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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Functional Equations Marathon March 2025
Levieee   0
a minute ago
1. before posting another problem please try your best to provide the solution to the previous solution because we don't want a backlog of many problems
2.one is welcome to send functional equations involving calculus (mainly basic real analysis type of proofs) as long it is of the form $\text{"find all functions:"}$
0 replies
+1 w
Levieee
a minute ago
0 replies
Another NT FE
nukelauncher   60
N 14 minutes ago by pi271828
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
60 replies
nukelauncher
Sep 22, 2020
pi271828
14 minutes ago
Perfect Squares, Infinite Integers and Integers
steven_zhang123   4
N 28 minutes ago by ohiorizzler1434
Source: China TST 2001 Quiz 5 P1
For which integer \( h \), are there infinitely many positive integers \( n \) such that \( \lfloor \sqrt{h^2 + 1} \cdot n \rfloor \) is a perfect square? (Here \( \lfloor x \rfloor \) denotes the integer part of the real number \( x \)?
4 replies
steven_zhang123
Yesterday at 12:06 PM
ohiorizzler1434
28 minutes ago
another geometry problem with sharky-devil point
anyone__42   11
N 30 minutes ago by zhenghua
Source: The francophone mathematical olympiads P1
Let $ABC$ be an acute triangle with $AC>AB$, Let $DEF$ be the intouch triangle with $D \in (BC)$,$E \in (AC)$,$F \in (AB)$,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$, and $X=(ABC)\cap (AEF)$.
Prove that $B,D,G$ and $X$ are concylic
11 replies
anyone__42
Jun 27, 2020
zhenghua
30 minutes ago
Line Perpendicular to Euler Line
tastymath75025   55
N an hour ago by ohiorizzler1434
Source: USA TSTST 2017 Problem 1, by Ray Li
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
55 replies
tastymath75025
Jun 29, 2017
ohiorizzler1434
an hour ago
Foot from vertex to Euler line
cjquines0   31
N an hour ago by pUssydestroyer777
Source: 2016 IMO Shortlist G5
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.
31 replies
cjquines0
Jul 19, 2017
pUssydestroyer777
an hour ago
Inequality => square
Rushil   12
N 2 hours ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
Rushil
Oct 7, 2005
ohiorizzler1434
2 hours ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 2 hours ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
2 hours ago
H not needed
dchenmathcounts   44
N 3 hours ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
3 hours ago
IZHO 2017 Functional equations
user01   51
N 3 hours ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
3 hours ago
chat gpt
fuv870   2
N 3 hours ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
3 hours ago
fuv870
3 hours ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 3 hours ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
3 hours ago
Waiting for a dm saying me again "old geometry"
drmzjoseph   0
4 hours ago
Source: Idk easy
Given $ABCD$ a tangencial quadrilateral that is not a rhombus, let $a,b,c,d$ be lengths of tangents from $A,B,C,D$ to the incircle of the quadrilateral which center is $I$. Let $M,N$ be the midpoints of $AC,BD$ resp. Prove that
\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]
0 replies
drmzjoseph
4 hours ago
0 replies
Finally hard NT on UKR MO from NT master
mshtand1   2
N 4 hours ago by IAmTheHazard
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 11.4
A pair of positive integer numbers \((a, b)\) is given. It turns out that for every positive integer number \(n\), for which the numbers \((n - a)(n + b)\) and \(n^2 - ab\) are positive, they have the same number of divisors. Is it necessarily true that \(a = b\)?

Proposed by Oleksii Masalitin
2 replies
mshtand1
Mar 13, 2025
IAmTheHazard
4 hours ago
Yet another classical inequality
Valentin Vornicu   43
N Dec 7, 2019 by sqing
Source: Balkan MO 2006, Problem 1
Let $ a,b,c>0$ be real numbers. Prove that
\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]
Greece

EDIT BY DARIJ:

Please use this topic to discuss the sources and history of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).
43 replies
Valentin Vornicu
Apr 29, 2006
sqing
Dec 7, 2019
Yet another classical inequality
G H J
G H BBookmark kLocked kLocked NReply
Source: Balkan MO 2006, Problem 1
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Valentin Vornicu
7301 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Let $ a,b,c>0$ be real numbers. Prove that
\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]
Greece

EDIT BY DARIJ:

Please use this topic to discuss the sources and history of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).
This post has been edited 1 time. Last edited by Valentin Vornicu, May 3, 2006, 3:18 PM
Z K Y
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silouan
3952 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
See this http://www.mathlinks.ro/Forum/viewtopic.php?t=161059

and I have no comments . How some people give a problem very-very-well-known in a big contest as BMO
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dule_00
246 posts
#3 • 3 Y
Y by Adventure10, Mango247, and 1 other user
yeah, like functional eqaution which was twice on BMO?
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mzdanowicz
2 posts
#4 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Now I know, where I saw this nice method...
What a pity, it's just the same problem ;/

[Moderator edit: The "nice method" referred to here is Mzdanovicz's solution in http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 .]
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Arne
3660 posts
#5 • 2 Y
Y by Adventure10 and 1 other user
I can't believe they give this problem at a BMO. It's just so incredibly well known :huh:
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Valentin Vornicu
7301 posts
#6 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Arne wrote:
I can't believe they give this problem at a BMO. It's just so incredibly well known :huh:
It was problem 1. I assume they knew it was kind of known but they put it there to be the easy problem in the competition ... :?
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Arne
3660 posts
#7 • 3 Y
Y by Adventure10, Mango247, and 1 other user
What makes it even worse is the fact that this is VERY hard for people who've never seen it before.

It's just unfair!
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Peter
3615 posts
#8 • 2 Y
Y by Adventure10 and 1 other user
That can be said of 30% of the IMC problems...
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Valentin Vornicu
7301 posts
#9 • 2 Y
Y by Adventure10 and 1 other user
Arne wrote:
What makes it even worse is the fact that this is VERY hard for people who've never seen it before.

It's just unfair!
Yeah, that can also be said about usage of ISL problems in TSTs around the world, given that some of the students might know some of the problems ahead ... :?

I belive we got very close to that point where almost all problems are known, and those which aren't are sent to the IMO :D
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pvthuan
844 posts
#10 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Arne wrote:
What makes it even worse is the fact that this is VERY hard for people who've never seen it before.

It's just unfair!

I can NOT believe that. Why don't they hire some inequality experts from Mathlinks? :lol: ;)
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Arne
3660 posts
#11 • 3 Y
Y by Adventure10, Mango247, and 1 other user
I guess all team leaders knew this problem. Why didn't they stop it from being selected? :(
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pvthuan
844 posts
#12 • 3 Y
Y by Adventure10, Mango247, and 1 other user
That can tell us something about Balkan Olympiad. In the IMO, this way is forbidden.
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Lovasz
868 posts
#13 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Is not this problem from Old and new inequalities?
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Arne
3660 posts
#14 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Might be, although the problem is quite old.

I think the BMO is making a joke of itself this way...
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esti
83 posts
#15 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Arne wrote:
Might be, although the problem is quite old.

I think the BMO is making a joke of itself this way...
I got feeling long time ago, that competition among only a few countries, like BMO ,has been a joke ever since.
However,reasons like making a friends and visiting beutiful parts of Mediteranian make it sufficient to keep it going every year. :)
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Centy
260 posts
#16 • 2 Y
Y by Adventure10 and 1 other user
I think when the same question in the BMO appeared twice in different years, eyebrows must have been raised.

This question appeared last year on the UK Correspondence Course and I assume it therefore a well publicised inequality.

Still I guess these regional Olympiads should also be about making friends and meeting new people. If it were just about the questions, then why isn't the whole thing done in two days?
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Arne
3660 posts
#17 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Of course, but I just can't believe the problem committee cannot find any better or new questions.
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Peter
3615 posts
#18 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Maybe... *starts writing the plot of a new detective novell*

this question just got proposed

all team leaders knew that their students knew the question

but none knew it was well-known to the other countries too

so they let is pass :diablo: :D
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Michael
99 posts
#19 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Valentin Vornicu wrote:
Let $a,b,c>0$ be real numbers. Prove that \[ \frac 1{a(1+b)} + \frac 1{b(1+c)} + \frac 1{c(1+a)} \geq \frac 3 { 1+abc}.  \]

This problem appeared some years ago in Crux Mathematicorum. It was proposed by Mohammed Aassila.
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Arne
3660 posts
#20 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Might be, it appeared like everywhere. I'm sure you can find at least 10 threads on this forum about this problem.
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stergiu
1648 posts
#21 • 3 Y
Y by Adventure10, Mango247, and 1 other user
I have this inequality in my book on Inequalities(written by Skombris and me, in Greek). As source we give there << Komal 1993 >> , so it was known before Aassila sent it to CRUX , vol 1998. This nice solution, sent by Silouan , is to be found there, too(by a chinese).Of course the solution was discused in ML very wide.

My opinion is that the problem should not be proposed in the BMO 2006 , although very good. problem. Competition problems must be original( this is very difficult) or clever and nice variations of nice problems.

Babis
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Tiks
1144 posts
#22 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Guys if you want to know,this problem even was proposed on Armenian National Olimpiad 1996 :lol: ,I think they have found most pupular ineq. in the world(my be after AM-GM :D ) and proposed it for BMO :oops: .
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Michael
99 posts
#23 • 2 Y
Y by Adventure10, Mango247
stergiu wrote:
I have this inequality in my book on Inequalities(written by Skombris and me, in Greek). As source we give there << Komal 1993 >> , so it was known before Aassila sent it to CRUX , vol 1998. This nice solution, sent by Silouan , is to be found there, too(by a chinese).Of course the solution was discused in ML very wide.

My opinion is that the problem should not be proposed in the BMO 2006 , although very good. problem. Competition problems must be original( this is very difficult) or clever and nice variations of nice problems.

Babis

Dear Babis Stergiou,

I have your book, and it unfortunately contains several imprecisions. The inequality appeared later on in Komal (but not in 1993). This inequality was first proposed by the Russian mathematician D. P. Mavlo in the Bulgarian journal Mathematica (number 4, 1987). A little later, it was published in Kvant (problem M1136, number 11-12, 1988). It was also a problem given at the training of the USA IMO team in 1993, and also was used in the training of the South African IMO team in 1994,....etc..etc
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ambros
301 posts
#24 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Guys i wish to see if this solution is correct and tell me as soon as possible please
thank you
Let $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq x$ where x is the best real .

Let $a(b+1)+b(c+1)+c(a+1)\geq y$ where y is the best contsnt .

By multipliing these we have and using B-C-S
$[\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}][a(b+1)+b(c+1)+c(a+1)]\geq xy\geq 9$

From Cauchy -AM-GM we have $a(b+1)+b(c+1)+c(a+1)\geq 6\sqrt{abc}$

so $y=6\sqrt{abc}$ and so $6x\sqrt{abc}\geq 9$ and so

$x\geq\frac{3}{2\sqrt{abc}}\geq\frac{3}{abc+1}$
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Albanian Eagle
1693 posts
#25 • 3 Y
Y by Adventure10, Mango247, and 1 other user
actually the problem appears also in the book "mathematical miniatures" by T.Andreescu

The rules were broken since the moment when the leaders allowed two problems from the same country!

(If you want more check the 4th problem)
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DuLac
54 posts
#26 • 2 Y
Y by Adventure10 and 1 other user
To: Albanian Eagle

The Leaders do not know the origin of the problems. This is a rule of IMO too. The proposing countries are revealed after the final selsction. In this BMO there was a real danger of having three problems proposed by the same country since the only Combinatorics problem was proposed by Greece! The Problem finally was eliminated with the intervention of the Selection Committee for exactly this reason. I got all this on the grapevine. I was not there!
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ambros
301 posts
#27 • 2 Y
Y by Adventure10 and 1 other user
Dear albanian eagle
1 congratulations
2 which problem is written in this book that you mention cause i have it ?
3 could you please see and think what i've written above and answer to me as soon as possible?
thank you
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Cezar Lupu
1906 posts
#28 • 2 Y
Y by Adventure10, Mango247
I'm quite sure that this inequality was proposed by Sotiris Louridas. What I do not understand is the fact that why inequalities are proposed by someone at BMO and JBMO and by someone at IMO??? :?
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DuLac
54 posts
#29 • 1 Y
Y by Adventure10
Wrong. The real author told me that he started from another known inequality and then transformed it by a to 1/a, b to 1/b and c to 1/c and arrived to this (known) inequality! The Problem Selection Committee and the Jury detected it after the Competition! (the problem was being used in training at MIT).
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stergiu
1648 posts
#30 • 2 Y
Y by Adventure10, Mango247
Michel ,

:) thank you for your nice information.You re better informed. I will use the history of the problem in the future.It is sure now that my source was not the older. I think , I have seen Bulgarian Olympiads 1987 and the name Mavlo is known to me.He must have proposed an ieq. problem in Bulgarian olympiad. Any way , probably - as Dulac said - the greek proposer created the same problem trying to create a new problem. This hapens often. If Louridas is the proposer(Lupu wrote this) , then I'm sure that he gave it as a new problem - in his opinion. He never likes - is a friend - to propose discussed proplems in competitions.
Babis
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DuLac
54 posts
#31 • 1 Y
Y by Adventure10
Louridas is definitely not the author of the problem! The perpetrator is well-known to the Greek Math Competitions scene.
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Peter
3615 posts
#32 • 1 Y
Y by Adventure10
DuLac wrote:
Source: Deep throat
lol @ hidden message in post #33 :D
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exalibur
17 posts
#33 • 2 Y
Y by Adventure10, Mango247
Also by eding 3/1+abc to both parts of the inequality and regrouping by AM-GM you can found the solution
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ambros
301 posts
#34 • 2 Y
Y by Adventure10, Mango247
HELLO GUYS HAS ANYBODY SEEN MY SOLUTION THAT I VE WRITTEN ABOVE? PLEASE SEE IF IT IS CORRECT AND SEN ME A PERSONAL MESSAGE OK? THANK YOU FOR YOUR UNDERSTANDING
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Peter
3615 posts
#35 • 2 Y
Y by Adventure10, Mango247
Please do not shout.

Your proves looks not entirely rigorous to me, although it might be easy to fix (no time to investigate).

You define y as the best constant, and you conclude $y=y=6\sqrt{abc}$, which is not a constant.
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ambros
301 posts
#36 • 2 Y
Y by Adventure10, Mango247
MY FRIEND
FISRTLY I DO NO SHOUT I WRITE THIS WAY BECAUSE I AM USED TO
SECONDLY WHAT DOES RIGOROUS MEAN?
THIRDLY I DID NOT SAY THAT WE TRY TO FIND Y AND AFTERWARDS TO FIND THE LEAST POSSIBLE VALUE FOR X WHICH IS GREATER THAN THE NUMBER ASKED IN THE PROBLEM OK?
THANK YOU PLEASE ANSWER TO ME WHEN YOU SEE THAT AND FIND IF THIS IS CORRECT OR WRONG OK?
THANK YOU
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Arne
3660 posts
#37 • 3 Y
Y by Adventure10, Mango247, and 1 other user
If you ever make another post using capitals only, it will be deleted. :mad:
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ambros
301 posts
#38 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Sorry my friend you do not need to bwe so angry :blush: now could you please see my solution and tellme if it is correct ? i have a kind of battle with someone and i have to know from people like you if my solution is correct :D
Thank you
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paladin8
3237 posts
#39 • 2 Y
Y by Adventure10 and 1 other user
ambros wrote:
Guys i wish to see if this solution is correct and tell me as soon as possible please
thank you
Let $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq x$ where x is the best real .

Let $a(b+1)+b(c+1)+c(a+1)\geq y$ where y is the best contsnt .

By multipliing these we have and using B-C-S
$[\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}][a(b+1)+b(c+1)+c(a+1)]\geq xy\geq 9$

From Cauchy -AM-GM we have $a(b+1)+b(c+1)+c(a+1)\geq 6\sqrt{abc}$

so $y=6\sqrt{abc}$ and so $6x\sqrt{abc}\geq 9$ and so

$x\geq\frac{3}{2\sqrt{abc}}\geq\frac{3}{abc+1}$

I don't think it works. Let $A = \frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}$ and $B = a(b+1)+b(c+1)+c(a+1)$.

Just because $AB \ge xy$ and $AB \ge 9$ does not imply that $xy \ge 9$. Even if you specify that $x,y$ are the "best" constants, you have $y = 6 \sqrt{abc}$, which is not a constant, so I don't think it holds.
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Peter
3615 posts
#40 • 3 Y
Y by Adventure10, Mango247, and 1 other user
So now the test for paladin, what if we want to know it for all numbers for which $abc=K$ for a certain constant $K$? ;)

[ because if that's true for all constants K, we have proven the original problem ]
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paladin8
3237 posts
#41 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Peter VDD wrote:
So now the test for paladin, what if we want to know it for all numbers for which $abc=K$ for a certain constant $K$? ;)

[ because if that's true for all constants K, we have proven the original problem ]

Hmm... well we still have to show that $y = 6\sqrt{K}$ would be the best constant, right? Which is equivalent to showing that there is always an equality case on the AM-GM for any $K$. But I don't think this is true, because the AM-GM has equality only when $a = b = c = 1$ so it only works for $K = 1$. Correct me if I'm wrong, but the proof doesn't seem to work.
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ambros
301 posts
#42 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Excuse me guys but i want to clarify two things
frist we do not imply xy>=9 we claim it and we search for which x,y is in charge that xy>=9
secondsice y is got form cauchy we know that for each a,b,c $y=6\sqrt{abc}$ so we get that for wach a,b,c $6x\sqrt{abc}\geq 9$ and so on
now that i was clarified could you please resee that and tell me ? thank you for all the time that you spend for this i am obliged
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sqing
41023 posts
#43 • 1 Y
Y by Adventure10
Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(1+a_2)}+\frac{1}{a_2(1+a_3)}+\cdots+\frac{1}{a_{n-1}(1+ a_n)}+\frac{1}{a_n( 1+a_1)}\geq\frac{n}{1+a_1a_2\cdots  a_n}.$$h
This post has been edited 1 time. Last edited by sqing, Aug 24, 2019, 2:09 AM
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sqing
41023 posts
#45 • 2 Y
Y by Adventure10, Mango247
sqing wrote:
Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(1+a_2)}+\frac{1}{a_2(1+a_3)}+\cdots+\frac{1}{a_{n-1}(1+ a_n)}+\frac{1}{a_n( 1+a_1)}\geq\frac{n}{1+a_1a_2\cdots  a_n}.$$
https://artofproblemsolving.com/community/c6h161059p10340314

Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(n-2+a_2)}+\frac{1}{a_2(n-2+a_3)}+\cdots+\frac{1}{a_{n-1}(n-2+ a_n)}+\frac{1}{a_n( n-2+a_1)}\geq\frac{n}{n-2+a_1a_2\cdots  a_n}.$$
This post has been edited 1 time. Last edited by sqing, Dec 7, 2019, 12:42 PM
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