what is "Buffalo Way"

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shaoxinyu

158 posts

shaoxinyu

#1 h Feb 23, 2013, 5:17 PM • 7 Y

Y by polynomiii, Kgxtixigct, Adventure10, Mango247, rightways, cubres, and 1 other user

what is "Buffalo Way"? Someone please help me.

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arqady

30171 posts

arqady

#2 h Feb 23, 2013, 10:17 PM • 109 Y

Y by shaoxinyu, dan23, hctb00, mister_ady, Hantaehee, War-Hammer, 125907, james400, fractals, spartan3223, lukasm, lumia, Unknown-6174, Binomial-theorem, andrejilievski, smy2012, Arangeh, Ashutoshmaths, AwesomeToad, vyfukas, csmath, AkshajK, chezbgone, Lord.of.AMC, Amir Hossein, kprepaf, bobthesmartypants, EpicSkills32, joybangla, nikoma, yugrey, amatysten, nik9796, rachitgoel, minimario, tensor, Ygg, MathLearner01, Einstein314, Konigsberg, john111111, bcp123, niraekjs, fz0718, TheMaskedMagician, johnkwon0328, mihajlon, rjiang16, AKAL3, mjuk, bitrak, ssk9208, acegikmoqsuwy2000, baopbc, El_Ectric, baladin, ThisIsASentence, CeuAzul, champion999, jlammy, polynomiii, GeneralCobra19, enhanced, Mr.Chem-Mathy, samoha, opptoinfinity, bel.jad5, veehz, valsidalv007, Toinfinity, mudok, Kgxtixigct, IAmTheHazard, vsamc, Havister, Lyte188, leibnitz, ilovepizza2020, tigerzhang, Loxing, Dragunov, Scrutiny, mijail, Quidditch, 407420, ThisNameIsNotAvailable, samrocksnature, Number1048576, NO_SQUARES, Adventure10, ehz2701, kiyoras_2001, TestX01, hectorleo123, cubres, and 14 other users

For example, prove that $\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0$ for non-negatives $x$ , $y$ and $z$ .
A proof by BW:
Let $x=\min\{x,y,z\}$ , $y=x+u$ and $z=x+v$ .
Hence, $\sum_{cyc}(x^3-x^2y-x^2z+xyz)=(u^2-uv+v^2)x+(u+v)(u-v)^2\geq0$ .

Sometimes it is better to use $y=x+u$ , $z=x+u+v$ or $y=x+u+v$ , $z=x+u$ , where $u$ and $v$ are non-negatives.

If $x$ , $y$ and $z$ are sides-lengths of triangle and $z=\max\{x,y,z\}$ , we can use the following substitution:
$x=a+u$ , $y=a+v$ and $z=a+u+v$ , where $a>0$ , $u\geq0$ and $v\geq0$ .
For example:
For all triangle prove that:
$a^3b^2+b^3c^2+c^3a^2\geq(a^2+b^2+c^2)abc$
Proof:
Let $c=\max\{a,b,c\}$ , $a=x+u$ , $b=x+v$ and $c=x+u+v$ , where $x>0$ , $u$ and $v$ are non-negatives.
Hence, $a^3b^2+b^3c^2+c^3a^2-(a^2+b^2+c^2)abc=(u^2-uv+v^2)x^3+$
$+3(u^3-uv^2+v^3)x^2+(3u^4+2u^3v-3u^2v^2-2uv^3+3v^4)x+u^5+u^4v-2u^2v^3+v^5\geq0$ .

Another example.
For non-negatives $a$ , $b$ , $c$ and $d$ prove that:
$a^4+b^4+c^4+d^4+4abcd\geq2(a^2bc+b^2cd+c^2da+d^2ab)$
A proof:
Let $a=\min\{a,b,c,d\}$ , $b=a+u$ , $c=a+v$ and $d=a+w$ .
Hence, $a^4+b^4+c^4+d^4+4abcd-2(a^2bc+b^2cd+c^2da+d^2ab)=$
$=2(2u^2+2v^2+2w^2-uv-uw-vw)a^2+$
$+2(2u^3+2v^3+2w^3-u^2v-u^2w-v^2w-w^2u)a+u^4+v^4+w^4-2u^2vw\geq0$ .

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mathh97

32 posts

mathh97

#3 h Dec 26, 2013, 6:42 PM • 2 Y

Y by Adventure10, cubres

@arqady How is it obvious that all the final inequalities hold?

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arqady

30171 posts

arqady

#4 h Dec 27, 2013, 10:38 PM • 10 Y

Y by amatysten, flamefoxx99, rjiang16, GoJensenOrGoHome, yiwen, Sidier, Adventure10, Mango247, cubres, and 1 other user

mathh97 wrote:

@arqady How is it obvious that all the final inequalities hold?

Because $\sum_{cyc}(u^2-uv)\geq0$ , $\sum_{cyc}(2u^3-u^2v-u^2w)\geq0$ and
$u^4+v^4+w^4=\frac{1}{2}u^4+\frac{1}{2}u^4+v^4+w^4\geq2u^2vw$ .

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cobbler

2180 posts

cobbler

#5 h Jan 28, 2014, 2:42 PM • 3 Y

Y by Adventure10, cubres, and 1 other user

I have a question:

What are some advantages of using BW in an inequality instead of other methods?

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SMOJ

2663 posts

SMOJ

#6 h Jan 28, 2014, 2:48 PM • 2 Y

Y by Adventure10, cubres

You dont need lots of thinking. I think I used this for Nesbitt's Inequality

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Ashutoshmaths

976 posts

Ashutoshmaths

#7 h Jan 28, 2014, 2:49 PM • 2 Y

Y by Adventure10, cubres

arkanm wrote:

I have a question:

What are some advantages of using BW in an inequality instead of other methods?

According to me, you never know that the step that you performed will give you a strong result or not.
But if you apply BW then the sharpness is preserved and after expansion, manipulations are easy.

At the moment, arqady's post has rating 5..... why isn't it 6.

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mavropnevma

15142 posts

mavropnevma

#8 h Jan 29, 2014, 6:25 PM • 4 Y

Y by amatysten, Adventure10, Mango247, cubres

Check this extremely recent one http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=573480 for a typical application.

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bobthesmartypants

4337 posts

bobthesmartypants

#9 h Jan 30, 2014, 4:55 AM • 3 Y

Y by Adventure10, Mango247, cubres

Can someone prove the following? I don't see how it is trivial because of my lack of inequality practice:

$u^3-uv^2+v^3\ge 0$

$3u^4+2u^3v-3u^2v^2-2uv^3+3v^4\ge 0$

$2u^3+2v^3+2w^3-u^2v-u^2w-v^2w-w^2u\ge 0$

All restrictions outlined by arqady are still applied on these.

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ssilwa

5451 posts

ssilwa

#10 h Jan 30, 2014, 4:57 AM • 3 Y

Y by Adventure10, Mango247, cubres

All applications of Muirhead.

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nikoma

1976 posts

nikoma

#11 h Jan 30, 2014, 8:05 PM • 3 Y

Y by Adventure10, Mango247, cubres

May I ask for the origin of the Buffalo way? Or is that just one of those things that's well known in mathematical folklore?

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ssilwa

5451 posts

ssilwa

#12 h Jan 30, 2014, 8:42 PM • 17 Y

Y by RocketSingh, AKAL3, fractals, jlammy, polynomiii, Vrangr, Kayak, OlympiadIneqByBruteForce, Kanep, Toinfinity, ilovepizza2020, Adventure10, Mango247, navier3072, cubres, and 2 other users

Some say it was the love child of Thor and Zeus, other point to the fall of the iron curtain.

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rsa365

165 posts

rsa365

#13 h Jan 30, 2014, 10:33 PM • 2 Y

Y by Adventure10, cubres

It was most likely created by someone named Buffalo.

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mavropnevma

15142 posts

mavropnevma

#14 h Jan 30, 2014, 11:00 PM • 3 Y

Y by Adventure10, Mango247, cubres

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2746276#p2746276. arqady says

mudok wrote:

Yes,the Buffalo way is very useful. Why it is called "Buffalo"?

I think, because it sweeps away and all of the way, like a buffalo.

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nikoma

1976 posts

nikoma

#15 h Jan 30, 2014, 11:18 PM • 3 Y

Y by Adventure10, Mango247, cubres

Here's also interesting solution by Buffalo Way http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1871465#p1871465

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borislav_mirchev

1525 posts

borislav_mirchev

#16 h Jan 30, 2014, 11:41 PM • 3 Y

Y by Adventure10, Mango247, cubres

I have a problem that can be solved in Buffalo way.

Let $a$ , $b$ , $c$ are positive reals such that $abc=1$ . Prove that: $\frac{a^2}{b^2+c}+\frac{b^2}{c^2+a}+\frac{c^2}{a^2+b} \geq \frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}$ .

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SMOJ

2663 posts

SMOJ

#17 h Jan 31, 2014, 8:35 AM • 3 Y

Y by PhantomR, Adventure10, cubres

I shall show everyone how I buffaloed Nessbit's when I did not know C-S.

Let $a\ge b\ge c$ and $a-x=b=c+y$

$\frac{b+x}{2b-y}+\frac{b}{2b+x-y}+\frac{b-y}{2b+x}=(\frac{1}{2}+\frac{x+\frac{1}{2}y}{2b-y})+(\frac{1}{2}+\frac{\frac{1}{2}y-\frac{1}{2}x}{2b+x-y})+(\frac{1}{2}+\frac{-\frac{1}{2}x-y}{2b+x})$ .

But then $\frac{x+\frac{1}{2}y}{2b-y}+\frac{\frac{1}{2}y-\frac{1}{2}x}{2b+x-y}+\frac{-\frac{1}{2}x-y}{2b+x}$
$\ge \frac{x+\frac{1}{2}y}{2b+x-y}+\frac{\frac{1}{2}y-\frac{1}{2}x}{2b+x-y}+\frac{-\frac{1}{2}x-y}{2b+x}$
$=\frac{y+\frac{1}{2}x}{2b+x-y}+\frac{-\frac{1}{2}x-y}{2b+x}\ge 0$ .

So yep, we are done with a lot of LaTeX

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borislav_mirchev

1525 posts

borislav_mirchev

#18 h Jan 31, 2014, 4:21 PM • 4 Y

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

The problem, proposed by me is not easy at all. @arqady mentioned it can be solved using Buffalo way and some calculations made by a computer may be used. Also I suppose - it can be solved without Buffalo Way, but it may be even harder.

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nikoma

1976 posts

nikoma

#19 h Feb 1, 2014, 11:56 AM • 6 Y

Y by borislav_mirchev, Wizard_32, FISHMJ25, Adventure10, Mango247, cubres

borislav_mirchev wrote:

I have a problem that can be solved in Buffalo way.

Let $a$ , $b$ , $c$ are positive reals such that $abc=1$ . Prove that: $\frac{a^2}{b^2+c}+\frac{b^2}{c^2+a}+\frac{c^2}{a^2+b} \geq \frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}$ .

Here's proof if $x \leq y \leq z$ , I can't manage to prove the case $x \leq z \leq y$ .
Click to reveal hidden text

Attachments:

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nikoma

1976 posts

nikoma

#20 h Feb 1, 2014, 4:56 PM • 2 Y

Y by Adventure10, cubres

I used mathematica

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borislav_mirchev

1525 posts

borislav_mirchev

#21 h Feb 1, 2014, 6:04 PM • 3 Y

Y by Adventure10, Mango247, cubres

It will be very interesting if someone manage to prove the remaining part of this inequality. I haven't ever seen a complete proof. It is probably a "new" problem not appearing in math competitions, magazines or books.

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nikoma

1976 posts

nikoma

#22 h Jun 7, 2014, 11:52 PM • 4 Y

Y by borislav_mirchev, Adventure10, Mango247, cubres

borislav_mirchev wrote:

I have a problem that can be solved in Buffalo way.

Let $a$ , $b$ , $c$ are positive reals such that $abc=1$ . Prove that: $\frac{a^2}{b^2+c}+\frac{b^2}{c^2+a}+\frac{c^2}{a^2+b} \geq \frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}$ .

Hi everyone, I managed to prove this crazy inequality by buffalo way and easy AM-GM inspired by this post http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3509698#p3509698 by math_science (I don't know why I didn't think of this before, I am so stupid, oh my god.). First of all we use well-known substitution $a = \frac{x}{y}$ , $b = \frac{y}{z}$ , $c = \frac{z}{x}$ , then we consider two cases.
1) If $x \leq y \leq z$ , then refer to this post http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3378796#p3378796
2) If $x \leq z \leq y$ , then let $z = x + u + v$ , $y = x + u$ , where $u,v \geq 0$ , then the inequality expands into the attached picture in which the expression needs to be non-negative, there are only two negative terms we need to take care of, namely $-x^4u^6v^{14}$ and $-5x^4u^7v^{13}$ , but this can be done easily by these two AM-GMs $x^4u^6v^{14} \leq \frac{x^5u^4v^{15} + x^3u^8v^{13}}{2}$ $5x^4u^7v^{13} \leq \frac{5(x^5u^5v^{14} + x^3u^9v^{12})}{2}$ and we are done!

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YESMAths

829 posts

YESMAths

#23 h Aug 9, 2014, 2:05 AM • 3 Y

Y by Adventure10, Mango247, cubres

nikoma wrote:

borislav_mirchev wrote:

I have a problem that can be solved in Buffalo way.

Let $a$ , $b$ , $c$ are positive reals such that $abc=1$ . Prove that: $\frac{a^2}{b^2+c}+\frac{b^2}{c^2+a}+\frac{c^2}{a^2+b} \geq \frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}$ .

Here's proof if $x \leq y \leq z$ , I can't manage to prove the case $x \leq z \leq y$ .
Click to reveal hidden text

Hmm.. yes sometimes using Buffalo Way can be too tedious, like the one shown in the picture, and that's what Dr. Sonnhard has been famous for these days.

But then the BW doesn't find much application in contests and olympiads, in its straight form, right?

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arqady

30171 posts

arqady

#24 h Aug 11, 2014, 1:05 PM • 4 Y

Y by Carl2015, Adventure10, Mango247, cubres

YESMAths wrote:

But then the BW doesn't find much application in contests and olympiads, in its straight form, right?

See here a first and a second examples:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2942645#p2942645
These problems from IMO.

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YESMAths

829 posts

YESMAths

#25 h Aug 11, 2014, 1:20 PM • 3 Y

Y by Adventure10, Mango247, cubres

arqady wrote:

YESMAths wrote:

But then the BW doesn't find much application in contests and olympiads, in its straight form, right?

See here a first and a second examples:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2942645#p2942645
These problems from IMO.

Yeah! But sometimes it becomes tedious without the theory, right? Like this one?

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Blitzkrieg97

236 posts

Blitzkrieg97

#26 h Oct 25, 2014, 8:34 PM • 2 Y

Y by Adventure10, cubres

SMOJ wrote:

I shall show everyone how I buffaloed Nessbit's when I did not know C-S.

Let $a\ge b\ge c$ and $a-x=b=c+y$

$\frac{b+x}{2b-y}+\frac{b}{2b+x-y}+\frac{b-y}{2b+x}=(\frac{1}{2}+\frac{x+\frac{1}{2}y}{2b-y})+(\frac{1}{2}+\frac{\frac{1}{2}y-\frac{1}{2}x}{2b+x-y})+(\frac{1}{2}+\frac{-\frac{1}{2}x-y}{2b+x})$ .

But then $\frac{x+\frac{1}{2}y}{2b-y}+\frac{\frac{1}{2}y-\frac{1}{2}x}{2b+x-y}+\frac{-\frac{1}{2}x-y}{2b+x}$
$\ge \frac{x+\frac{1}{2}y}{2b+x-y}+\frac{\frac{1}{2}y-\frac{1}{2}x}{2b+x-y}+\frac{-\frac{1}{2}x-y}{2b+x}$
$=\frac{y+\frac{1}{2}x}{2b+x-y}+\frac{-\frac{1}{2}x-y}{2b+x}\ge 0$ .

So yep, we are done with a lot of LaTeX

does it need C-S? it can be easily solved by Am-Gm.(Beautiful solution by buffalo btw.)

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RopuToran

609 posts

RopuToran

#27 h Jul 23, 2018, 4:21 PM • 3 Y

Y by Adventure10, Mango247, cubres

arqady wrote:

mathh97 wrote:

@arqady How is it obvious that all the final inequalities hold?

Because $\sum_{cyc}(u^2-uv)\geq0$ , $\sum_{cyc}(2u^3-u^2v-u^2w)\geq0$ and
$u^4+v^4+w^4=\frac{1}{2}u^4+\frac{1}{2}u^4+v^4+w^4\geq2u^2vw$ .

Your $u,v,w$ solution seems interesting. What is the idea ?

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arqady

30171 posts

arqady

#28 h Jul 23, 2018, 4:44 PM • 3 Y

Y by Adventure10, Mango247, cubres

The last inequality it's AM-GM.

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Kgxtixigct

555 posts

Kgxtixigct

#30 h Jun 27, 2020, 3:30 AM • 1 Y

Y by cubres

arqady wrote:

For example, prove that $\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0$ for non-negatives $x$ , $y$ and $z$ .
A proof by BW:
Let $x=\min\{x,y,z\}$ , $y=x+u$ and $z=x+v$ .
Hence, $\sum_{cyc}(x^3-x^2y-x^2z+xyz)=(u^2-uv+v^2)x+(u+v)(u-v)^2\geq0$ .

Sometimes it is better to use $y=x+u$ , $z=x+u+v$ or $y=x+u+v$ , $z=x+u$ , where $u$ and $v$ are non-negatives.

If $x$ , $y$ and $z$ are sides-lengths of triangle and $z=\max\{x,y,z\}$ , we can use the following substitution:
$x=a+u$ , $y=a+v$ and $z=a+u+v$ , where $a>0$ , $u\geq0$ and $v\geq0$ .
For example:
For all triangle prove that:
$a^3b^2+b^3c^2+c^3a^2\geq(a^2+b^2+c^2)abc$ Proof:
Let $c=\max\{a,b,c\}$ , $a=x+u$ , $b=x+v$ and $c=x+u+v$ , where $x>0$ , $u$ and $v$ are non-negatives.
Hence, $a^3b^2+b^3c^2+c^3a^2-(a^2+b^2+c^2)abc=(u^2-uv+v^2)x^3+$
$+3(u^3-uv^2+v^3)x^2+(3u^4+2u^3v-3u^2v^2-2uv^3+3v^4)x+u^5+u^4v-2u^2v^3+v^5\geq0$ .

Another example.
For non-negatives $a$ , $b$ , $c$ and $d$ prove that:
$a^4+b^4+c^4+d^4+4abcd\geq2(a^2bc+b^2cd+c^2da+d^2ab)$ A proof:
Let $a=\min\{a,b,c,d\}$ , $b=a+u$ , $c=a+v$ and $d=a+w$ .
Hence, $a^4+b^4+c^4+d^4+4abcd-2(a^2bc+b^2cd+c^2da+d^2ab)=$
$=2(2u^2+2v^2+2w^2-uv-uw-vw)a^2+$
$+2(2u^3+2v^3+2w^3-u^2v-u^2w-v^2w-w^2u)a+u^4+v^4+w^4-2u^2vw\geq0$ .

But this is schurs as well...

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