Let $(X,+)$ be an abelian group and $d$ a metric on $X$. Suppose $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences. What conditions on the relation between the group operation and the metric are sufficient to show that $\{a_n+b_n\}$ is Cauchy? The proof in $\mathbb{R}$ simply relies on the fact that the Euclidean metric is induced by a norm: $d(a_m+b_m,a_n+b_n)=\|(a_m+b_m)-(a_n+b_n)\|\leq\|a_m-a_n\|+\|b_m-b_n\|$ I believe this proof works so long as the triangle inequality relates $+$ and the norm which induces $d$.
I think a weaker assumption which still works is if the metric is translation invariant. Then,
$\begin{align*} d(a_m+b_m,a_n+b_n) &\leq d(a_m+b_m,a_n+b_m) + d(a_n+b_m,a_n+b_n) \\ &= d(a_m,a_n) + d(b_m,b_n) \end{align*}$
Are there weaker assumptions under which the sum is Cauchy? What is an example in this case? Is there an example of a space and Cauchy sequences whose sum is not Cauchy?
This is probably not a standard way of phrasing the question. I'm just trying to strip down assumptions to clarify my understanding. I thought about this for a while but was having trouble coming up with examples.