<s>[INST] 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.
 [/INST] $$a+b+c=6\tag{1}$$
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{47}{60}\tag{2}$$
$$S = \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$
$S = \left(\frac{a+b+c}{b+c}-1\right)+\left(\frac{a+b+c}{c+a}-1\right)+\left(\frac{a+b+c}{a+b}-1\right) =6\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3=6\cdot\frac{47}{60}-3$
$S =\frac{47}{10}-3 =\frac{17}{10}$
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