Mathematics Olympiad Question $a+b+c=7$, ...

Question

Given $a+b+c=7$ and $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} = 0.7$, need to find $\frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{a+c}$.

I have noted that these two differ by a factor of $10$. So I divided the first equation by $10$ and equated the two. But that did not lead me anywhere. I have also tried to multiply one by the other, but the result obtained is identical to just multiplying the second expression by $7$.

Answer 1

The trick it's just to add and subtract $1$ from each fraction and then to factorize:

$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a+b+c}{b+c}-1+\frac{a+b+c}{c+a}-1+\frac{a+b+c}{a+b}-1=(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})-3=7\cdot 0.7 -3=1.9$$

Answer 2

$$\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}$$ $$=\frac{7-(a+b)}{a+b}+\frac{7-(b+c)}{b+c}+\frac{7-(a+c)}{a+c}$$ $$=\frac{7}{a+b}+\frac{7}{b+c}+\frac{7}{a+c}-3$$ $$=7\cdot 0.7-3=1.9$$