If $V,W$ are valuation domains, then $V+W$ is a ring. I am stuck at solving the following exercize:

Let $V,W$ be two valuation domains with the same fraction field $Q$. Then $V+W$ is a subring of $Q$ (hence it is the smallest subring of $Q$ containing both $V,W$).

There is also a hint: use the fact that $ab = (a+b)(a^{-1}+b^{-1})^{-1}$.
Here is what I have done so far:
it is sufficient to show that for all $a \in V$, $b \in W$ we have $ab \in V+W$. So, argue by contradiction and suppose that $ab \notin V + W$. Then $ab \notin V$ and $ab \notin W$. This implies that $(ab)^{-1} \in V \cap W$.
Moreover we have $a \notin W$, and $b \notin V$: this implies that $a^{-1} \in W$ and $b^{-1} \in V$. Hence $a^{-1}+b^{-1} \in V+W$.
But then I got stuck. There must be some easy algebraic manipulation to conclude, but I don't get it.
 A: It suffices to show that $(a^{-1}+b^{-1})^{-1}\in V\cap W$ if $a\in V\setminus W$ and $b\in W\setminus V$.  If, say, $(a^{-1}+b^{-1})^{-1}\not\in V$, then $a^{-1}+b^{-1}$ is in the maximal ideal of $V$.  But since $b\not\in V$, $b^{-1}$ is also in the maximal ideal of $V$, so this would imply that $a^{-1}$ is in the maximal ideal of $V$.  This contradicts the assumption that $a\in V$.
You may get some better intuition for this if you imagine that $Q$ consists of some sort of "meromorphic functions" on a set $X$, and there are points $x,y\in X$ such that $V$ is the set of functions that do not have a pole at $x$ and $W$ is the set of functions that do not have a pole at $y$.  Then if $a\in V\setminus W$, $a$ has a pole at $y$ but not $x$, so $a^{-1}$ vanishes at $y$ but not at $x$.  Similarly, if $b\in W\setminus V$, $b^{-1}$ vanishes at $x$ but not at $y$.  It is then clear that $a^{-1}+b^{-1}$ cannot vanish at either $x$ or $y$, so $(a^{-1}+b^{-1})^{-1}$ does not have a pole at either $x$ or $y$.
