Question 25
Three real numbers $a$, $b$ and $c$ are such that
$$a+b+c=4\quad\text{and}\quad\frac1{a+b}+\frac1{b+c}+\frac1{c+a}=5.$$
Then $\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}$ is equal to
$$a+b+c=4\quad\text{and}\quad\frac1{a+b}+\frac1{b+c}+\frac1{c+a}=5.$$
Then $\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}$ is equal to