Timeline for Reverse Inequality
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2020 at 22:27 | comment | added | მამუკა ჯიბლაძე | Using A220883 one may revert the series for $y(1-y)^\alpha$ where $y=1-x$, obtaining a version for $f$ as follows:$$f(\alpha,C)=1-C-\alpha C^2-...-\prod_{k=0}^{n-2}(k+n\alpha)\frac{C^n}{n!}-...$$ | |
| Nov 23, 2020 at 19:15 | comment | added | მამუკა ჯიბლაძე | In any case there is no $x$ with $x^\alpha(1-x)>C$ for $C>\frac{\alpha^\alpha}{(1+\alpha)^{1+\alpha}}$. | |
| Nov 23, 2020 at 19:10 | comment | added | მამუკა ჯიბლაძე | You are not able to formulate an explicit requirement on $f$? I mean, like $f$ must minimize something and maximize some other thing or something like that? | |
| Nov 20, 2020 at 13:17 | comment | added | Valentino | Unfortunately, it isn't, @IlyaBogdanov. I got a better result applying AM-GM to $x^{\alpha}(1-x)$, but not even that seems to be good enough. | |
| Nov 20, 2020 at 9:26 | comment | added | Ilya Bogdanov | Isn't $f(\alpha,C)=1-C$ enough for you? | |
| Nov 20, 2020 at 5:15 | review | Close votes | |||
| Nov 29, 2020 at 13:05 | |||||
| Nov 19, 2020 at 22:14 | history | asked | Valentino | CC BY-SA 4.0 |