Timeline for Ratio of Sequences Sum Inequality
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Apr 5, 2016 at 15:48 | history | suggested | CommunityBot | CC BY-SA 3.0 | Should be $\sum a_i/b_i \ge \sum_i \left(a_i/\sum_j b_j\right) = \sum_i a_i / \sum_i b_i$. |
| Apr 5, 2016 at 15:18 | review | Suggested edits | |||
| S Apr 5, 2016 at 15:48 | |||||
| May 16, 2013 at 19:21 | comment | added | Dan Stahlke | Shouldn't that be $\ge$ rather than $\le$? | |
| Dec 15, 2011 at 16:24 | comment | added | Michael Biro | Yes, I have an instance of $\sum_i a_i/(\sum_i b_i - (n-1))$, but I haven't been able to show that it's tight. I'm going to edit the question to show this. | |
| Dec 15, 2011 at 16:24 | comment | added | Emil Jeřábek | Well, that’s tight if $\sum_ib_i< 2n-1$. If $\sum_ib_i$ is larger, it is more tricky. | |
| Dec 15, 2011 at 16:18 | comment | added | Emil Jeřábek | I think the tight bound is $\sum_ia_i/(\sum_ib_i-(n-1))$. | |
| Dec 15, 2011 at 16:08 | comment | added | Michael Biro | Thanks for the answer. I have that bound already, but I don't think it's tight. One of the bounds is that $b_i \geq 1$, so we can't have $b_i = \epsilon$ and $\epsilon$ close to $0$. | |
| Dec 15, 2011 at 15:55 | history | answered | Anthony Quas | CC BY-SA 3.0 |