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.
Multiplying the given expressions together:

\begin{align}
\frac{47}{10} &amp;= (a+b+c)\bigg(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\bigg)   \\ \\
&amp;= \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a} \\ \\
&amp;=3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} \\ \\
\end{align}

$$\Longrightarrow \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} = \frac{17}{10}$$
The only possible numbers a, b, c that add up to 6 are 1, 2, 3.

Substituting these in the first equation gives you 47/60 so they are the correct values and there is no repeated number for example a and b both being 1 and c being 4.

So simply substituting in the second equation gives you 17/10 immediately.

The three fractions in each equation are all the available permutations.

There is no need for any other manipulations.