\_{1\le k\le n}s\_k}$.


