Timeline for Polynomial inequality $n^2\sum_{i=1}^na_i^3\geq\left(\sum_{i=1}^na_i\right)^3$
Current License: CC BY-SA 4.0
31 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2020 at 23:53 | comment | added | user6976 | @MaxAlekseyev: True. I have removed my suggestion. | |
| Apr 18, 2020 at 22:20 | comment | added | Fedor Petrov | @jcdornano yes: $6\sum a_i^3=(\sum a_i)^3-3(\sum a_i)(\sum a_i a_j)+3\sum a_i a_j a_k$. | |
| Apr 18, 2020 at 22:19 | answer | added | Conrad | timeline score: 0 | |
| Apr 18, 2020 at 22:02 | comment | added | Conrad | $111$ taken $9$ times and $-199$ gives a counterexample for $n=10$ with a positive sum of cubes (more generally $a+\frac{1}{10}$ taken $9$ times and $-9a+\frac{1}{10}$ for $a > \frac{3}{80}$ and close to it - $a=\frac{3}{80}+\frac{1}{800}$ and normalizing to integers gives the above | |
| Apr 18, 2020 at 20:36 | comment | added | jcdornano | Is it obvioussly true if we take $a_0+a_1....+a_n=0$? | |
| Apr 18, 2020 at 20:17 | comment | added | Michael Rozenberg | @Max Alekseyev Or Holder, or Jensen, or the Vasc's EV, or more and more... | |
| Apr 18, 2020 at 20:15 | comment | added | Max Alekseyev | For nonnegative arguments, the inequality in question follows from the power mean inequality. | |
| Apr 18, 2020 at 20:00 | comment | added | Michael Rozenberg | Thank you very much, Darij !!! | |
| Apr 18, 2020 at 19:52 | comment | added | user6976 | @darijgrinberg: This contradicts a comment from the OP, but it is true. | |
| Apr 18, 2020 at 19:35 | comment | added | Fedor Petrov | It could be much more harder than olympiad problem if it were true. | |
| Apr 18, 2020 at 19:29 | answer | added | Fedor Petrov | timeline score: 10 | |
| Apr 18, 2020 at 19:20 | answer | added | Per Alexandersson | timeline score: 2 | |
| Apr 18, 2020 at 19:11 | comment | added | darij grinberg | SageMath (actually just Python) gives $(-2, 1, 1, 1, 1, 1, 1, 1, 1)$ as a counterexample for $n=9$. Can you check? | |
| Apr 18, 2020 at 18:21 | comment | added | Fedor Petrov | With Rolle (applied $n-3$ times to $\prod (x-a_i)$) it gives something which is wrong for large $n$. But I think Rolle does not reduce an inequality to the equivalent one: the antiderivative of a polynomial with real roots only may fail to have real roots only. I suggested something less elegant and straightforward: is three variables are mutually distinct, we may vary them so that the difference LHS-RHS increases. It reduces the problem to the situation when $a_i$'s take only two different values. | |
| Apr 18, 2020 at 17:25 | comment | added | Michael Rozenberg | @Fedor Petrov I proved it for $3\leq n\leq 8$ by using of the $uvw$'s method and Rolle For $n\geq9$ it does not work, I think. Maybe I don't see something. | |
| Apr 18, 2020 at 17:21 | comment | added | Fedor Petrov | Did you try $pqr$-method? (I mean, fix all variables except $a_1,a_2,a_3$, also fix $\sum a_i$ and $\sum\limits_{1\leq i<j<k\leq n}a_ia_ja_k$.) | |
| Apr 18, 2020 at 17:00 | comment | added | Michael Rozenberg | @Mark Sapir When I was in high school, I was taught that nothing does not help, when we need to prove inequalities. | |
| Apr 18, 2020 at 16:55 | comment | added | user6976 | Now you can try taking derivative with repspect to $a_2$. When I was in high school, I was taught to prove inequalities that way. | |
| Apr 18, 2020 at 16:16 | comment | added | Michael Rozenberg | I checked that for $n=9$ and $a_2=...=a_9$ it's true. | |
| Apr 18, 2020 at 16:09 | comment | added | user6976 | The inequality is always true if all summands are equal (it becomes an equality). Did you try to take derivatives with respect to $a_i$ to show that the inequality becomes strict for different summands? (For some reason I think that the inequality is true for every $n$.) | |
| Apr 18, 2020 at 15:47 | comment | added | Michael Rozenberg | Than you, Mark! It's OK. For me more interesting, how we can approach to this task. For exponent $2$ it's obvious, of course. | |
| Apr 18, 2020 at 15:43 | comment | added | user6976 | Note that if you decrease all exponents by 1 (replace 3 by 2 and 2 by 1), it becomes a corollary of Cauchy-Schwartz without any restriction on $a_i$. | |
| Apr 18, 2020 at 15:12 | history | edited | YCor | CC BY-SA 4.0 |
edited title
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| Apr 18, 2020 at 15:07 | comment | added | YCor | Well, it's confusing, as it conveys some wrong information. I edited your post; of course you can write that your guess is that it fails for large $n$, but "it seems" suggested that you have a serious reason to believe so. | |
| Apr 18, 2020 at 15:05 | history | edited | YCor | CC BY-SA 4.0 |
Rephrased
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| Apr 18, 2020 at 14:16 | comment | added | Michael Rozenberg | @YCor I just think so because I solved during my live one problem or maybe two. Can I think so? You do not allow me? :) | |
| Apr 18, 2020 at 14:11 | comment | added | YCor | But in your post you say "it seems that it's wrong for a big value of $n$". What makes you believe that it's wrong for some large $n$? you wrote you have no counterexample for $n=9$. | |
| Apr 18, 2020 at 14:10 | comment | added | Michael Rozenberg | @YCor I have no a counterexample. See please better my post. | |
| Apr 18, 2020 at 14:09 | comment | added | YCor | Can you be more precise on "it seems that it's wrong for a big value of $n$"? you have an explicit counterexample? for some explicit $n$? | |
| Apr 18, 2020 at 14:03 | history | edited | Michael Rozenberg | CC BY-SA 4.0 |
added 88 characters in body
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| Apr 18, 2020 at 13:56 | history | asked | Michael Rozenberg | CC BY-SA 4.0 |