If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$

Question

I could not approach the problem at all though I think I could have done something by using Cauchy-Schwarz inequality, but could not pull this together. Please help.

Shoudn't you have the constraint here such that $a=b=c\ne 0$? — homegrown, Feb 27 '14 at 19:53

score 8 · Accepted Answer · answered Feb 27 '14 at 20:26

8

Assuming

$$\frac a{b-c}+\frac b{c-a}+\frac c{a-b}=0$$

we also have

$$\frac a{(b-c)^2}+\frac b{(c-a)(b-c)}+\frac c{(a-b)(b-c)}=0$$

as well as

$$\frac a{(b-c)(a-b)}+\frac b{(c-a)(a-b)}+\frac c{(a-b)^2}=0$$

and

$$\frac a{(b-c)(c-a)}+\frac b{(c-a)^2}+\frac c{(a-b)(c-a)}=0$$

These three sum together as

$$\frac a{(b-c)^2}+\frac b{(c-a)^2}+\frac c{(a-b)^2}\\+\frac {a+c}{(a-b)(b-c)}+\frac {a+b}{(c-a)(b-c)}+\frac{b+c}{(c-a)(a-b)}=0\tag 1$$

With the three later fractions, we can multiply by $1$ as follows:

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

Doing this to each yields

$$\frac {c^2-a^2}{(a-b)(b-c)(c-a)}+\frac {b^2-c^2}{(a-b)(b-c)(c-a)}+\frac {a^2-b^2}{(a-b)(b-c)(c-a)}=0$$

This equivalence should be clear by inspection, which transforms $(1)$ into

$$\frac a{(b-c)^2}+\frac b{(c-a)^2}+\frac c{(a-b)^2} = 0$$

answered Feb 27 '14 at 20:26

abiessu

8,115

1

Very nice explanation +1 – homegrown Feb 27 '14 at 20:30
@jnh: thank you; note that after I posted this I realized that !ElThor also happens to have recommended the same approach without quite so much white space... – abiessu Feb 27 '14 at 20:31
@abiessu Yes, but the better the explanation behind the idea, the better the solution. – homegrown Feb 27 '14 at 20:33
Thank you...the most simplest approach...but an elegant solution. – Hawk Feb 28 '14 at 04:24

ElThor · Answer 2 · 2014-02-27T21:10:43.147

7

You only need to prove that $$ \begin{multline} \left(\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b}\right)\left(\frac{1}{b - c} + \frac{1}{c - a} + \frac{1}{a - b}\right) \\ = \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \end{multline}$$

Where did that come from?
Well, consider this simple equality $$ \frac{a}{b - c}\left(\color{green}{\frac{1}{c - a} + \frac{1}{a - b}} + \frac{1}{b - c}\right) = \color{green}{\frac{ac - ab}{(a - b)(b - c)(c - a)}} + \frac{a}{(b - c)^2} $$

When you take cyclic sum of left and right hand sides, the first fraction in RHS dissapears.

In other words, we can write 2 similar ones: $$ \begin{align} \frac{b}{c - a}\left(\frac{1}{b - c} + \frac{1}{a - b} + \frac{1}{c - a}\right) = \frac{ab - bc}{(a - b)(b - c)(c - a)} + \frac{b}{(c - a)^2} \\ \frac{c}{a - b}\left(\frac{1}{b - c} + \frac{1}{c - a} + \frac{1}{a - b}\right) = \frac{bc - ac}{(a - b)(b - c)(c - a)} + \frac{c}{(a - b)^2} \end{align} $$

Three last equations add up to the first equation since those big fractions right after the equal sign cancel out.

edited Feb 27 '14 at 21:10

answered Feb 27 '14 at 20:07

ElThor

912

Are you sure about this? – Hawk Feb 27 '14 at 20:11
@Hawk Absolutely sure. – ElThor Feb 27 '14 at 20:12
1

@ElThor What is the intuition behind this idea? – homegrown Feb 27 '14 at 20:16
@ElThor How does it help answer the original question ? – Ewan Delanoy Feb 27 '14 at 20:30
@EwanDelanoy Actually I made a mistake. Now it should be correct. – ElThor Feb 27 '14 at 20:35
2

The worst solution so far suddenly becomes the best solution ! Unexpected development ... – Ewan Delanoy Feb 27 '14 at 20:40

score 3 · Answer 3 · answered Feb 27 '14 at 20:26

Let

$$ T_1=\sum_{cyc} \frac{a}{b-c} $$

By hypothesis we have $T_1=0$, but if we put $T_2=T_1(ab+ac+bc-a^2-b^2-c^2)$, we have

$$ T_2=\sum_{cyc} \frac{a(ab+ac+bc-a^2-b^2-c^2)}{b-c} $$

$$ T_2=\sum_{cyc} \frac{a^2b}{b-c}+ \sum_{cyc} \frac{a^2c}{b-c}+ \sum_{cyc} \frac{abc}{b-c}- \sum_{cyc} \frac{a^3}{b-c}- \sum_{cyc} \frac{ab^2}{b-c}- \sum_{cyc} \frac{ac^2}{b-c} $$

$$ T_2=\sum_{cyc} \frac{a^2b}{b-c}+ \sum_{cyc} \frac{a^2b}{c-b}+ \sum_{cyc} \frac{abc}{b-c}- \sum_{cyc} \frac{a^3}{b-c}- \sum_{cyc} \frac{ab^2}{b-c}- \sum_{cyc} \frac{ab^2}{c-b} $$

$$ T_2= \sum_{cyc} \frac{abc}{b-c}- \sum_{cyc} \frac{a^3}{b-c} =\sum_{cyc} \frac{a(bc-a^2)}{b-c} $$

$$ T_2=\sum_{cyc} \frac{a(bc-a^2)}{b-c} +\sum_{cyc} \frac{-a^2b}{b-c} +\sum_{cyc} \frac{-a^2c}{b-c} $$

and hence

$$ T_2=\sum_{cyc} \frac{a(b-a)(c-a)}{b-c}= -\sum_{cyc} \frac{a(a-b)(c-a)}{b-c} $$

So if we put $T_3=\frac{T_2}{(a-b)(b-c)(c-a)}$, we see that

$$ T_3=-\sum_{cyc} \frac{a}{(b-c)^2} $$

and we are done.

Thank you...this is a very good solution, but a little complicated (+1) — Hawk, Feb 28 '14 at 04:23

If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$

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