Prove that hat if $a,b \ge 2$ then $ab \ge a+b$ Prove that if $a,b \ge 2$ then $a+b \le ab$
so if $a \ge 2$ and $b \ge 2$ then $a-1 \ge 1$ and $b-1 \ge 1$
$(a-1)(b-1) \ge b-1$
$(a-1)(b-1) \ge 1$
$(a-1)(b-1) - 1 \ge 0$
$ab -a -b\ge 0$
$ab \ge a + b$
Thanks in advance, is this valid?

Let $a\equiv x+ 2,\,b\equiv y+ 2$
$$\therefore\,ab- a- b= xy+ x+ y\geqq 0$$
$$\because\,x,\,y\geqq 0$$
