What I've done.
I figured that to make the rectangles perimeter greater than $4r$, I would have to choose a length that is possible of that. Let the area of a rectangle equal $r$.
$$(l*w)= r$$
then we know the perimeter has to be greater than $4r$ so
we let $p 4r$, and substitute for $p$ and $r$ to get
$2(w+l) > 4(w*l)$ solve for $l$ to get
$$w+l > 2(w*l)$$
$$w > 2wl-l$$
$$w > l(2w-1)$$
$$ \frac wl > 2w-1$$
$$ \frac1l > \frac{(2w-1)}{w} $$
$$l < - \frac{w}{(2w-1)} $$
therefore for any real number $r$ the area of a rectangle will be eqaual to $r$ and its perimeter will be greater than $4r$ when $l < - \frac{w}{(2w-1)} $
Just trying to see if this proof makes sense.
Let $w$ and $h$ be the width of the rectangle. You are requested to find suitable values for $w$ and $h$ so that \begin{gather} w\cdot h = R \\ 2 w + 2 h > 4 \cdot R \end{gather}
Inserting the first equation into the second inequation, you obtain : $$ 2 w + 2 \frac{R}{w} > 4 R $$ Assuming the non degenerate case that $w>0$. \begin{gather} w + \frac{R}{w} > 2 R\\ w^2-2Rw + R > 0 \end{gather}
Solving, you obtain : $$ w = \frac{2R \pm \sqrt{ 4R^2 - 4R}}{2} = R \pm \sqrt{R^2-R} $$
The term of degree 2 having a positive sign, the second degree polynomial is positive at the extreme ("outside" of the roots), thus you should pick $w< R-\sqrt{R^2-R}$ or $w> R-\sqrt{R^2-R}$.