I just started learning Discrete Math and I have stumbled upon a question which has got me stumped and I honestly have no clue where to start except with the assumption that $a < x$ and $b < y$.
The question is:
If $a < x$ and $b < y$, then the rectangle with corners at $(x,y)$, $(x,0)$, $(0,y)$, and $(0,0)$ has an area greater than $ab$.
What are the length of the sides of the rectangle? If you plot them, it will be easy to see that the lengths are $x$ and $y$.
The area of a rectangle is base x height so the area of this rectangle is $xy$. We know $a < x$ and $b < y$. Let x = $a$ + $\alpha$ and y = $b$ + $\beta$ for some $\alpha$ and $\beta$ positive real numbers. Then we substitute the area $xy$ into ($a$ + $\alpha$)($b$ + $\beta$). Multiply this out and get that this is greater than $ab$.