Is this true!
Given $a,b>0$, real numbers, then $a+b\leq ab$ If not, when this could be true?
Is this true!
Given $a,b>0$, real numbers, then $a+b\leq ab$ If not, when this could be true?
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!
You could show the inequality graphically as below
Note: White space correspond to x, y values not satisfying the inequality
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 $b\geq a/(a-1)$.