a^2b(a-b) + b^2c(b-c) + c^2a(c-a) >= 0 for a triangle

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ashwath.rabindranath

437 posts

ashwath.rabindranath

#1 h Oct 9, 2005, 5:47 AM • 9 Y

Y by Adventure10, samrocksnature, jhu08, Mango247, Tamam, cubres, and 3 other users

Let $a$ , $b$ and $c$ be the lengths of the sides of a triangle. Prove that
$a^{2}b(a - b) + b^{2}c(b - c) + c^{2}a(c - a)\ge 0.$
Determine when equality occurs.

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Cezar Lupu

1906 posts

Cezar Lupu

#2 h Oct 9, 2005, 9:05 AM • 8 Y

Y by Vietjung, Adventure10, jhu08, Mango247, cubres, and 3 other users

Come on, not this again. it has been posted many times before, I think. From the subtitutions $a=y+z, b=z+x, c=x+y$ and the inequality is equivalent with $\sum\frac{x^{2}}{y}\geq\sum x.$ From here it is clear how to proceed.

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Gibbenergy

640 posts

Gibbenergy

#3 h Oct 10, 2005, 12:21 AM • 18 Y

Y by Obunaga, di_duc, urun, bedwarmer, kilua, xlzq, LJQ, JasperL, meanperson792, Adventure10, megarnie, EHHSW, Mango247, cubres, and 4 other users

Hi Cezar Lupu!
As I do this IMO again, I found the very interesting solution for it and it is only 1 line . Can u believe it?
He he . Ok ! Here is mine ( Any please don't ask me why I found it!);
IMO inequality is equivalent to this "obvious inequality":
$\frac12((a-b)^2(a+b-c)(b+c-a)+(b-c)^2(b+c-a)(a+c-b)+(c-a)^2(a+c-b)(a+b-c))\geq 0$
Right !

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ashwath.rabindranath

437 posts

ashwath.rabindranath

#4 h Oct 10, 2005, 5:01 AM • 5 Y

Y by Adventure10, EHHSW, Mango247, cubres, and 1 other user

Sure enough, that's what i did.I was wondering if you could use Schur to solve it?
Anyone?
Ashwath

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jarod

184 posts

jarod

#5 h Oct 10, 2005, 6:43 AM • 7 Y

Y by john10, Wizard_32, Adventure10, ehuseyinyigit, cubres, and 2 other users

Gibbenergy wrote:

Hi Cezar Lupu!
As I do this IMO again, I found the very interesting solution for it and it is only 1 line . Can u believe it?
He he . Ok ! Here is mine ( Any please don't ask me why I found it!);
IMO inequality is equivalent to this "obvious inequality":
$\frac12((a-b)^2(a+b-c)(b+c-a)+(b-c)^2(b+c-a)(a+c-b)+(c-a)^2(a+c-b)(a+b-c))\geq 0$
Right !

It's not new and obviously it's not your solution, Gibb . The ineq is homogeneous , so i can easily change the ineq into the fomula: $A(b-c)^2+B(c-a)^2+C(a-b)^2 \geq 0$

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Cezar Lupu

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Cezar Lupu

#6 h Oct 10, 2005, 5:17 PM • 6 Y

Y by Adventure10, megarnie, Mango247, cubres, and 2 other users

Hey Gibb this one-line solution had been given by an IMO contestant.

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Gibbenergy

640 posts

Gibbenergy

#7 h Oct 11, 2005, 5:06 PM • 16 Y

Y by bedwarmer, Obunaga, di_duc, kilua, xlzq, Wizard_32, Adventure10, megarnie, Mango247, cubres, and 6 other users

Hey guys!
That is my sol and I have found it by myself. I don't care who has posted it before.
I found my sols because as I know the IMO US team at that time was also use the trick that put $x=u+v...$ It is obviously unnatural. So it is not important with the way they showed. The question is how can the author generate it? I believed that the author of IMO 83 used the same technique that I have discovered now to make it. So it doesn't matter that my sols now is posted before.
If you say you can make $A (b-c)^2$ i believed it is the theory. The important thing is how can you find $A$ ? $(b-c)^2....$ and also their permutation. If you can show me that technique, I would admire you as my best teacher.
I feel that I don't want to share any my technique to you guys either. I will keep it as my secrect.
Thanks.

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Soarer

2589 posts

Soarer

#8 h Oct 11, 2005, 5:35 PM • 6 Y

Y by AlastorMoody, Adventure10, cubres, and 3 other users

Gibbenergy wrote:

Hey guys!
That is my sol and I have found it by myself. I don't care who has posted it before.
I found my sols because as I know the IMO US team at that time was also use the trick that put $x=u+v...$ It is obviously unnatural. So it is not important with the way they showed. The question is how can the author generate it? I believed that the author of IMO 83 used the same technique that I have discovered now to make it. So it doesn't matter that my sols now is posted before.
If you say you can make $A (b-c)^2$ i believed it is the theory. The important thing is how can you find $A$ ? $(b-c)^2....$ and also their permutation. If you can show me that technique, I would admire you as my best teacher.
I feel that I don't want to share any my technique to you guys either. I will keep it as my secrect.
Thanks.

Ha! It's really funny. If you are willing to search on the forum, you can see that many people here prove inequalities by rewriting the inequality into sum of squares. Obviously they have a technique, or at least a general way to write inequalities in that form. Although it may not be able to be written rigorously, a general approach does exist I think.
Of course you have the right not to share your technique and keep it as your secret. But I just wonder, if you don't even want to share, why do you hope others are willing to share with you? It's just like you want others to share something with you while you are reluctant to hand out anything. What a joke.

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Arne

3660 posts

Arne

#9 h Oct 11, 2005, 6:25 PM • 7 Y

Y by Wizard_32, Adventure10, Mango247, ehuseyinyigit, cubres, and 2 other users

What a ridiculous conversation. Well, there are plenty of people willing to share NEW and interesting techniques with us, so we shouldn't worry at all.

And, oh yes, topic locked.

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lotharek

44 posts

lotharek

#10 h Nov 11, 2005, 9:35 PM • 6 Y

Y by LJQ, Adventure10, cubres, and 3 other users

I use Ravi's substitution.

Let $a=x+y, b=y+z, c=z+x$ .

Some computation give...:

$x^3z+xy^3+yz^3\geq xyz(x+y+z),$
$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq x+y+z,$ what is true by rearrangement inequality.

Equality holds, when $x=y=z$ , so when $a=b=c$ .

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darij grinberg

6555 posts

darij grinberg

#11 h Nov 11, 2005, 9:58 PM • 3 Y

Y by Adventure10, Mango247, cubres

Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=55406 and http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 (posts #8-#10) .

darij

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oocal13

4 posts

oocal13

#12 h Mar 23, 2006, 9:13 AM • 5 Y

Y by Adventure10, Mango247, cubres, and 2 other users

The inequality equivalent to:
$a^{3}b+b^{3}c+c^{3}a \ge a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$
And this is a result of Moorhead Inequality
$\left({3,1,0}\right) > \left({2,2,0}\right)$

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tim1234133

523 posts

tim1234133

#13 h Feb 22, 2007, 11:33 AM • 5 Y

Y by Adventure10, Mango247, cubres, and 2 other users

This is an old topic but I think that it's worth pointing out that the LHS of your inequality is cyclic, not symmetric. Muirhead only works on symmetric sums, so the rearrangement method is needed.

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GacKiem

178 posts

GacKiem

#14 h Feb 22, 2007, 3:44 PM • 4 Y

Y by Adventure10, cubres, and 2 other users

We can use the Entirely Mixing variable method to solve this problem (but this method is so strong to solve such an easy problem). It's sufficient to consider one case $a=b+c$ and the rest is easy.

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Wikipedian1337

62 posts

Wikipedian1337

#15 h Dec 29, 2008, 7:16 AM • 3 Y

Y by Adventure10, Mango247, cubres

GacKiem wrote:

We can use the Entirely Mixing variable method to solve this problem (but this method is so strong to solve such an easy problem). It's sufficient to consider one case $a = b + c$ and the rest is easy.

Sorry about reviving an old thread, but after reading http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf, I still don't get the method. I can't show that replacing $(a,b,c)$ with $(a-c,b-c,0)$ will necessarily decrease LHS.

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JustKeepRunning

2958 posts

JustKeepRunning

#48 h Aug 11, 2021, 9:46 PM • 1 Y

Y by cubres

@above what do you mean? The symemtric is just $2$ times the cyclic...

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samrocksnature

8791 posts

samrocksnature

#49 h Aug 11, 2021, 9:48 PM • 1 Y

Y by cubres

JustKeepRunning wrote:

@above what do you mean? The symemtric is just $2$ times the cyclic...

I don't know why - I just remarked that some of the other posts stated that rearrangement/muirhead don't work.

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jasperE3

11395 posts

jasperE3

#50 h Aug 11, 2021, 11:43 PM • 1 Y

Y by cubres

JustKeepRunning wrote:

@above what do you mean? The symemtric is just $2$ times the cyclic...

For instance $\sum_{\text{sym}}a^2b=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$ but $2\sum_{\text{cyc}}a^2b=2a^2b+2b^2c+2c^2a$ .

Making Muirhead work isn't trivial as far as I can tell.

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jasperE3

11395 posts

jasperE3

#51 h Aug 11, 2021, 11:46 PM • 1 Y

Y by cubres

With Ravi ( $a=x+y,b=y+z,c=z+x$ ) we have to prove:
$(x+y)^2(y+z)(x-z)+(y+z)^2(z+x)(y-x)+(z+x)^2(x+y)(z-y)\ge0$ $\Leftrightarrow2x^3z+2xy^3+2yz^3\ge2x^2yz+2xy^2z+2xyz^2,$ which is true since $\frac{x^2}y+\frac{y^2}z+\frac{z^2}x\ge x+y+z$ by C-S.

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#52 h Apr 23, 2022, 5:48 AM • 1 Y

Y by cubres

Since sides of triangle it is true if we WLOG, $a\geq b\geq c \implies a^2+bc \geq b^2+ca\geq c^2+ab.$ By rearrangement inequality, $\begin{align*} \sum_{\text{cyc}} \frac{a^2+bc}{c} \geq \sum_{\text{cyc}} \frac{a^2+bc}{a}\\ \implies \sum_{\text{cyc}} \frac{a^2}{c} \geq \sum_{\text{cyc}} \frac{bc}{a} \\ \implies \sum_{\text{cyc}} a^3b \geq \sum_{\text{cyc}} a^2b^2.\end{align*}$ Equality when $a=b=c.$ And the claim follows. $\square$

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#53 h Apr 23, 2022, 5:53 AM • 1 Y

Y by cubres

Ravi inequality seems cool, could I have a introductory note on it?

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samrocksnature

8791 posts

samrocksnature

#54 h Apr 23, 2022, 5:53 AM • 3 Y

Y by ZETA_in_olympiad, Turtle09, cubres

ZETA_in_olympiad wrote:

Ravi inequality seems cool, could I have a introductory note on it?

here

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john0512

4191 posts

john0512

#55 h Mar 21, 2023, 4:22 AM • 1 Y

Y by cubres

Rewrite the inequality as $\sum_{cyc}a^2b(a-b)\geq 0.$ Let $a=y+z,b=x+z,c=x+y$ . Then, this becomes $\sum_{cyc}(y+z)^2(x+z)(y-x)\geq 0.$ Expanding, this is $\sum_{cyc}(y^2+2yz+z^2)(xy-x^2+yz-xz)\geq 0$ $\sum_{cyc}xy^3+y^3z+yz^3-xz^3-2x^2yz+xy^2z-xyz^2-x^2y^2-x^2z^2+2y^2z^2\geq 0.$ Note that $\sum_{cyc}xy^2z-xyz^2=\sum_{cyc} -x^2y^2-x^2z^2+2y^2z^2=\sum_{cyc}y^3z-xz^3=0,$ so it suffices to show that $\sum_{cyc}xy^3+yz^3-2x^2yz\geq 0.$ However, this is just $\sum_{sym}x^3y\geq \sum_{sym} x^2yz,$ which is just Muirhead's since $(3,1,0)\succ (2,1,1)$ so we are done. Equality occurs when $x=y=z$ or equivalently $a=b=c$ from the Muirhead step.

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huashiliao2020

1292 posts

huashiliao2020

#56 h Jun 10, 2023, 6:58 PM • 1 Y

Y by cubres

ugh.. i just had to expand it bruh

After my attachment, the inequality follows from C-S since $LHS(y+z+x)\ge (x+y+z)(y+z+x). \blacksquare$

Attachments:

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huashiliao2020

1292 posts

huashiliao2020

#57 h Jun 12, 2023, 9:11 PM • 1 Y

Y by cubres

oops other sol less expanding but much harder to find

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AgentC

43 posts

AgentC

#58 h May 13, 2024, 5:42 PM • 1 Y

Y by cubres

abasse wrote:

a²b(a-b)>=a²(a-c) (a-b) because b>=a-c
the make the same thing
We are a²b(a-b)+b²c(b-c)+c²a(c-a)>= a²(a-b) (a-c) +b²(b-a) (b-c) +c²(c-b)(c-a) >=0 schur

but then what about the equality case, since b>a-c.
I think if a-b is negative then "a²b(a-b)>=a²(a-c) (a-b)" is not correct

This post has been edited 1 time. Last edited by AgentC, May 13, 2024, 5:54 PM
Reason: 1

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eg4334

636 posts

eg4334

#59 h Sep 4, 2024, 6:34 PM • 1 Y

Y by cubres

Ravi Sub:
$\sum_{\text{cyc}}(y+z)^2(x+z)(y-x) \geq 0$ Expand: $\sum_{\text{cyc}} xy^3+xy^2z+y^3z+2y^2z^2+yz^3 \geq \sum_{\text{cyc}} x^2y^2 + 2x^2yz + x^2z^2 + xyz^2 + xz^3$ Cancel terms: $\sum_{\text{cyc}} xy^3+yz^3 \geq \sum_{\text{cyc}} 2x^2yz$ or equivalently $\sum_{\text{sym}} x^3y \geq \sum_{\text{sym}} x^2yz$ which by $\{ 3, 1, 0\} \succ \{ 2, 1, 1\}$ we are done. Equality holds when the triangle is equilateral.

This post has been edited 1 time. Last edited by eg4334, Sep 4, 2024, 6:36 PM
Reason: equality

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Tony_stark0094

69 posts

Tony_stark0094

#60 h Apr 20, 2025, 7:48 AM • 1 Y

Y by cubres

With Ravi sub and some algebraic manipulations it is equivalent to prove $\sum \frac{x^2}{y} \geq \sum x$ which is trivial by AM-GM

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Ilikeminecraft

677 posts

Ilikeminecraft

#61 h Apr 25, 2025, 6:54 AM • 1 Y

Y by cubres

Susbtitute $a = y + z, b = x + z, c = x + y.$ Simplifies to $x^3z + yz^3 + y^3x \geq x^2yz + xy^2z + xyz^2.$ However, $x^3y + xy^2z\geq2x^2yz,$ and so we are done.