If $a,b,c\in \mathbb R$ and, $$\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\right)=-27$$ Evaluate $$\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$ I noticed that you can reduce the question to the following:
If $(1-2x)(1-2y)(1-2z)=-27$, then evaluate $x+y+z$ where $x=\frac{b}{a+b}, \ y=\frac{c}{b+c}, \ z=\frac{a}{c+a}$
I'm not sure how to solve this. I think it may use inequalities or pure algebra.
Please help me out here. Thanks a lot :)
Let's call $\begin{cases} U=\frac a{a+b}+\frac b{b+c}+\frac c{c+a}\\V=\frac b{a+b}+\frac c{b+c}+\frac a{c+a}\end{cases}$
We have trivially $U+V=3$
Now $U-V=\frac {a-b}{a+b}+\frac {b-c}{b+c}+\frac {c-a}{c+a}=\cdots=-\frac {a-b}{a+b} \times\frac {b-c}{b+c}\times\frac {c-a}{c+a}=27$
Therefore $U=15$ and $V=-12$.
let me continue from your work.
Note that $$\frac{1}{x}-1=\frac{a}{b} $$ so $$\frac{(1-x)(1-y)(1-z)}{xyz}=1$$
$$\Rightarrow 1-(x+y+z)+xy+yz+xz-2xyz=0$$
$$\Rightarrow xy+xz+yz-2xyz=(x+y+z)-1$$
Also $$(1-2x)(1-2y)(1-2z)=-27$$
$$\Rightarrow 1-2(x+y+z)+4(xy+yz+xz)-8xyz=-27$$
$$1-2(x+y+z)+4(x+y+z)-4=-27$$
$$\Rightarrow x+y+z=-12$$
