Solution
= Solution
Evaluate the <absolute values> on the three intervals separated by one and two:
$$
|x-2|+|x-1|=\begin{cases}3-2x,&x<1,\\1,&1\leq x\leq2,\\2x-3,&x>2.\end{cases}
$$
The strict inequality therefore holds precisely for
$$
\boxed{x<1\quad\text{or}\quad x>2.}
$$