Can area of rectangle be greater than the square of its diagonal?

The area of the square built on the diagonal must be at least twice the area of the rectangle:

$\hskip 4 cm$

Another proof without words, at the suggestion of Semiclassical:

The dark rectangle has some fixed diagonal $d$. The large square has area $d^2$.

A simple explanation without proof or pictures:

The diagonal of a rectangle is at least as long as each of its sides, so the square of the diagonal must be at least the product of the sides.

In fact, the squared diagonal must be at least twice the area, i.e. $a^2+b^2\ge 2ab$ if orthogonal sides' lengths are $a,\,b$. Why? Because the difference is $(a-b)^2\ge 0$.