This feels like a way too easy question, but I've been sitting on this problem for a few days now, and can't make much progress. We define a valuation ring by two equivalent properties: For an integral domain A and its field of fractions K we have
a) for every $x \in K^*$, one has $x\in A$ or $x^{-1}\in A$ and
b) the partially ordered group $(\Gamma, \geq) = (K^*/A^*, x \geq y$ iff $ xy^{-1} \in A)$ is totally ordered and abelian.
Then we call A a valuation ring. We define a valuation $v: K \rightarrow \Gamma \cup \infty$ , $ x \mapsto xA^*$ and $0 \mapsto \infty$. I have already shown that $v(x+y) \geq min(v(x),v(y))$. Now I want to show that v is additive, ie that $v(xy) = v(x) + v(y)$. We can assume that either $v(xy) \geq v(x) + v(y)$, or the converse, since $\Gamma$ is totally ordered, and then try to prove the other direction. So I started with assuming $v(xy) \geq v(x) + v(y)$. Now, I have trouble formulating this relation in terms of the definition of the $\geq$ - relation. I tried writing this as $xy(x+y)^{-1} \in A$, but that didn't help me in proving that $(x+y)y^{-1}x^{-1}\in A$. So I figured I'd made a mistake: because if I assume $v(xy) \geq v(x+y)$, THAT should be translated to $xy(x+y)^{-1} \in A$, and $v(x+y)$ and $v(x)+v(y)$ are two different things.
So now how do I go about this? What would be the correct translation of my assumption that $v(xy) \geq v(x) + v(y)$?
Turning my comment into an answer to get it off the unanswered list. The group structure on $\Gamma$ that we denote $+$ is in fact not defined as $xA^* + yA^* = (x + y)A^*$, but as $xA^* + yA^* = (xy)A^*$. With that definition, the fact that $v(xy) = v(x) + v(y)$ is fairly straightforward.
As to whether multiplicative notation would be more sensible, I guess that makes sense. One source of valuations (though they're typically Archimedean) is by taking an absolute value $|\cdot|$ on your ring, and writing $v(x) = e^{-|x|}$. (I suppose this is why valuations are also called exponential valuations.) The group $\Gamma$ is then $\mathbb{R}_{>0}$ with the multiplicative structure. Surely a $\cdot$ would be more sensible than a $+$ here.
But of course, it's all just convention, anyway. :)