Algebra OSN

$\chi := \frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} = \frac{2010}{2011}$

$\chi$ is an improper fraction, where both the numerator and the denominator are polynomials of degree $3$ . So,

$\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} \equiv P + \frac{A}{a+b} + \frac{B}{b+c} + \frac{C}{c+a}$

To get $A$ , multiply the above equation by $(a+b)$ and set $b = -a$ .

$\begin{aligned} P(a+b) + A + \frac{B(a+b)}{b+c} + \frac{C(a+b)}{c+a} &= \frac{(a-b)(b-c)(c-a)}{(b+c)(c+a)} \\ \implies A &= \frac{(a+a)(-a-c)(c-a)}{(-a+c)(c+a)} \\ \therefore A &= -2a \end{aligned}$

Similarly, $B = -2b, \> \> C = -2c$

To get $P$ , multiply the equation by $(a+b)(b+c)(c+a)$ and set $a=b=c$ .

$\begin{aligned} P(a+b)(b+c)(c+a) + A (b+c)(c+a) + B (a+b)(c+a) + C (a+b)(b+c) &= (a-b)(b-c)(c-a) \\ P(a+b)(b+c)(c+a) - 2a(b+c)(c+a) - 2b(a+b)(c+a) - 2c(a+b)(b+c) &= (a-b)(b-c)(c-a) \\ P(8c) - 8c - 8c - 8c &= 0\\ \implies P &= 3 \end{aligned}$

Therefore,

$\begin{aligned} \frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} &= 3 - \frac{2a}{a+b} - \frac{2b}{b+c} - \frac{2c}{c+a} \\ \chi &= 3 - 2 \Bigg( \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} \Bigg) \\ \implies \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} &= \frac{3 - \chi}{2} \\ &= \frac{3-\frac{2010}{2011} }{2} \\ &= \boxed{\frac{4023}{4022}} \end{aligned}$

Given that

$\begin{aligned} \frac {(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} & = \frac {2010}{2011} \\ \frac {ab^2+bc^2+ca^2 - a^2b-b^2c-c^2a}{ab^2+bc^2+ca^2 + a^2b+b^2c+c^2a+2abc} & = \frac {2010}{2011} \\ \frac {\sum ab^2 - \sum a^2b} {\sum ab^2 + \sum a^2b +2abc} & = \frac {2010}{2011} \\ 2011\sum ab^2 - 2011 \sum a^2b & = 2010 \sum ab^2 + 2010 \sum a^2b +4020 abc \\ \implies \sum ab^2 & = 4021 \sum a^2b +4020 abc \end{aligned}$

Then we have:

$\begin{aligned} \frac a{a+b} + \frac b{b+c} + \frac c{c+a} & = \frac {\sum ab^2 + 2\sum a^2b + 3abc} {\sum ab^2 + \sum a^2b + 2abc} \\ & = \frac {4023\sum a^2b + 4023abc}{4022\sum a^2b + 4022abc} \\ & = \frac {4023}{4022} \\ & \approx \boxed{1.00025} \end{aligned}$

Note: This is a cheeky solution:) Pick $a=0$ , so we have $\frac{c-b}{b+c}=\frac{2010}{2011}\implies c=4021b$ then it is suffices to compute $\frac{b}{b+c}+1=\frac{1}{4022}+1=\boxed{\frac{4023}{4022}}$

There are in fact three solutions to the first equation. Unfortunately, none of solutions determine fixed values for the variables. Fortunately, all three solutions, when substituted into the second equation and the resultant expression is simplified, the same determined value results: $\frac{4023}{4022}\approx 1.0002486325211337642963699651914470412729985082049$ .

The three solutions:

$\{b\to 4021 a,c\to 0\}, \\ \left\{c\to \frac{-\sqrt{16184526 a^2 b^2-64681804 a^3 b+a^4+32328836 a b^3+16168441 b^4}+a^2-4020 a b-4021 b^2}{2 (4021 a-b)}\right\}, \\ \left\{c\to \frac{\sqrt{16184526 a^2 b^2-64681804 a^3 b+a^4+32328836 a b^3+16168441 b^4}+a^2-4020 a b-4021 b^2}{2 (4021 a-b)}\right\}$

When solving the expressions, the variables were assumed to be complex.