Assume $R$ is a ring, I am to show that $1+rs$ is invertible if and only if $1+sr$ is invertible.
I am not sure what to do. Someone gave me a hint that $s(1+rs)=(1+sr)s$, however I've not been able to use this tip at all.
So far my only intuition is that I must find that using the inverse of the left element, I can make the second element equal $1$, but the arithmetic doesn't simplify itself.
HINT:
The question may well be a duplicate, apologies in that case.
Learned this trick a long time ago from a professional. Formally, one can write $$(1+sr)^{-1}=1 - s r + sr sr -\cdots= 1- s( 1- r s + r s r s-\cdots )r$$ so formally $$(1+sr)^{-1} = 1-s(1+rs)^{-1}r$$
So now one checks directly that if $(1+rs)$ is invertible then the inverse of $(1+sr)$ is indeed $1-s(1+rs)^{-1}r$.
ADDED:
For my own use, let's check that.
$$(1+sr)( 1-s(1+rs)^{-1}r ) = 1 + sr -s(1+rs)^{-1}r-sr s(1+rs)^{-1}r= \\ = 1 + s(1-(1+rs)(1+rs)^{-1})r = 1 + s(1-1)r =1 + 0 = 1$$