TITLE: Does $a(a^{-1} \bmod b) + b(b^{-1} \bmod a) < ab + 2$ hold?
QUESTION [5 upvotes]: Here we assume $(a,b) = 1$ and $a^{-1}$ and $b^{-1}$ are canonical (e.g.  $0 < a^{-1} < b)$.
Counter example or proof are appreciated!

REPLY [3 votes]: Hint. Let $x=a(a^{-1} \bmod b) + b(b^{-1} \bmod a) $.
Then $x\equiv 1\pmod{ab}$ and $x<2ab$.