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

Multiplying the given expressions together:

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

$$\Longrightarrow \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} = \frac{17}{10}$$

$$a+b+c=6\tag{1}$$

$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{47}{60}\tag{2}$$

$$S = \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$

$S = \left(\frac{a+b+c}{b+c}-1\right)+\left(\frac{a+b+c}{c+a}-1\right)+\left(\frac{a+b+c}{a+b}-1\right) =6\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3=6\cdot\frac{47}{60}-3$

$S =\frac{47}{10}-3 =\frac{17}{10}$

Hint:$$\frac{a+b+c}{b+c}=1+\frac{a}{b+c}$$