A+b+c=0

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Source: Lithuanian Mathematical Olympiad 2006

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297 posts

Xixas

#1 h Apr 12, 2006, 2:31 PM • 2 Y

Y by Adventure10, Mango247

Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$ .

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99 posts

#2 h Apr 12, 2006, 3:15 PM • 3 Y

Y by Mathskidd, Adventure10, Mango247

Xixas wrote:

Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$ .

The following identity is easily verified:
${{y-z}\over{x}}+{{z-x}\over{y}}+{{x-y}\over{z}}=-{{(y-z)(z-x)(x-y)}\over{xyz}}\quad\hbox{for}\quad x,y,z\neq 0.\leqno (1)$
Set $x=b-c,y=c-a,z=a-b$ in (1). In view of $a+b+c=0$ we have $y-z=b+c-2a=-3a,z-x=-3b,x-y=-3c$ , and (1) becomes
${{-3a}\over{b-c}}+{{-3b}\over{c-a}}+{{-3c}\over{a-b}}=-{{(-3a)(-3b)(-3c)}\over{(b-c)(c-a)(a-b)}},$
i.e.,
${{a}\over{b-c}}+{{b}\over{c-a}}+{{c}\over{a-b}}=-{{9abc}\over{(b-c)(c-a)(a-b)}}.$
Now set $x=a,y=b,z=c$ in (1):
${{b-c}\over{a}}+{{c-a}\over{b}}+{{a-b}\over{c}}=-{{(b-c)(c-a)(a-b)}\over{abc}}.$
Multiplying the last two equalities we get the claimed one.

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1648 posts

#3 h Apr 12, 2006, 5:53 PM • 2 Y

Y by Adventure10, Mango247

Nice work !

The problem comes from the Austrian- Polish competition.I have recently read a very similar - almost the same - solution in an older issue of Crux.

Babis

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398 posts

#4 h Apr 13, 2006, 9:12 AM • 2 Y

Y by Adventure10, Mango247

Indeed

http://www.kalva.demon.co.uk/aus-pol/ap85.html

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3952 posts

#5 h Apr 13, 2006, 4:49 PM • 2 Y

Y by Adventure10, Mango247

The problem indeed is classic and quite well-known (at least in Greece ).
Another solution is . Let $S$ the given product

Notice that $\frac{a}{b-c}(\frac{c-a}{b}+\frac{a-b}{c})=\frac{2a^2}{bc}$

Similar for the other two .

We expand the given product and we use he above .so
$S=3+\frac{2(a^3+b^3+c^3)}{abc}$ .But $a^3+b^3+c^3=3abc$ when $a+b+c=0$

so $S=9$ and we are done

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46 posts

#6 h Apr 22, 2006, 12:19 PM • 2 Y

Y by Adventure10, Mango247

Another solution..

Click to reveal hidden text

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297 posts

Xixas

#7 h Apr 22, 2006, 3:42 PM • 3 Y

Y by Kunihiko_Chikaya, Adventure10, Mango247

Ghm

Can we apply AM-HM to potentialy negative $x, y, z$ ?

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46 posts

#8 h Apr 22, 2006, 4:18 PM • 2 Y

Y by Adventure10, Mango247

Hm I have to admit I really didn't consider that restriction. Isn't it only for the inequalities with geometric mean?

Actually it is also possible to proof this inequality using Cauchy-Schwarz: (Though it seems that this time we really have a restriction because of those d $a$ mn sqrt{} things.

)

$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq (\sqrt{x}\cdot\sqrt{\frac{1}{x}} +\sqrt{y}\cdot\sqrt{\frac{1}{y}}+ \sqrt{z}\cdot\sqrt{\frac{1}{z}})^2=9$

[EDIT]

Ok, we just have to try $(-10, 1, 1)$ and see that my solution sucks.

This post has been edited 1 time. Last edited by ArcticMonkey, Apr 22, 2006, 4:23 PM

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297 posts

Xixas

#9 h Apr 22, 2006, 4:20 PM • 2 Y

Y by Adventure10, Mango247

That's the reason that prevented me from using Cauchy-Schwarz on the contest.

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1896 posts

#10 h Mar 29, 2009, 8:36 PM • 2 Y

Y by Adventure10, Mango247

Since $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
It follows that if $a + b + c = 0$ , then $a^3 + b^3 + c^3 = 3abc$
The numerator of the left factor is
$- \sum (a^3 - a^2b - a^2c + abc) = - \sum (a^3 - a^2(b + c) + abc) = - \sum (2a^3 + abc)$
Since $3abc = a^3 + b^3 + c^3$ , the left factor is
$\frac { - 3 \sum a^3}{(a - b)(b - c)(c - a)}$
The right factor is
$\frac {b^2c - bc^2 + c^2a - ca^2 + a^2b - ab^2}{abc}$
But the denominator, $abc = \frac {1}{3} \sum a^3$ . Everything cancels, and we're left with 9.

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#11 h Feb 12, 2012, 5:15 PM • 2 Y

Y by Adventure10, Mango247

silouan wrote:

Notice that $\frac{a}{b-c}(\frac{c-a}{b}+\frac{a-b}{c})=\frac{2a^2}{bc}$

How did you come up with that ?

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42043 posts

sqing

#12 h Aug 9, 2014, 7:34 AM • 3 Y

Y by boss_1998, Adventure10, Mango247

Xixas wrote:

Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$ .

$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=-\frac{9abc}{(a-b)(b-c)(c-a)},$

$\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=-\frac{(a-b)(b-c)(c-a)}{abc}.$
here

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Kunihiko_Chikaya

14514 posts

Kunihiko_Chikaya

#13 h Aug 3, 2015, 3:29 AM • 2 Y

Y by Adventure10, Mango247

The problem is very classic. In Japanese univesity entrance exam, Nara University entrance exam/Medicine has posed te same problem in early 1960 or later 1950.

I would like to know the original problem or articles regarding the problem.

Are there anyone having the information ？

Any information would be appreciated.

Thanks in advance.

kunny

This post has been edited 2 times. Last edited by Kunihiko_Chikaya, Aug 3, 2015, 8:17 AM

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30238 posts

arqady

#14 h Aug 3, 2015, 5:41 AM • 3 Y

Y by Kunihiko_Chikaya, Adventure10, Mango247

I have found this problem in the following book.
В.Кречмар, "Задачник по алгебре", (1937 year, page 15, problem 18).

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1 post

#15 h Aug 6, 2015, 4:34 PM • 2 Y

Y by Adventure10, Mango247

Since a+b+c=0, c=-a-b
Therefore, substituting this in you get the attached image.
Take out 9 from the numerator, cancel 2a^4 + 5(a^3)b - 5a(b^3) - 2b^4 from the top and bottom... and you're done!

Attachments:

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sqing

#16 h Mar 13, 2017, 10:41 PM • 3 Y

Y by Adventure10, Mango247, alice211

Xixas wrote:

Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$ .

Kyrgyzstan national 2016

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42043 posts

sqing

#17 h Sep 4, 2017, 12:27 AM • 3 Y

Y by Adventure10, Mango247, AlexCenteno2007

Let $a,b,c$ be real numbers such that $a+b+c=0$ .Prove that $\frac{a^2+b^2+c^2}{2}\cdot \frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}$

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AlexCenteno2007

150 posts

AlexCenteno2007

#18 h Apr 17, 2025, 12:06 AM

Y by

sqing wrote:

Let $a,b,c$ be real numbers such that $a+b+c=0$ .Prove that $\frac{a^2+b^2+c^2}{2}\cdot \frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}$

Solution:
By elementary symmetric polynomials we have S1= $a+b+c=0$
S2=-2 $\sigma(2)$ and
S3=3 $\sigma(3)$
By substituting in a direct way we have to
$\frac{S2}{2}\cdot \frac{S2}{2}\cdot\frac{S3}{3}=\frac{a^7+b^7+c^7}{7}$ But I know that S7= $7\sigma(3)\cdot \sigma(2)^2$
From this the result is immediate.

This post has been edited 3 times. Last edited by AlexCenteno2007, Apr 17, 2025, 12:08 AM
Reason: Error

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42043 posts

sqing

#19 h Apr 17, 2025, 2:18 AM

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Thanks.

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