8
$\begingroup$

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

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.

$\endgroup$
  • $\begingroup$ Shoudn't you have the constraint here such that $a=b=c\ne 0$? $\endgroup$ – homegrown Feb 27 '14 at 19:53
  • $\begingroup$ @jnh Yes, I missed it. Thank you for identification. $\endgroup$ – Hawk Feb 27 '14 at 19:54
8
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Very nice explanation +1 $\endgroup$ – homegrown Feb 27 '14 at 20:30
  • $\begingroup$ @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... $\endgroup$ – abiessu Feb 27 '14 at 20:31
  • $\begingroup$ @abiessu Yes, but the better the explanation behind the idea, the better the solution. $\endgroup$ – homegrown Feb 27 '14 at 20:33
  • $\begingroup$ Thank you...the most simplest approach...but an elegant solution. $\endgroup$ – Hawk Feb 28 '14 at 4:24
7
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Are you sure about this? $\endgroup$ – Hawk Feb 27 '14 at 20:11
  • $\begingroup$ @Hawk Absolutely sure. $\endgroup$ – ElThor Feb 27 '14 at 20:12
  • 1
    $\begingroup$ @ElThor What is the intuition behind this idea? $\endgroup$ – homegrown Feb 27 '14 at 20:16
  • $\begingroup$ @ElThor How does it help answer the original question ? $\endgroup$ – Ewan Delanoy Feb 27 '14 at 20:30
  • 2
    $\begingroup$ The worst solution so far suddenly becomes the best solution ! Unexpected development ... $\endgroup$ – Ewan Delanoy Feb 27 '14 at 20:40
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you...this is a very good solution, but a little complicated (+1) $\endgroup$ – Hawk Feb 28 '14 at 4:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.