Find $S = \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b}$ if values of $a+b+c$ and $\frac1{a+b}+\frac1{b+c}+\frac1{a+c}$ are given I just stumbled upon a contest question from last year's city olympiad math contest:
Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{47}{60}$, find the value of $S = \dfrac{a}{b+c}+\dfrac{b}{c+a} + \dfrac{c}{a+b}$.
Since I just saw it from an online forum "elsewhere", I thought I'd want to hear from other more skilled and experienced MSE members about your tactics and approaches to the solution of this interesting question.

Even if the Joshua's answer is probably the most elegant solution, let me offer you an alternative way of solving it, that is through simplification due to arbitrary values.
You have three unknowns and only two equations. This allows you to get rid of one variable, for example, c = 0. (Note that only one can be zero).
Hence, the problem can be rewritten as:
1) $a + b = 6$
2) $\dfrac{1}{a + b} + \dfrac{1}{b} + \dfrac{1}{a} = \dfrac{1}{a + b} + \dfrac{a + b}{ab} = \dfrac{47}{60}$
$S = \dfrac{a}{b} + \dfrac{b}{a} = \dfrac{a^2 + b^2}{ab} = \dfrac{(a + b)^2 - 2ab}{ab}$.
At this point, you can clearly either find both the values of $a$ and $b$ (annoying) or trick a little more.
Substituting $x = a + b$ and $y = ab$, we have the perfectly equivalent problem
1) $x = 6$
2) $\dfrac{1}{x} + \dfrac{x}{y} = \dfrac{47}{60}$
$S = \dfrac{x^2 - 2y}{y}$.
Substituting the first equation to the second, we get $\dfrac{6}{y} = \dfrac{47}{60} - \dfrac{1}{6} = \dfrac{37}{60}$ and $y = \dfrac{10\cdot6^2}{37}$.
Finally, $S = \dfrac{x^2 - 2y}{y} = \left(6^2-2\cdot\dfrac{10\cdot6^2}{37}\right)\dfrac{37}{10\cdot6^2} = \left(1-\dfrac{20}{37}\right)\dfrac{37}{10} = \dfrac{37-20}{37} \cdot \dfrac{37}{10} = \dfrac{17}{10}$.
Answer $S = \dfrac{17}{10}$.
