Solve for real $a$ and $b$:

$$
a+b < ab+1
$$
The inequality is equivalent to $a b-a-b+1>0$, or $(a-1)(b-1)>0$. Thus, $(a-1)$ and $(b-1)$ must be non-zero and of the same sign. The solutions to this inequality are then the pairs $(a, b)$ such that either $a>1$ and $b>1$, or $a<1$ and $b<1$.