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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
IMO Shortlist 2012, Geometry 3
lyukhson   75
N 31 minutes ago by numbertheory97
Source: IMO Shortlist 2012, Geometry 3
In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
75 replies
lyukhson
Jul 29, 2013
numbertheory97
31 minutes ago
Diophantine
TheUltimate123   31
N 44 minutes ago by SomeonecoolLovesMaths
Source: CJMO 2023/1 (https://aops.com/community/c594864h3031323p27271877)
Find all triples of positive integers \((a,b,p)\) with \(p\) prime and \[a^p+b^p=p!.\]
Proposed by IndoMathXdZ
31 replies
TheUltimate123
Mar 29, 2023
SomeonecoolLovesMaths
44 minutes ago
Cyclic ine
m4thbl3nd3r   1
N an hour ago by arqady
Let $a,b,c>0$ such that $a^2+b^2+c^2=3$. Prove that $$\sum \frac{a^2}{b}+abc \ge 4$$
1 reply
m4thbl3nd3r
5 hours ago
arqady
an hour ago
Non-homogenous Inequality
Adywastaken   7
N an hour ago by ehuseyinyigit
Source: NMTC 2024/7
$a, b, c\in \mathbb{R_{+}}$ such that $ab+bc+ca=3abc$. Show that $a^2b+b^2c+c^2a \ge 2(a+b+c)-3$. When will equality hold?
7 replies
Adywastaken
4 hours ago
ehuseyinyigit
an hour ago
An FE. Who woulda thunk it?
nikenissan   117
N 3 hours ago by EpicBird08
Source: 2021 USAJMO Problem 1
Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]
117 replies
nikenissan
Apr 15, 2021
EpicBird08
3 hours ago
9 Will I make JMO?
EaZ_Shadow   18
N 4 hours ago by mathkidAP
will I be able to make it... will the cutoffs will be pre-2024
18 replies
EaZ_Shadow
Feb 7, 2025
mathkidAP
4 hours ago
Past USAMO Medals
sdpandit   1
N Today at 12:27 PM by CatCatHead
Does anyone know where to find lists of USAMO medalists from past years? I can find the 2025 list on their website, but they don't seem to keep lists from previous years and I can't find it anywhere else. Thanks!
1 reply
sdpandit
May 8, 2025
CatCatHead
Today at 12:27 PM
usamOOK geometry
KevinYang2.71   106
N Yesterday at 11:54 PM by jasperE3
Source: USAMO 2025/4, USAJMO 2025/5
Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$.
106 replies
KevinYang2.71
Mar 21, 2025
jasperE3
Yesterday at 11:54 PM
Geo #3 EQuals FReak out
Th3Numb3rThr33   106
N Yesterday at 10:56 PM by BS2012
Source: 2018 USAJMO #3
Let $ABCD$ be a quadrilateral inscribed in circle $\omega$ with $\overline{AC} \perp \overline{BD}$. Let $E$ and $F$ be the reflections of $D$ over lines $BA$ and $BC$, respectively, and let $P$ be the intersection of lines $BD$ and $EF$. Suppose that the circumcircle of $\triangle EPD$ meets $\omega$ at $D$ and $Q$, and the circumcircle of $\triangle FPD$ meets $\omega$ at $D$ and $R$. Show that $EQ = FR$.
106 replies
Th3Numb3rThr33
Apr 18, 2018
BS2012
Yesterday at 10:56 PM
Aime ll 2022 problem 5
Rook567   4
N Yesterday at 9:02 PM by Rook567
I don’t understand the solution. I got 220 as answer. Why does it insist, for example two primes must add to the third, when you can take 2,19,19 or 2,7,11 which for drawing purposes is equivalent to 1,1,2 and 2,7,9?
4 replies
Rook567
Thursday at 9:08 PM
Rook567
Yesterday at 9:02 PM
USAJMO problem 2: Side lengths of an acute triangle
BOGTRO   59
N Yesterday at 4:41 PM by ostriches88
Source: Also USAMO problem 1
Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_1, a_2, \hdots, a_n$ with $\text{max}(a_1,a_2,\hdots,a_n) \leq n \cdot \text{min}(a_1,a_2,\hdots,a_n)$, there exist three that are the side lengths of an acute triangle.
59 replies
BOGTRO
Apr 24, 2012
ostriches88
Yesterday at 4:41 PM
high tech FE as J1?!
imagien_bad   60
N Yesterday at 3:03 PM by SimplisticFormulas
Source: USAJMO 2025/1
Let $\mathbb Z$ be the set of integers, and let $f\colon \mathbb Z \to \mathbb Z$ be a function. Prove that there are infinitely many integers $c$ such that the function $g\colon \mathbb Z \to \mathbb Z$ defined by $g(x) = f(x) + cx$ is not bijective.
Note: A function $g\colon \mathbb Z \to \mathbb Z$ is bijective if for every integer $b$, there exists exactly one integer $a$ such that $g(a) = b$.
60 replies
imagien_bad
Mar 20, 2025
SimplisticFormulas
Yesterday at 3:03 PM
k Can I make the IMO team next year?
aopslover08   26
N Yesterday at 12:14 PM by steve4916
Hi everyone,

I am a current 11th grader living in Orange, Texas. I recently started doing competition math and I think I am pretty good at it. Recently I did a mock AMC8 and achieved a score of 21/25, which falls in the top 1% DHR. I also talked to my math teacher and she says I am an above average student.

Given my natural talent and the fact that I am willing to work ~3.5 hours a week studying competition math, do you think I will be able to make IMO next year? I am aware of the difficulty of this task but my mom says that I can achieve whatever I put my mind to, as long as I work hard.

Here is my plan for the next few months:

month 1-2: finish studying pre-algebra and learn geometry
month 3-4: learn pre-calculus
month 5-6: start doing IMO shortlist problems
month 7+: keep doing ISL/IMO problems.

Is this a feasible task? I am a girl btw.
26 replies
1 viewing
aopslover08
May 8, 2025
steve4916
Yesterday at 12:14 PM
Isosceles everywhere
reallyasian   28
N Yesterday at 12:02 PM by MATHS_ENTUSIAST
Source: 2020 AIME I #1
In $\triangle ABC$ with $AB=AC$, point $D$ lies strictly between $A$ and $C$ on side $\overline{AC}$, and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC$. The degree measure of $\angle ABC$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
28 replies
reallyasian
Mar 12, 2020
MATHS_ENTUSIAST
Yesterday at 12:02 PM
2020 EGMO P2: Sum inequality with permutations
alifenix-   28
N Apr 29, 2025 by math-olympiad-clown
Source: 2020 EGMO P2
Find all lists $(x_1, x_2, \ldots, x_{2020})$ of non-negative real numbers such that the following three conditions are all satisfied:

[list]
[*] $x_1 \le x_2 \le \ldots \le x_{2020}$;
[*] $x_{2020} \le x_1  + 1$;
[*] there is a permutation $(y_1, y_2, \ldots, y_{2020})$ of $(x_1, x_2, \ldots, x_{2020})$ such that $$\sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \sum_{i = 1}^{2020} x_i^3.$$[/list]

A permutation of a list is a list of the same length, with the same entries, but the entries are allowed to be in any order. For example, $(2, 1, 2)$ is a permutation of $(1, 2, 2)$, and they are both permutations of $(2, 2, 1)$. Note that any list is a permutation of itself.
28 replies
alifenix-
Apr 18, 2020
math-olympiad-clown
Apr 29, 2025
2020 EGMO P2: Sum inequality with permutations
G H J
G H BBookmark kLocked kLocked NReply
Source: 2020 EGMO P2
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alifenix-
1547 posts
#1 • 5 Y
Y by anser, itslumi, v4913, megarnie, Rounak_iitr
Find all lists $(x_1, x_2, \ldots, x_{2020})$ of non-negative real numbers such that the following three conditions are all satisfied:
  • $x_1 \le x_2 \le \ldots \le x_{2020}$;
  • $x_{2020} \le x_1  + 1$;
  • there is a permutation $(y_1, y_2, \ldots, y_{2020})$ of $(x_1, x_2, \ldots, x_{2020})$ such that $$\sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \sum_{i = 1}^{2020} x_i^3.$$

A permutation of a list is a list of the same length, with the same entries, but the entries are allowed to be in any order. For example, $(2, 1, 2)$ is a permutation of $(1, 2, 2)$, and they are both permutations of $(2, 2, 1)$. Note that any list is a permutation of itself.
This post has been edited 1 time. Last edited by alifenix-, Apr 18, 2020, 10:04 PM
Z K Y
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alifenix-
1547 posts
#2 • 6 Y
Y by wu2481632, karitoshi, v4913, Pluto04, Mango247, khina
By rearrangement, we need $$\sum_{i = 1}^{2020} ((x_i + 1)(x_{2021 - i} + 1))^2 \le 8 \sum_{i = 1}^{2020} x_i^3.$$
Consider the inequality $$(a + 1)^2(b + 1)^2 \le 4(a^3 + b^3),$$with $a \le b$. It can be shown that it is only true when $(a, b) = (0, 1), (1, 2)$; we do it using calculus. Evidently $b \in [a, a + 1]$, so $(a + 1)^2(b + 1)^2 - 4(a^3 + b^3)$, taking as a function of $b$, is minimized either when the derivative $$\frac{\partial}{\partial b} (a + 1)(b + 1)^2 - 4(a^3 + b^3) = 2(b + 1)(a + 1)^2 - 12b^2 = 0$$or at the endpoints of the interval. The derivative is a quadratic with roots at $$b = \frac{(a + 1)^2 \pm \sqrt{(a + 1)^4 + 24(a + 1)^2}}{12},$$and since $b \ge 0$, we must take the larger root. However, since we need a minimum, the second derivative at this root, $$\frac{\partial}{\partial b} 2(b + 1)(a + 1)^2 - 12b^2 = 2(a + 1)^2 - 24b,$$must be positive and thus $b < \frac{(a + 1)^2}{12}$. However $\frac{(a + 1)^2 + \sqrt{(a + 1)^4 + 24(a + 1)^2}}{12} > \frac{(a + 1)^2}{12}$, so there are no minimums inside the interval.

If $b = a$, then we get $$(a + 1)^2 (b + 1)^2 - 4(a^3 + b^3) = (a + 1)^4 - 8a^3 = (a - 1)^4 + 8a,$$which is always positive, contradiction. If $b = a + 1$, then we get $$(a + 1)^2(a + 2)^2 - 4(a^3 + (a + 1)^3) = a^2(a - 1)^2,$$so as desired the only pairs $(a, b)$ where this is true are $(0, 1)$ and $(1, 2)$, and furthermore they are equality cases.

Thus we must have either $$x_1 = x_2 = \ldots = x_{1010} = 0, x_{1011} = x_{1012} = \ldots = x_{2020} = 1,$$or $$x_1 = x_2 = \ldots = x_{1010} = 1, x_{1011} = x_{1012} = \ldots = x_{2020} = 2.$$
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v_Enhance
6877 posts
#3 • 12 Y
Y by tapir1729, alifenix-, Inconsistent, karitoshi, Illuzion, itslumi, v4913, mijail, Pluto04, megarnie, HamstPan38825, khina
alifenix- wrote:
Consider the inequality $$(a + 1)^2(b + 1)^2 \le 4(a^3 + b^3),$$with $a \le b$. It can be shown that it is only true when $(a, b) = (0, 1), (1, 2)$; we do it using calculus.

Seems the following is cleaner (remembering that $|a-b| \le 1$ is given):
\[ a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)[(a-b)^2 + ab] \]\[ \le (a+b)(1+ab) \le \left( \frac{(a+b)+(1+ab)}{2} \right)^2 = \frac{(a+1)^2(b+1)^2}{4}. \]
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naman12
1358 posts
#4 • 2 Y
Y by Mango247, Mango247
v_Enhance wrote:
alifenix- wrote:
Consider the inequality $$(a + 1)^2(b + 1)^2 \le 4(a^3 + b^3),$$with $a \le b$. It can be shown that it is only true when $(a, b) = (0, 1), (1, 2)$; we do it using calculus.

Seems the following is cleaner:
\[ a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)[(a-b)^2 + ab] \]\[ \le (a+b)(1+ab) \le \left( \frac{(a+b)+(1+ab)}{2} \right)^2 = \frac{(a+1)^2(b+1)^2}{4}. \]

Oh yeah. Thanks. I forgot about that!
This post has been edited 1 time. Last edited by naman12, Apr 18, 2020, 10:53 PM
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v_Enhance
6877 posts
#5 • 3 Y
Y by v4913, HamstPan38825, asdf334
Isn't it given that $|a-b| \le 1$ for any $a$ and $b$ in the problem statement? (I suppose the quote I had didn't mention that :) )
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IndoMathXdZ
691 posts
#6 • 7 Y
Y by CyclicConcaveTriangle, karitoshi, GorgonMathDota, Sugiyem, itslumi, Pluto04, ILOVEMYFAMILY
Personally, i think this is a lot easier than 01, solved #01 much slower than this?
Since there exists a permutation $\{y_i\}$ that satisfies such condition. By rearrangement Inequality, we have
\[ \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021 - i} + 1))^2 \le 8 \sum_{i = 1}^{2020} x_i^3 \]$\textbf{Main Claim.}$ For any nonnegative reals $x \ge y$ such that $x - y \le 1$, then
\[ 4(x^3 + y^3) \le ((x+1)(y+1))^2 \]$\textit{Proof.}$ Notice that
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x+y)((x-y)^2 + xy) \le (x+y)(xy + 1) \le \frac{((x+1)(y+1))^2}{4} \]Equality holds if and only if $xy + 1 = x + y$ and $x = y + 1$, which gives us $(2,1)$ and $(1,0)$.
Therefore, applying this to the original statement, we have that $x_i \{ 0, 1, 2 \}$ for each $i \in [1,2020]$.
Now, suppose that $x_1 = 0$, this gives us $x_{2020} = 1$, forcing all the others $(x_i, x_{2020 - i})$ to be $(0,1)$ as well, giving $(0,0,0,\dots,0,1,1,\dots,1)$ as a solution.
Suppose that $x_1 = 1$, this gives $x_{2020} = 2$, forcing all the others $(x_i, x_{2020 - i})$ to be $(1,2)$ as well, giving $(1,1,1,\dots,1,2,2,\dots,2)$ as a solution.

Therefore, the solutions are
\[ x_1 = x_2 = \dots = x_{1010} = 1 \ \text{and} \ x_{1011} = x_{1012} = \dots = x_{2020} = 2 \]and
\[ x_1 = x_2 = \dots = x_{1010} = 0 \ \text{and} \ x_{1011} = x_{1012} = \dots = x_{2020} = 1 \]
Remark. At first i tried to use the fact $|x - y| \le 1$ to prove $(x-1)^4 + (y - 1)^4 + 8(x+y) \ge (x+y+2)^2$, but when i square the initial condition (lol) and look back at the identity $x^3 + y^3$, solved this really fast accidentally.
This post has been edited 2 times. Last edited by IndoMathXdZ, Apr 19, 2020, 3:47 AM
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naman12
1358 posts
#7 • 5 Y
Y by Inconsistent, vsamc, user1729, Aryan-23, ehuseyinyigit
Yeah. I didn't solve this very fast. Took me wayyyyy too much time (like > 30 minutes):
We have the following
Claim. For $0\leq x\leq y\leq x+1$, we have $4(x^3+y^3)\leq (x+1)^2(y+1)^2$. Furthermore, equality holds for $(x,y)=(0,1),(1,2)$.
Proof. We start with defining the function
\[f(y)=(x+1)^2(y+1)^2-4x^3-4y^3=-4y^3+(x+1)^2(y^2+2y+1)-4x^3\]Now, we expand to get
\[f(y)=-4y^3+(x+1)^2y^2+2(x+1)^2y+(x+1)^2-4x^3\]Now, we note that
\[f'(y)=-12y^2+2(x+1)^2y+2(x+1)^2\]We note that
\[f'(x)=-12x^2+2(x+1)^2x+2(x+1)^2=2x^3-6x^2+6x+2=2(x-1)^3+4\]and
\[f'(x+1)=-12(x+1)^2+2(x+1)^3+2(x+1)^2=2(x+1)^2(x-4)=2x^3-4x^2-14x-8\]Now, if $x\geq 4$, then $f(x)$ is decreasing, so it suffices to check $f(x)$ on its bounds. For $x<4$, we get that there is exactly one root in $[x,x+1]$ of $f'(x)$. We can check this is a maximum as
\[f'\left(\dfrac{(x+1)^2}{12}\right)=-12\left(\dfrac{(x+1)^2}{12}\right)^2+2(x+1)^2\left(\dfrac{(x+1)^2}{12}\right)+2(x+1)^2\]Simplifying the first two, we get
\[f'\left(\dfrac{(x+1)^2}{12}\right)=\dfrac{(x+1)^4}{12}+2(x+1)^2>0\]so thus we get that the root of $f'(x)$ is greater than $\dfrac{(x+1)^2}{12}$. Now, we get that if this root is $r$, it is a maximum if and only if $f''(r)>0$. However, we have
\[f''(r)=-24r+(x+1)^2>0\]as $r>\dfrac{(x+1)^2}{12}$. Thus, in both cases where $x\geq 4$ and $x<4$, we have that the minimum of $f(x)$ occur on the bounds. Now, the rest is easy to check. We get
\[f(x)=(x+1)^4-8x^3=(x-1)^4+8x>0\]as we can easily check that $(x+1)^4-(x-1)^4=8x^3+8x$. Thus, we also have
\[f(x+1)=(x+1)^2(x+2)^2-4x^3-4(x+1)^3=(x+1)^2(x^2)-4x^3=(x-1)^2x^2\geq 0\]with equality only for $x=0,1$. Thus, the claim has been proved. $\blacksquare$.

Now, we note by the Rearrangement Inequality, we get
\[4\sum_{i=1}^{2020}(x_i^3+x_{2021-i}^3)=8\sum_{i=1}^{2020}x_i^3=\sum_{i=1}^{2020}(x_i+1)^2(y_i+1)^2\geq\sum_{i=1}^{2020}(x_i+1)^2(x_{2021-i}+1)^2\]However, we have that applying our claim on $x_i,x_{2020-i}$ for all $1\leq i\leq 2020$, we get
\[4\sum_{i=1}^{2020}(x_i^3+x_{2021-i}^3\leq \sum_{i=1}^{2020}(x_i+1)^2(x_{2021-i}+1)^2\]with equality only for $x_1=x_2=\cdots=x_{1010}=0,1$ and $x_{1011}=x_{1012}=\cdots=x_{2020}$. Thus, we can define it as:
\[\boxed{x_k=\begin{cases}0,1&k=1\\x_1&2\leq k\leq 1010\\x_1+1&1011\leq k\leq 2020\end{cases}}\]I hate not being able to see like v_enhance.
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MarkBcc168
1595 posts
#8 • 1 Y
Y by Tintarn
Here is a cleaner way to prove the inequality.

The answer is $(0,0,\hdots,0,1,1,\hdots,1)$ and $(1,1,\hdots,1,2,2,\hdots,2)$ where each numbers appear $1010$ times. This is easily seen to work so we prove that these are all solutions. Consider the following.

Claim: For any reals $a,b$ such that $|a-b|\leq 1$, we have $(a+1)^2(b+1)^2\geq 4(a^3+b^3)$, with equality at $(0,1)$, $(1,2)$.

Proof: We first reduce this to one-var. Set $a=t+\delta$ and $b=t-\delta$ where $\delta\in (0,\tfrac 12]$. We have
$$\text{LHS} = ((t+1)^2-\delta^2)^2 \geq \left((t+1)^2-\tfrac{1}{4}\right)^2.$$Moreover,
$$\text{RHS} = 4((t+\delta)^3 + (t-\delta)^3)= 8t(t^2+3\delta^2) \leq 8t\left(t^2+\tfrac{3}{4}\right)$$Hence it suffices to prove that $\left((t+1)^2-\tfrac{1}{4}\right)^2\geq 8t^3+6t$, which could be easily expanded to $\left(t-\tfrac{1}{2}\right)^2\left(t-\tfrac{3}{2}\right)^2\geq 0$. The equality case can easily be reverse-engineered. $\blacksquare$

Applying the claim with $(x_i, y_i)$ and sum up, we get that
$$\sum_{i=1}^{2020}((x_i+1)(y_i+1))^2\geq 8\sum_{i=1}^{2020}x_i^3.$$The equality case occurs if and only if $(x_i, y_i) = (0,1), (1,2)$ for each $i$. It's easy to see that they must be $(0,1)$ or $(1,2)$ for all $i$. Moreover, each number appears in $x_1,x_2,\hdots,x_{2020}$ exactly $1010$ times. Hence there are two such lists.
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arqady
30244 posts
#9 • 1 Y
Y by Mango247
It's enough to prove that $(a+1)^2(b+1)^2\geq4(a^3+b^3),$ where $a$ and $b$ are non-negatives such that $(a-b)^2\le1.$
Indeed, let $a+b=2u$ and $ab=v^2$.
Thus, $$4u^2-4v^2\leq1$$and we need to probe that $$(v^2+2u+1)^2\geq4(8u^3-6uv^2)$$or
$$v^4+(28u+2)v^2+(2u+1)^2\geq32u^3,$$for which it's enough to prove that
$$v^4+(28u+2)v^2+(2u+1)^2\geq8u(4v^2+1)$$or
$$v^4-2(2u-1)v^2+(2u-1)^2\geq0$$or $$(v^2-2u+1)^2\geq0$$and from here we can get the equality cases occurring.
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Th3Numb3rThr33
1247 posts
#10
Y by
Solved with TheUltimate123 and eisirrational.

The answer is $(\underbrace{0,\dots,0}_{1010 \ \text{0's}}, \underbrace{1,\dots,1}_{1010 \ \text{1's}})$ and $(\underbrace{1,\dots,1}_{1010 \ \text{1's}}, \underbrace{2,\dots,2}_{1010 \ \text{2's}})$. These can be easily checked to work, so we now show uniqueness.

By the Rearrangement Inequality, we have that
\[8 \sum_{i = 1}^{2020} x_i^3 = \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 \geq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2.\]
We, in fact, show the reverse inequality in the following claim.

Claim. For all reals $a \leq b \leq a+1$, we have $8(a^3+b^3) \leq 2(a+1)^2(b+1)^2$. Moreover, equality holds when $(a,b) = (0,1)$ or $(1,2)$.

Proof. Verify that
\[(a+1)^2(b+1)^2 - 4(a^3+b^3) = (a-1)^2(b-1)^2 + 4(a+b)(1-(a-b)^2) \geq 0\]as desired.

Finally, apply the claim to $(a,b) = (x_i, x_{2021-i})$ and sum the resulting inequalities to show that
\[8 \sum_{i = 1}^{2020} x_i^3 \leq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2.\]So equality must hold, as desired.
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pad
1671 posts
#11
Y by
Claim: If $0\le b-a\le 1$, then $(a+1)^2(b+1)^2 \ge 4(a^3+b^3)$, with equality iff $(a,b)=(0,1),(1,2)$.

Proof: We see that
\begin{align*}
    4(a^3+b^3) &= 4(a+b)(a^2-ab+b^2) = 4(a+b)[(b-a)^2 + ab] \\
    &\le 4(a+b)(1+ab)  = (2a+2b)(2+2ab) \le (a+b+1+ab)^2 \\
    &=(a+1)^2(b+1)^2. 
\end{align*}Equality is achieved when $b-a=1$ and $a+b=1+ab$ simultaneously, i.e. $(a,b)=(0,1),(1,2)$. $\square$


By the Rearrangement inequality, the minimum LHS is achieved when we pair the smallest elements with the largest available:
\[ \sum_{i=1}^{2020} ((x_i+1)(y_i+1))^2 \ge \sum_{i=1}^{2020} ((x_i+1)(x_{2021-i}+1))^2. \]Since all the $x_i$'s are within 1 of each other, we apply the claim to get
\[ ((x_i+1)(x_{2021-i}+1))^2 \ge 4(x_i^3+x_{2021-i}^3). \]Sum the above cyclically to conclude that the LHS is at least the RHS in the third condition. Equality is achieved when $(x_i,x_{2021-i}) = (0,1)\text{ or }(1,2)$ for all $i$. Note that we cannot have both combinations in a single working tuple, since all the $x_i$'s are within 1 of each other. Hence, the two equality cases are
\[ (x_1,\ldots,x_{2020}) = (0,\ldots,0,1,\ldots,1), \ (1,\ldots,1,2,\ldots,2), \]where each of the two tuples above has 1010 of each number.

Remarks
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GeronimoStilton
1521 posts
#12
Y by
The two lists satisfying these criteria are $(1,1,\dots, 1,2,2,\dots, 2)$ where there are $1010$ of each of $1$ and $2$ and the similarly defined $(0,0,\dots, 0,1,1,\dots, 1)$. It is easy to check that both of these work. Now, consider some satisfactory sequence.

By Rearrangement Inequality, we have
\[\sum_{i=1}^{2020} (x_i+1)^2(y_i+1)^2 \ge \sum_{i=1}^{2020} (x_i+1)^2(x_{2021-i}+1)^2 = 2\sum_{i=1}^{1010} (x_i+1)^2(x_{2021-i}+1)^2.\]Clearly, there exists some $i$ with
\[(x_i+1)^2(x_{2021-i}+1)^2 \le 4(x_i^3+x_{2021-i}^3).\]Let $x_i+1=a,x_{2021-i}+1=b$. Note that $b \le 1+a$. We have
\[a^2b^2\le 4(a^3-3a^2+3a-1+(b-1)^3).\]Observe that the function
\[f(b) = a^2b^2-(b-1)^3\]has
\[f’(b) = 2a^2b - 3(b-1)^2 \ge 2a^2b-3a^2 = a^2(2b-3),\]so $f$ certainly has nonnegative slope for $b \ge 3/2$. For $b \le 3/2$, we get $a^2b^2 \ge 9/4$ and $4(a-1)^3 + 4(b-1)^3 \le 1$, a contradiction.

Thus, we ought to check this inequality at just $b=a+1$ and $b=a$. At $b=a$ we get
\[a^4\le 8(a-1)^3 = (2a-2)^3.\]Note that the function $f(a) = a^4-(2a-2)^3$ is convex because $f''(a) = 12a^2 - 48(a-1) = 12(a^2-4a+4)$ is nonnegative. Thus, this inequality can never be satisfied because $1^4 > 8(1-1)^3$ and $f’(1) = 4\cdot 1^3-24(1-1)^2 > 0$. That is, we now only need to check $b=a+1$. Plug in $b=a+1$ to obtain
\[a^4+2a^3+a^2 = a^2(a+1)^2 \le 4(a-1)^3 + 4a^3 = 8a^3 - 12a^2+12a-4.\]Rearrange to get
\[0 \ge a^4-6a^3+13a^2-12a+4 = (a-2)(a^3-4a^2+5a-2) = (a-2)^2(a^2-2a+1) = (a-1)^2(a-2)^2.\]Note that this inequality is always at most an equality; this means that each $x_i+1$ is one of $1$ or $2$. But clearly the information about $b$ forces either all $x_i$ with $1\le i \le 1010$ to be $0$ or all $1$. This yields the two claimed solutions above.
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ftheftics
651 posts
#13 • 1 Y
Y by Gerninza
My solution is same as everyone done . BUT MY Solution to $4(x^3+y^3)\le (x+1)^2(y+1)^2)$ is bit different . Here it is -

$(x+1)^2(y+1)^2$


$=(xy+x+y+1)^2$

$\ge (xy+x+y +(x-y)^2)^2$ [cause $|x-y|\le 1$ ].

$=(x+y +(x^2+y^2-xy)^2$

$\ge (x+y)(x^2+y^2-xy).4$[AM-GM]

$=4(x^3+y^3)$ .


Obviously equality holds for $(x-y)^2=1$ and $x^2+y^2-xy =x+y$ .
This post has been edited 1 time. Last edited by ftheftics, May 13, 2020, 1:56 AM
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Aryan-23
558 posts
#14 • 4 Y
Y by AlastorMoody, Pluto04, kamatadu, Mango247
So troll :mad:

We claim that the only solutions are \[\boxed{x_k=\begin{cases}0,1&k=1\\x_1&2\leq k\leq 1010\\x_1+1&1011\leq k\leq 2020\end{cases}}\]
Invoking Rearrangement inequality, we have \[8 \sum_{i = 1}^{2020} x_i^3 = \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 \geq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2.\].

We now present a crucial claim .

$\mathbf{Lemma}$ Consider non-negative reals $a \leq b$ with $a-b \leq 1$ . We claim the following inequality holds
$$(a + 1)^2(b + 1)^2 \geq 4(a^3 + b^3)$$
with equality iff $(a,b)=(1,0) \text {  or} (2,1)$

$\mathbf{Proof :}$ We have
$$\left((a+1)(b+1) \right)^2 = \left((a+b) +(ab+1)\right)^2 \geq 4(a+b)(ab+1) \geq 4(a+b)(ab+ (a-b)^2) = 4(a+b)(a^2+ab+b^2) = 4(a^3+b^3) $$Equality occurs when $a=b+1$ and $a+b=ab+1$ .

Now we are ready to finish .
Note that , from the claim we have,
$$  \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2 \geq 8\sum_{i = 1}^{2020} x_i^3 $$.

So we have the forementioned solutions as desired as equality must hold in all the estimates we used .
$\blacksquare$ .
This post has been edited 2 times. Last edited by Aryan-23, Jul 5, 2020, 7:24 PM
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Sohil_Doshi
141 posts
#15
Y by
Solution
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IvoBucata
46 posts
#16
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We know that $(x_1+1)^2\leq (x_2+1)^2 \leq \cdots \leq (x_{2020}+1)^2 $ . Now Rearrangement inequality gives us $$\sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 \geq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2020-i} + 1))^2$$Now I'll prove that for each $1\leq i \leq 1010$ e have $$((x_i + 1)(x_{2020-i} + 1))^2+((x_{2020-i} + 1)(x_i + 1))^2 \geq 8(x_i^3+x_{2020-i}^3) (*)$$and when summing all of these up we would get $$\sum_{i = 1}^{2020} ((x_i + 1)(x_{2020-i} + 1))^2 \geq 8 \sum_{i = 1}^{2020} x_i^3 $$so the only solutions will be the equality cases in $(*)$.

Now let's for brevity have $x_i=x$ and $x_{2020-i}=y$ and we also know that $x_{2020-i}-x_i=y-x\leq 1$. Now let $y-x=k$, so we know that $0\leq k\leq 1$. Now $(*)$ transforms into proving $$(x+1)^2(x+1+k)^2 \geq 4(x^3+(x+k)^3) \Leftrightarrow $$$$(x+1)^4+2(x+1)^3k+(x+1)^2k^2 \geq 4(x^3+x^3+3x^2k+3xk^2+k^3) \Leftrightarrow $$$$x^4+4x^3+6x^2+4x+1+2x^3k+6x^2k+6xk+2k+x^2k^2+2xk^2+k^2\geq $$$$\geq 8x^3+12x^2k+12xk^2=4k^3 \Leftrightarrow $$$$x^4+2x^3(k-2)+x^2(6-6k+k^2)+x(4+6k-10k^2)+(k^2+2k+1-4k^3)\geq 0$$$$x^2(x^2+2x(k-2)+(k-2)^2)+x^2(6-6k+k^2-(k-2)^2)+x(4+6k-10k^2)+(k^2+2k+1-4k^3)\geq 0$$$$x^2(x+(k-2))^2+2x^2(1-k)+x(k-1)(-10k-4)+(k-1)(-4k^2-3k-1)\geq 0$$Now we have the following inequalities (for $0\leq k\leq 1$)
$$x^2(x+(k-2))^2\geq 0$$$$2x(1-k)\geq 0$$$$x(k-1)(-10k-4)\geq 0$$$$(k-1)(-4k^2-3k-1)\geq 0$$There is an equality in the last one only when $k=1$, so we get that we should have $k=1$.Now $(*)$ transforms into $$x^2(x-1)^2\geq 0$$and equality holds only when $x=0;1$.

We got that in $$((x_i + 1)(x_{2020-i} + 1))^2+((x_{2020-i} + 1)(x_i + 1))^2 \geq 8(x_i^3+x_{2020-i}^3)$$equality holds only when $x_i=1$ and $x_{2020-i} = 2$ or when $x_i=0$ and $x_{2020-i}=1$. Together with the constraints in the statement we get that the only solutions are $$x_1=x_2=\cdots =x_{1010}=0 ; x_{1011}=\cdots = x_{2020}=1 $$$$x_1=x_2=\cdots =x_{1010}=1 ; x_{1011}=\cdots = x_{2020}=2 $$
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IAmTheHazard
5001 posts
#17 • 1 Y
Y by centslordm
Yet another proof of the main inequality

The answer is $(0,\ldots,0,1,\ldots,1)$ and $(1,\ldots,1,2,\ldots,2)$, where both tuples contain exactly $1010$ ones. We can verify that these work by letting $(y_1,\ldots,y_{2020})$ be the reverse of $(x_1,\ldots,x_{2020})$.

The key inequality is the following bound:
$$\sum_{i=1}^{2020}(x_i+1)^2(x_{2020-i}+1)^2\geq 8\sum_{i=1}^{2020} x_i^3.$$It suffices to "break up" this summation and prove
$$(x_i+1)^2(x_{2020-i}+1)^2\geq 4x_i^3+4x_{2020-i}^3.$$Let $a=x_i$ and $b=x_{2020-i}$, so $a \leq b \leq a+1$. The inequality becomes $(a+1)^2(b+1)^2 \geq 4a^3+4b^3 \iff (a+1)^2(b+1)^2-4a^3-4b^3 \geq 0$. Fix $b$ and let $f(a)$ denote the value of the expression that we want to prove is nonnegative over the interval $I=[0,1] \cap [b-1,b]$.
We can compute $f'(a)=2(b+1)^2(a+1)-12a^2$. Since $f'(0)=2(b+1)^2>0$ and $f'(-\infty)=-\infty<0$, it follows that there is exactly one nonnegative root of $f'$ (Descartes' rule of signs works too!). Hence there is at most one turning point of $f$ in $I$, where $f$ must go from increasing to decreasing. As such, $f$ attains its minimum at an endpoint of $I$. We have
\begin{align*}
f(b)&=(b+1)^4-8b^3=b^4-4b^3+6b^2+4b+1=(b-1)^4+8b^3>0\\
f(b-1)&=b^2(b+1)^2-4b^3-4(b-1)^3=(b-2)^2(b-1)^2\geq 0\\
f(0)&=(b+1)^2-4b^3=(1-b)(4b^2+3b+1)\geq 0~\forall b \leq 1,\\
\end{align*}hence $f(a)$ is indeed nonnegative for $a \in I$, and is zero only if $(a,b)=(0,1)$ or $(a,b)=(1,2)$, by looking at the equality cases.
Then by rearrangement, for any permutation $(y_1,\ldots,y_{2020})$, we have
$$\sum_{i=1}^{2020}(x_i+1)^2(y_i+1)^2\geq \sum_{i=1}^{2020}(x_i+1)^2(x_{2020-i}+1)^2\geq 8\sum_{i=1}^{2020} x_i^3.$$It is clear that we can't have both of our equality cases hold at the same time (since that makes $x_1<x_{2020}+1$), so we extract the desired solution set. $\blacksquare$
This post has been edited 3 times. Last edited by IAmTheHazard, Jan 9, 2023, 3:38 PM
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HamstPan38825
8863 posts
#18 • 1 Y
Y by Mango247
The key is to use the following equaltiy:

Claim. $$(x+1)^2(y+1)^2 \geq 4(x^3+y^3)$$as long as $|x-y| \leq 1$ with equality at $(0, 1)$ and $(1, 2)$.

We can use the inequality $(x-y)^2 \leq 1$, and substitute $a = x+y, b = xy$, so that the given equation reduces to $$(a+b+1)^2 \geq 4a(b+1).$$This can just be done by considering $\Delta_a$ for the resulting quadratic in $a$, details omitted. $\blacksquare$

For the final part, we can use Rearrangement in the form $$\sum_{i=1}^{2020} (x_i+1)^2(y_i+1)^2 \geq \sum_{i=1}^{2020} (x_i+1)^2(x_{2021-i}+1)^2 \geq 8\sum_{i=1}^{1010} (x_i^3+x_{2021-i}^3).$$Thus, equality must hold everywhere, which implies that $(x_i, x_{2021-i}) = (0, 1)$ or $(1, 2)$. Obviously the solutions must all come from one of these combinations, so only $(0, 0, \cdots, 0, 1, 1, \cdots, 1)$ and $(1, 1, \cdots, 1, 2, 2, \cdots, 2)$, where there are 1010 of each number, work.
This post has been edited 1 time. Last edited by HamstPan38825, Jan 7, 2023, 4:03 AM
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shendrew7
795 posts
#19
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Our given condition can be rewritten as
\[\sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 4 \sum_{i = 1}^{2020} (x_i^3 + y_i^3).\]
Due to the order we are given, Rearrangement inequality implies
\[\sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2 \leq 4 \sum_{i = 1}^{2020} (x_i^3 + x_{2021-i}^3).\]
However, comparing each individual summand, we notice the inequality
\[\left((a+1)(b+1)\right)^2 \ge \left(2 \sqrt{(ab+1)(a+b)}\right)^2 = 4(a+b)(ab+1) \ge 4(a+b)(ab+(a-b)^2) = 4(a^3+b^3),\]
where $a = x_i$ and $b = x_{2021-i}$ for simplicity, and $(a-b)^2 \leq 1$ by our second condition. Hence we must have the equality case, or
\begin{align*}
(a-b)^2 = 1 &\implies a = b \pm 1 \\
ab+1 &= a+b,
\end{align*}
from which we get the pairs $(0, 1)$ and $(1, 2)$. Thus our possible sequences are
\[\boxed{a_1 = a_2 = \ldots = a_{1010} = 0, \quad a_{1011} = a_{1012} = \ldots = a_{2020} = 1}\]\[\boxed{a_1 = a_2 = \ldots = a_{1010} = 1, \quad a_{1011} = a_{1012} = \ldots = a_{2020} = 2}\]
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joshualiu315
2534 posts
#20
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The answer is $(\underbrace{0,\dots,0}_{1010 \ \text{0's}}, \underbrace{1,\dots,1}_{1010 \ \text{1's}})$ and $(\underbrace{1,\dots,1}_{1010 \ \text{1's}}, \underbrace{2,\dots,2}_{1010 \ \text{2's}})$. These can be easily checked to work, so now we prove that these are the only solutions

Claim: For reals $0 \le x \le y \le x+1$, we have $2(x+1)^2(y+1)^2 \ge 8(x^3+y^3)$, equality at $(x,y)=(0,1), (1,2)$.

Proof: Expansion gives

\begin{align*}
&(x+1)^2(y+1)^2-4(x^3+y^3) \\
= \ &(x-1)^2(y-1)^2+4(x+y)(xy+1)-4(x^3+y^3) \\
= \ &(x-1)^2(y-1)^2 + 4(x+y)(xy+1-(x^2+y^2-xy)) \\
= \ &(x-1)^2(y-1)^2 + 4(x+y)(1-(x-y)^2) \ge 0. \ \square
\end{align*}
Substituting in $(x,y)=(x_i, x_{2021-i})$ and summing, we get

\[\sum_{i=1}^{2020} ((x_i-1)(x_{2021-i}-1))^2 \ge 8 \sum_{i = 1}^{2020} x_i^3\]
However, by the Rearrangement Inequality, we have

\[\sum_{i=1}^{2020} ((x_i-1)(x_{2021-i}-1))^2 \le \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \sum_{i = 1}^{2020} x_i^3\]
Thus, we must have the equality case, meaning $(x_i, x_{2021-i})=(0,1), (1,2)$ for $1 \le 1 \le 1010$, giving us our two solutions.
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spectator01
60 posts
#21
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Note that
\[\sum_{i=1}^{2020}{(x_i+1)^2(x_{2021-i}+1)^2} \leq \sum_{i=1}^{2020}{(x_i+1)^2(y_i+1)^2}\]from rearrangement inequality. We claim that
\[8(x^3+y^3) \leq 2(x+1)^2(y+1)^2\]with equality case when they're both $1$, if $(x-y)^2\leq 1$. From, $(x-y)^2\leq 1$, we get
\[(x-y)^2\leq 1 \implies x^2+y^2 \leq 1+2xy \implies x^3+y^3 \leq (1+xy)(x+y)\]Note that
\[0\leq(x-1)^2(y-1)^2 \implies 4(1+xy)(x+y) \leq (x+1)^2(y+1)^2\]completing the proof. The equality case of such inequality is when one of them is $1$ and the other is $1$ away, giving $(0,1)$ and $(1,2)$. Thus, our only two sequences are $x_{i} = (0,1)$, $i \in \{1, 2, \cdots, 1010\}$, $x_{i} = (1,2)$, $i \in \{1011, 1012, \cdots, 2020\}$.
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OronSH
1745 posts
#22
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We first prove that if nonnegative $x,y$ satisfy $(x-y)^2\le 1,$ then $((x+1)(y+1))^2 \ge 4(x^3+y^3).$ To do this, let $x+y=s,xy=p$ and our condition is $s^2\le 4p+1.$ We wish to show that $s^2+p^2+14sp+2s+2p+1 \ge 4s^3.$ From our condition we may instead consider $s^2+p^2+14sp+2s+2p+1 \ge 16sp+4s,$ which implies the result. However this simplifies to $(p-s+1)^2 \ge 0,$ which is clear, and equality holds iff both $(x-y)^2=1$ and $(xy-x-y+1)^2=0,$ that is, either $x=1$ or $y=1.$ Thus equality holds when $(x,y)=(0,1),(1,0),(1,2),(2,1).$

Now by Rearrangement Inequality on our original expression, we see that $\sum_{i=1}^{2020}((x_i+1)(y_i+1))^2 \ge \sum_{i=1}^{2020}((x_i+1)(x_{2021-i}+1))^2.$ Now letting $x_i=x$ and $x_{2021-i}=y,$ since we have $|x_i-x_{2021-i}| \le |x_{2020}-x_1|\le 1,$ we see that $((x_i+1)(x_{2021-i}+1))^2 \ge 4(x_i^3+x_{2021-i}^3).$ Summing this across all $1 \le i \le 2020,$ we see that $\sum_{i=1}^{2020}((x_i+1)(x_{2021-i}+1))^2 \ge 8\sum_{i=1}^{2020} x_i^3,$ and equality holds iff each pair of $(x_i,x_{2021-i})$ is one of the four we found above. From here we notice that we cannot have both a $0$ and a $2,$ and since the sequence is increasing we get that the only possibilities are $x_1=x_2=\cdots=x_{1010}=0,x_{1011}=x_{1012}=\cdots=x_{2020}=1$ and $x_1=x_2=\cdots=x_{1010}=1,x_{1011}=x_{1012}=\cdots=x_{2020}=2.$
This post has been edited 1 time. Last edited by OronSH, Jan 13, 2024, 5:40 PM
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dolphinday
1324 posts
#23 • 1 Y
Y by ehuseyinyigit
We will first show that $4(x^3 + y^3) \leq ((x + 1)(y + 1))^2$, if $|x-y| \leq 1$.

\[(xy + x + y + 1)^2 = ((x + 1)(y + 1))^2\]\[= (xy + x + y + 1)^2 \geq (xy + x + y + (x - y)^2)^2 \]\[= (x^2 - xy + x + y + y^2)^2 \geq 4(x^2 - xy + y^2)(x + y) \]\[=  4(x^3 + y^3) \]\[\implies  4(x^3 + y^3) \leq ((x + 1)(y + 1))^2 \]

Then by Rearrangement Inequality,
$\newline$
\[\sum_{i = 1}^{2020} ((x_i + 1)(x_{2020-i} + 1))^2 \leq \sum_{i=1}^{2020} ((x_i + 1)(y_i + 1))^2 = \sum_{i = 1}^{2020} x_i^3 \]Which implies that this is the equality case of the Rearrangement Inequality, so the list $x_i$ either consists of $1010$ $0$'s and $1010$ $1$'s, or $1010$ $1$'s and $1010$ $2$'s.
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blueprimes
353 posts
#24
Y by
Feels more like an equality problem instead of manipulation. This took way too long, why didn't I try the obvious thing to do first :stretcher:

We will prove that the only solutions $(x_1, x_2, \dots, x_{2020})$ is the multiset containing $1010$ copies of $0$ and $1$, and the multiset containing $1010$ copies of $1$ and $2$, which are easily shown to work. We begin with the following claim:

$\textbf{Claim 1.}$ If $|a - b| \le 1$, then $(a + 1)^2 (b + 1)^2 \le 4(a^3 + b^3)$.
Proof. Let $a + b = 2u$, $ab = v^2$, we have
\[|a - b| \le 1 \iff (a + b)^2 - 4ab \le 1 \iff v^2 \ge \dfrac{4u^2 - 1}{4}\]and we wish to show
\[(a + 1)^2 (b + 1)^2 \ge 4(a^3 + b^3) \iff (ab + a + b + 1)^2 \ge 4(a + b)[(a + b)^2 - 3ab] \iff (2u + v^2 + 1)^2 \ge 8u(4u^2 - 3v^2). \]Now utilizing the earlier condition we have $(2u + v^2 + 1)^2 \ge \left( \dfrac{4u^2 + 8u + 3}{4} \right)^2$ and $8u \left( \dfrac{4u^2 + 3}{4}\right) \ge 8u(4u^2 - 3v^2)$ so it suffices to prove that
\[\left( \dfrac{4u^2 + 8u + 3}{4} \right)^2 \ge 8u \left( \dfrac{4u^2 + 3}{4}\right) \iff 16u^4 - 64u^3 + 88u^2 - 48u + 9 \ge 0 \iff (2u - 1)^2 (2u - 3)^2 \ge 0\]which is true. Equality cases occur at $u = 1/2, 3/2$, and reverse-engineering the original pairs we get $(a, b) = (0, 1), (1, 2)$ and permutations.

Now returning to the original problem, note that by Chebyshev (also extended Rearrangement) we obtain
$$8 \sum_{i = 1}^{2020} x_i^3 \ge \sum_{i = 1}^{2020} (x_i + 1)^2 (x_{2021 - i} + 1)^2$$but summing $\textbf{Claim 1.}$ over all $i$ on ordered pairs $(x_i, x_{2021 - i})$ yields the same inequality with an inverted sign. Hence, equality must hold, which easily generates the solution sets claimed earlier.
This post has been edited 2 times. Last edited by blueprimes, Aug 27, 2024, 11:45 AM
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Mathandski
757 posts
#25
Y by
What a missed solve ;-;
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mariairam
8 posts
#26 • 1 Y
Y by vi144
mostly similar to other solutions but instead of rearrangements we use substitutions
This post has been edited 1 time. Last edited by mariairam, Dec 16, 2024, 7:29 PM
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Ilikeminecraft
627 posts
#27
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I claim the inequality $((x + 1)(y + 1))^2 \leq 4 (x^3 + y^3)$. We have that $x^3 + y^3 = (x + y)((x - y)^2 + xy) \leq (x + y)(1 + xy) \leq\left(\frac{(x + y)^2 + (1 + xy)^2}{2}\right)^2 = \frac{(x + 1)^2(y + 1)^2}{4}.$ Equality holds if and only if $x - y = 1, x + y = 1 + xy.$ Hence, $(x, y) = (0, 1), (1, 2).$

Hence,
$$8 \sum_{i = 1}^{2020} x_i^3 = \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 \geq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021 - i} + 1))^2 \geq 8\sum_{i = 1}^{2020} x_i^3$$where the 1st inequality is by rearrangement, and 2nd is by our claim.

Hence, equality can only occur when they are: $(0, ..., 0, 1, ..., 1), (1, ..., 1, 2, ..., 2),$ where each one appears 1010 times.
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Maximilian113
575 posts
#28
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Observe that $|x_i-y_i| \leq 1$ we have $$4x_i^3+4y_i^3 = 4(x_i+y_i)((x_i-y_i)^2+x_iy_i) \leq 4(x_i+y_i)(1+x_iy_i) \leq ((1+x_i)(1+y_i))^2$$by AM-GM. Summing yields equality holds, so $|x_i-y_i| = 1$ and $x_i+y_i=1+x_iy_i \implies (x_i-1)(y_i-1)=0.$ So $(x_i, y_i) = (1, 0), (1, 2).$ Hence the solutions are the first $1010$ being $0$ and the rest being $1,$ or the first $1010$ being $1$ and the rest being $2.$
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math-olympiad-clown
30 posts
#29
Y by
We first observe that the rearrangement inequality tells us the left-hand side is minimized when \((y_1, \ldots, y_{2020}) = (x_{2020}, x_{2019}, \ldots, x_1)\), i.e., when the permutation is in reverse order.

Let \(a = x_i\), \(b = x_{2021 - i}\), so each term becomes\[((a+1)(b+1))^2.\]The corresponding right-hand side terms are
\[4(a^3 + b^3).\]So we want to show: \[(a+1)^2(b+1)^2 \ge 4(a^3 + b^3).\]
Note that
\[4(a^3 + b^3) = 4(a + b)(a^2 - ab + b^2) = 4(a + b)((a - b)^2 + ab).\]On the other hand,
\[(a+1)^2(b+1)^2 = (ab + a + b + 1)^2.\]So it suffices to prove: \[(ab + a + b + 1)^2 \ge 4(a + b)(ab + 1),\]and this can be checked immediately by AM-GM.

Thus, for the original identity to hold with equality, we must have a-b=1and a+b=1+ab ,this implies b=0 a=1 or b=1 a=2
and then we can assume there are m 0's and 2020-m 1's or n 1's and 2020-n 2's and easily calculate that m=1010.
Hence, the only valid sequences are those where exactly half of the values are 1 and half are 2 (or 0 and 1), arranged such that \(x_i + x_{2021 - i}\) is constant across all \(i\).
This post has been edited 2 times. Last edited by math-olympiad-clown, Apr 29, 2025, 2:51 PM
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