lean_workbook_plus_70700

Note that $(a - 1)(b - 1) \ge 0$ since $-1 \le a \le 1$ and $-1 \le b \le 1$ (so that $a - 1 \le 0$ and $b - 1 \le 0$ ). Then $ab + 1 \ge a + b$ . But also $(a + 1)(b + 1) \ge 0$ by a similar argument, so $ab + 1 \ge -(a + b)$ . Then $ab + 1 \ge |a + b|$ . But $|ab + 1| \ge ab + 1$ , so we get $|ab + 1| \ge |a + b|$ . Note that equality holds if and only if $a = b = 1$ or $a = b = -1$ .