Timeline for Conjecture: $a^n+b^n+c^n\ge x^n+y^n+z^n$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2022 at 11:21 | vote | accept | math110 | ||
| S Apr 17, 2019 at 3:04 | history | bounty ended | CommunityBot | ||
| S Apr 17, 2019 at 3:04 | history | notice removed | CommunityBot | ||
| S Apr 9, 2019 at 6:03 | history | suggested | user64494 | CC BY-SA 4.0 |
A typo is corrected.
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| Apr 9, 2019 at 4:42 | review | Suggested edits | |||
| S Apr 9, 2019 at 6:03 | |||||
| S Apr 9, 2019 at 1:11 | history | bounty started | math110 | ||
| S Apr 9, 2019 at 1:11 | history | notice added | math110 | Authoritative reference needed | |
| Apr 3, 2019 at 14:22 | answer | added | Ilya Bogdanov | timeline score: 7 | |
| Apr 3, 2019 at 14:22 | answer | added | user100927 | timeline score: 5 | |
| Apr 3, 2019 at 7:20 | review | Close votes | |||
| Apr 3, 2019 at 14:25 | |||||
| Apr 3, 2019 at 5:42 | comment | added | user35593 | Maybe I was wrong. It turns out that if you fix some $s, p$ then $ab+bc+ca$ is a monotonically increasing in the maximum $a$. Maybe one can prove somithing similar for $a^n+b^n+c^n$ or one can prove that it is monoton in $ab+bc+ca$. | |
| Apr 3, 2019 at 5:25 | comment | added | math110 | Now,I can't any Counterexample | |
| Apr 3, 2019 at 5:21 | comment | added | user35593 | I dont think your conjecture is true. $a^n+b^n+c^n$ can be written in terms of $s=a+b+c$, $p=abc$ and $ab+bc+ca=a(s-a)+p/a$. If you fix $s$ and $p$ you can find $a>x$ with $a(s-a)+p/a=x(s-x)+p/x$ and therefore if you define $b,c, y,z$ s.t. $a+b+c=x+y+z=s$ and $abc=xyz=p$ that $a^n+b^n+c^n=x^n+y^n+z^n$ for all $n$. Now if you perturb $x$ slightly you can make $x^n+y^n+z^n$ larger while $a>x$ is still true | |
| Apr 3, 2019 at 4:54 | history | edited | math110 | CC BY-SA 4.0 |
added 109 characters in body
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| Apr 3, 2019 at 4:16 | history | edited | Venkataramana | CC BY-SA 4.0 |
edited title
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| Apr 3, 2019 at 3:50 | history | asked | math110 | CC BY-SA 4.0 |