Yet another classical inequality

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Valentin Vornicu

7301 posts

Valentin Vornicu

#1 h Apr 29, 2006, 4:19 PM • 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

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silouan

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silouan

#2 h Apr 29, 2006, 7:45 PM • 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

dule_00

#3 h Apr 29, 2006, 7:54 PM • 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

mzdanowicz

#4 h Apr 29, 2006, 8:00 PM • 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

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Arne

#5 h Apr 29, 2006, 10:08 PM • 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

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Valentin Vornicu

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Valentin Vornicu

#6 h Apr 29, 2006, 10:19 PM • 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

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

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Arne

#7 h Apr 29, 2006, 10:46 PM • 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

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Peter

#8 h Apr 29, 2006, 11:21 PM • 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

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Valentin Vornicu

#9 h Apr 29, 2006, 11:26 PM • 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

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pvthuan

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pvthuan

#10 h Apr 30, 2006, 10:26 AM • 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?

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Arne

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Arne

#11 h Apr 30, 2006, 10:38 AM • 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

pvthuan

#12 h Apr 30, 2006, 10:42 AM • 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

Lovasz

#13 h Apr 30, 2006, 11:49 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Is not this problem from Old and new inequalities?

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Arne

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Arne

#14 h Apr 30, 2006, 11:53 AM • 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

esti

#15 h Apr 30, 2006, 4:54 PM • 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

Centy

#16 h May 1, 2006, 9:25 AM • 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

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Arne

#17 h May 1, 2006, 10:34 AM • 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

Peter

#18 h May 1, 2006, 1:05 PM • 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

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Michael

99 posts

Michael

#19 h May 1, 2006, 8:37 PM • 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

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Arne

#20 h May 1, 2006, 8:48 PM • 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

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stergiu

#21 h May 3, 2006, 7:52 PM • 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

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Tiks

#22 h May 3, 2006, 8:08 PM • 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

,I think they have found most pupular ineq. in the world(my be after AM-GM

) and proposed it for BMO

.

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Michael

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Michael

#23 h May 4, 2006, 11:12 AM • 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

ambros

#24 h May 4, 2006, 2:08 PM • 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

Albanian Eagle

#25 h May 4, 2006, 2:29 PM • 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

DuLac

#26 h May 4, 2006, 6:31 PM • 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

ambros

#27 h May 4, 2006, 6:34 PM • 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

Cezar Lupu

#28 h May 4, 2006, 8:49 PM • 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

DuLac

#29 h May 5, 2006, 7:24 AM • 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

stergiu

#30 h May 5, 2006, 8:24 AM • 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

DuLac

#31 h May 5, 2006, 9:39 AM • 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

Peter

#32 h May 5, 2006, 9:43 AM • 1 Y

Y by Adventure10

DuLac wrote:

Source: Deep throat

lol @ hidden message in post #33

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exalibur

17 posts

exalibur

#33 h May 5, 2006, 4:46 PM • 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

ambros

#34 h May 6, 2006, 8:42 AM • 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

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Peter

#35 h May 6, 2006, 8:50 AM • 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

ambros

#36 h May 6, 2006, 10:08 AM • 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

Arne

#37 h May 6, 2006, 11:41 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

If you ever make another post using capitals only, it will be deleted.

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ambros

301 posts

ambros

#38 h May 7, 2006, 7:09 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Sorry my friend you do not need to bwe so angry

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

Thank you

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paladin8

3237 posts

paladin8

#39 h May 7, 2006, 7:35 PM • 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

Peter

#40 h May 7, 2006, 8:58 PM • 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

paladin8

#41 h May 8, 2006, 12:23 AM • 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

ambros

#42 h May 8, 2006, 8:05 PM • 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

41130 posts

sqing

#43 h Aug 24, 2019, 2:06 AM • 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

41130 posts

sqing

#45 h Dec 7, 2019, 12:22 PM • 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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