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Is this true!

Given $a,b>0$, real numbers, then $$a+b\leq ab$$ If not, when this could be true?

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Clearly, if $a=1$ or $b=1$, it can't be true. You then check the cases of, say, $a > 1$ and $a < 1$... – J. M. Oct 29 '11 at 2:22
Just an interesting comment. This can be extended to complex numbers too. See $(1+i)+(1-i)=(1+i)(1-i)$ and $2=2$. – GarouDan Oct 29 '11 at 2:32
$(a_1+a_2i)+(b_1+b_2i) \leq (a_1+a_2i)(b_1+b_2i)$ $(a_1+b_1)+(a_2+b_2)i \leq (a_1b_1-a_2b_2)+(a_1b_2+a_2b_1)i$ $(a_1+b_1-a_1b_1+a_2b_2)+(a_2+b_2-a_1b_2-a_2b_1)i \leq 0$ But, this inequation just holds if $(a_2+b_2-a_1b_2-a_2b_1)=0$ and we have too $(a_1+b_1-a_1b_1+a_2b_2) \leq 0$ – GarouDan Oct 29 '11 at 2:40

Rewrite the inequality as $(a-1)(b-1) \ge 1$. Now can you take it from here?

Comment: One could call the idea completing the rectangle. It is a lot less useful than completing the square!

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You could show the inequality graphically as belowplot

Note: White space correspond to x, y values not satisfying the inequality

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I assume $a$ and $b$ are both positive real numbers, $a+b\leq ab$.

Suppose $a>1$ then $a-1>0$ from $a+b\leq ab$, $ab-b\geq a$

So, $b(a-1)\geq a$. That is $b\geq a/(a-1)$.

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if $a=0.5, b=0.5$ then $1> ab=0.25$! – Monica Oct 29 '11 at 5:18
@Monica, check my proof, I have cleared my mistake. – Hassan Muhammad Oct 29 '11 at 6:04

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