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.

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

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

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

Another solution..

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

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.

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

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.

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 ?

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

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

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

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!

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

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