How can I prove this for $a+b+c=0$

Question

We wish to prove that

$$\left(\frac{a-b}{c}+\ \frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9,$$ for $a+b+c=0$ and $$abc \neq 0 , a\neq b , b\neq c, c \neq a.$$ I tried working with only $2$ fractions at a time, switching some of the variables (ex.$a=-b-c$) and then adding them up, and multiplying first but could not go further.

Thank you in advance.

Answer 1

You can expand directly, by noticing that the first parantheses is $$\frac{ab(a-b) + bc(b-c) + ca(c-a)}{abc} = -\frac{(a-b)(b-c)(c-a)}{abc}$$

So the expression becomes

$$\frac{a^3+b^3+c^3 + 3abc - a^2b - a^2c - b^2a -b^2c - c^2a - c^2b}{abc}$$

Collecting $a^3, -a^2b, - a^2c $ and similarly this becomes

$$\frac{a^2(a - b -c) + b^2(b-c-a) + c^2(c - a -b)}{abc} + 3$$

Now $a-b-c = 2a$ by the condition so this becomes just

$$2\frac{a^3+b^3+c^3}{abc} + 3$$

But there is an identity $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$, so we get $a^3+b^3+c^3 = 3abc$.

Thus we get that the initial expression is $9$.

Answer 2

Comment:As commented user376343 we have:

$$B=-A\times A=-[a+b+c-(\frac ab+\frac bc+\frac ca)]^2=-(\frac ab+\frac bc+\frac ca)^2$$

$$\frac ab+\frac bc+\frac ca=\frac{ca^2+ab^2+bc^2}{abc}\geq3$$

This inequality becomes equality if $a=1, b=2$ and $c=3$. I checked it for $a+b+c=0$ and it is not equal to 3. It is a negative value.I think correct constrain is : a, b and c positive and product must have a negative sign and the sign of inequality must be $\geq$.