Suppose that $a_n$ and $b_n$ are two real sequences satisfying $a_nb_n\to 1$ and $a_n+b_n\to 2$. Does it follow that $a_n$ and $b_n$ both converge to $1$?
I was working on this problem as one of the exercises from this online source of problems in analysis. I have not been able to find a counterexample, but I have deduced some necessary conditions on $a_n,b_n$:
Eventually $a_n$ and $b_n$ have the same sign, and this must be $>0$:
If they do not eventually share the same sign, then $a_nb_n$ can't converge to $1$. If they are not both positive, their sum can't converge to $2$.
There exists $t>0$ and $N$ such that $a_n>t$ for all $n\ge N$. [Hence the same conclusion holds for $b_n$.]:
If not, there is some subsequence $a_{n_k}$ with $a_{n_k}\to 0$. But then $b_{n_k}\to\infty$ as $a_{n_k}b_{n_k}\to1$. Then $a_{n_k}+b_{n_k}\to\infty$, absurd.
$a_n$ and $b_n$ are both bounded above:
Otherwise, there must be a subsequence of the other that tends to $0$, which is impossible by 2.
I imagine a counterexample would be very strange, but I feel like a proof is also not too far out of reach.
Thanks for any suggestions.
P.S. Does the conclusion change if we allow complex sequences $a_n,b_n$?