How to find $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$? For numbers $a,b,c$ we know that $\frac{a-c}{b+c}+\frac{b-a}{c+a}+\frac{c-b}{a+b}=1$ What is the value of $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$ ?
$1)3\qquad\qquad2)4\qquad\qquad3)6\qquad\qquad4)2$
Because it is a multiple choice question at first I tried to find $a,b,c$ so that each fraction be equal to $\frac13$ and their sum be $1$. but unfortunately I couldn't find such numbers.
I think adding the fractions by finding common denominator is not a good idea because the numerator will be so ugly. but I can rewrite it as:
$$\frac{a}{b+c}-\frac c{b+c}+\frac b{c+a}-\frac{a}{c+a}+\frac{c}{a+b}-\frac{b}{a+b}=1$$
But I can't continue from here.

$\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} = \frac{a+b+c}{b+c} - \frac{c}{b+c} +\frac{b+c+a}{c+a} -\frac{a}{c+a}+\frac{c+a+b}{a+b}  - \frac{b}{a+b}$
$= \frac{a}{b+c} + 1 - \frac{c}{b+c}+ \frac{b}{c+a} + 1 -\frac{a}{c+a} + \frac{c}{a+b} + 1 - \frac{b}{a+b}$
$= \frac{a-c}{b+c}+1 + \frac{b-a}{c+a}+1 + \frac{c-b}{a+b} +1$
$ = 4$
