Prove that hat if $a,b \ge 2$ then $ab \ge a+b$

Question

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?

Personally, I'd say "WLOG assume $a\geq b$. Then $a+b\leq a+a=a\times 2\leq ab$" where we used the fact that $b\geq 2$ in the last inequality. — JMoravitz, May 24 '19 at 15:07

score 2 · Answer 1 · answered May 24 '19 at 15:05

2

I would write $$a(b-1)\geq b$$ and this is $$a(b-1)\geq b-1+1$$ or $$(b-1)(a-1)\geq 1$$ this is true since $$a-1\geq 1$$ and $$b-1\geq 1$$

answered May 24 '19 at 15:05

score 2 · Accepted Answer · answered May 24 '19 at 15:12

2

$$ ab\ge 2a$$

$$ab\ge 2b$$

Add both sides

$$2ab \ge 2a+2b$$ $$ ab\ge a+b $$

answered May 24 '19 at 15:12

score 0 · Answer 3 · answered May 25 '19 at 09:12

0

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

answered May 25 '19 at 09:12

3 Answers3