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.

If $a and $b, then the rectangle with corners at (x,y), (x,0), (0,y), and (0,0) has an area greater than ab.
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Step 1
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 and $b. Let $x=a+\alpha$ and $y=b+\beta$ for some $\alpha$ and $\beta$ positive real numbers. Then we substitute the area xy into $\left(a+\alpha \right)\left(b+\beta \right)$. Multiply this out and get that this is greater than ab.