Let $\mu_1$ and $\mu_2$ and $\nu$ be two finite measures on the measurable space $(\Omega, \mathcal{S})$ such that $\nu \ll \mu_i$, $i = 1, 2$, and $f_i = \frac{d\nu}{d\mu_i}$. Show that then $\nu \ll \mu_1 + \mu_2$ and $$ \frac{d\nu}{d(\mu_1 + \mu_2)} = \frac{f_1 \cdot f_2}{f_1 + f_2} $$
I have managed to prove the first part of it, but i have some question for the second part. Just by writing the definition of the f i and transforming the eqution i get to. $$ \frac{d\nu}{d(\mu_1) + d(\mu_2)} $$ is this equal to the statment above?
Hint: $\int_E (\frac 1 {f_1}+\frac 1{f_2}) d\nu=\int_E \frac 1 {f_1} f_1d\mu_1+\int_E \frac 1 {f_1} f_1d\mu_2=\mu_1(E)+\mu_2(E)$ and this implies $\frac {d\nu}{d(\mu_1+\mu_2)}=\frac 1 {\frac 1 {f_1}+\frac 1{f_2}}=\frac {f_1f_2}{f_1+f_2}$.