<s>[INST] Given $\frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)}=\frac{1}{11}$. Find $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}.$ 
Given: $\{a,b,c\}\subset \Bbb R$, $\frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)}=\frac{1}{11}\ \ (1)$. Find the value of $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ \ (2).$

This question appeared in OBM 2005, the Brazilian Math Olympiad. The answer is $\frac{17}{11}$. But my solution is, perhaps, unnecessarily complicated as involves a lot of algebra. Is there an easy way to solve?
My attempt (outline): First, I developed expression (1), getting
$$abc=5a^2c+5ab^2+5bc^2-6a^2b-6b^2c-6c^2a\ \ (3)$$
Then, developed (2) to get
$$(2)=\frac{{3abc + 2a^2 b + 2ac^2 + 2b^2 c + a^2 c + ab^2 + bc^2 }}{{2abc + a^2 b + ac^2 + a^2 c +b^2 c + ab^2 + bc^2 }}\ (4)$$
Now, replace $abc$ definition from (3) into (4), and after some additional algebra, get 
$$\frac{{17a^2 c + 17ab^2 + 17bc^2 - 17a^2 b - 17ac^2 - 17b^2 c}}{{11a^2 c + 11ab^2 + 11bc^2 - 11a^2 b - 11ac^2 - 11b^2 c}}=\frac{17}{11}$$
Question: is there a simpler way to get to the result? Simpler answers/helpful hints are welcomed.
 [/INST] Set $$X = \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{a+c}$$ and $$Y = \frac{b}{a+b} + \frac{c}{b+c} + \frac{a}{a+c}.$$ Then evidently $X+Y = 3$ and we may compute that $$X - Y = \frac{a-b}{a+b} + \frac{b-c}{b+c} + \frac{c-a}{a+c} = \frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(a+c)} = \frac{1}{11}$$ Solving this system of linear equations, it follows that $$ (X,Y) = \left(\frac{17}{11},  \frac{16}{11}\right)$$
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