-3
$\begingroup$

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 $n \times \sqrt{\max_{1\leq k\leq n} s_k}$ ?

Improve this question
$\endgroup$
2
  • $\begingroup$ the bound you state is not given in terms of $S$, in what sense is it a bound "knowing the value of $S$"? $\endgroup$
    – Carlo Beenakker
    Commented Nov 24, 2021 at 12:09
  • 2
    $\begingroup$ You may use Cauchy-Schwarz inequality to find the best bound (attained for all $s_k = S/n$) However the question is not suitable for this site; try math.stackexchange.com instead $\endgroup$
    – Pietro Majer
    Commented Nov 24, 2021 at 13:46

1 Answer 1

Reset to default
2
$\begingroup$

Yes: by the generalized mean inequality (or, more specifically, by the AM--QM inequality), $\sqrt{nS}$ is an upper bound on $\sum_{k=1}^n\sqrt{s_k}$, which is better than $n\sqrt{\max_{1\le k\le n}s_k}$.

Improve this answer
$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged .