I was solving an exercise, so I realized that the one easiest way to do it is using a "weird", but nice identity below. I've tried to found out it on internet but I've founded nothingness, and I wondering how to show easily this identity ? As a matter of the fact, it's a beautiful identity.


The one way I think about this I'll let here :


Now, I'll take one of the term of RHS

$(a −b)(b +c)(c + a)\\ =(a −b)(b−c + 2c)(a −c + 2c)\\ =(a −b)(b −c)(a −c)+ 2c(a −b)(b−c + a −c + 2c) \\=(a −b)(b −c)(a −c)+ 2c(a −b)(a +b)$

Similarly, We'll do that with the remainder

$(b−c)(a+b)(c + a)+(c − a)(a +b)(b+ c)\\=(a +b)[(b −c)(c + a)+(c −a)(b +c)]\\=(a +b)(bc +ba −c^{2} −ca +cb+c^{2} −ab −ac)\\= (a +b)(2bc − 2ac)\\= −2c(a −b)(a +b)$

and we get : $$\boxed{\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}}$$

So it's too many work to show this identity. I just want to know if there's a simple way to show that or I don't know. I'm questing that because I didn't find anything on internet.

  • $\begingroup$ I doubt this has a nice proof and even if it has there is probably nothing illuminating about it. $\endgroup$ Dec 26 '13 at 23:36
  • $\begingroup$ @Anybody Please have a look at my answer also. $\endgroup$ Dec 27 '13 at 17:15
  • $\begingroup$ @Ewin Please read my solution also . $\endgroup$ Dec 28 '13 at 16:25

Let $x=\frac{a-b}{a+b}$ and so on. We then have: $(1-x)(1-y)(1-z)=(1+x)(1+y)(1+z)$

The terms with an even absolute degree will cancel, therefore: $x+y+z=-xyz$

  • $\begingroup$ +1. What prompted you to see the identity for $(x,y)$? $\endgroup$
    – Hans
    Feb 23 '19 at 1:48
  • $\begingroup$ @Hans I just noted that both $1+x$ and $1-x$ have a nice numerator. Then everything follows more or less naturally :) $\endgroup$
    – chubakueno
    Feb 15 at 1:22

Bringing all to one side and clearing denominators, we wish to prove zero

$$f(c) = (a\!-\!b)(b\!-\!c)(c\!-\!a)+(a\!-\!b)(b\!+\!c)(c\!+\!a)+(b\!-\!c)(a\!+\!b)(c\!+\!a) + (c\!-\!a)(a\!+\!b)(b\!+\!c)$$

Note $\ f(a) = (a\!-\!b)(b\!+\!a)2a\!+\!(b\!-\!a)(a\!+\!b)2a = 0.\ $ By symmetry $f(b) = 0.$

The coefficent of $c^2$ is $\ {-}(a\!-\!b) \!+\! (a\!-\!b) -(a\!+\!b)\!+\!(a\!+\!b) = 0.$

Thus $f(c)$ is a polynomial in $\,c\,$ of degree $\le 1$ with $\,2\,$ roots, so $f = 0$.



This equation is a cubic(degree 3) in "a".So at maximum it can have 3 roots if it is not an identity. But if you show that it has 4 roots , then it becomes an identity.

Roots to try :

b,c(easy ones),0(also easy),-a(think it over).

Now you have shown that it has 4 roots. Thus it becomes an identity.

If b,c are also variables same procedure can be applied for them also.

  • $\begingroup$ Could you please elaborate a little more what you have thought ? $\endgroup$
    – Ewin
    Dec 28 '13 at 17:05
  • $\begingroup$ An equation becomes an identity if it is valid for any value of the variable(not just specific values).Consider $x=x$. This is any identity since it is valid for all values of $x(1,3.25,\pi,etc.)$.Had it been an equation(which it is surely not) then at maximum it would have 1 root. But as soon as you show that it has 2 roots, then it becomes an identity.Please reread my answer as I have made some changes to it. $\endgroup$ Dec 29 '13 at 9:09
  • 1
    $\begingroup$ +1. Best solution. However, should it not be quadratic (degree 2) rather than cubic equation, after multiplying both sides by the denominator on the right hand side, in $a$? $\endgroup$
    – Hans
    Feb 16 at 0:03

I don't know if there is an easier solution but I think that an alternative solution for a,b,c positive real numbers, can be derived using the law of tangents. http://en.wikipedia.org/wiki/Law_of_tangents Of course this will not cover all cases since for a,b,c lengths of sides of a triangle we have $a+b>c$, $b+c>a$, $c+a>b$.

  • $\begingroup$ not all cases, when $a>b+c$, what can you do? $\endgroup$
    – chenbai
    Dec 27 '13 at 6:23

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