How to relate areas of circle, square, rectangle and triangle if they have same perimeter?? I was given a question which was like:

Suppose that a circle, square, rectangle and triangle have same perimeter. How area their areas related??

My work:
I broke the question in parts and tried to prove it seperately:
STEP $1$. Suppose the given perimeter is $P$.So the radius of circle will be ${r=P/2\pi}$.
And hence area of the circle will be:   ${P^2/4\pi}$.
Side of square will be $P/4$ and hence area of square will be $P^2/16$. 
It is obvious that ${P^2/4\pi}$  >  $P^2/16$. 
So Area of circle is more than area of square.
STEP $2$: Whether by using $AM>GM$ inequality or by using a bit of differentiation, we can get that product off two quantities is maximum when they are equal.
So, Area of square is more than Area of rectangle.
Here, the hardest part come (at least for me). I can't relate the area of triangle to any of Circle, Square or Rectangle. It would have been easy if the triangle was Equilateral, Right angle but the question is talking about a general triangle.
However, when I considered the situation with different examples (Assuming the perimeter and calculating the area, I got the relation:
Circle > Square > Rectangle > Triangle.
I shall be highly thankful if you guys can establish the relation in purely mathematical way.
EDIT: This question does not ask that whether circle have the largest area among all 2-D figures or not. The OP just want to relate the areas of Circle, Square, Rectangle, Triangle in a pure mathematical way.

To generalize: It is easily shown that any regular $n$-gon with side length $a$ the radius $r_n$ of its inscribed circle is
$$r_n=\frac{a/2}{\tan(\pi/n)},$$
hence
the ratio of its perimeter $P$ and the diameter of its inscribed circle is
$$\pi_n:=n\cdot\tan(\pi/n).$$
Let $r_n$ be the radius of its inscribed circle.  Now verify that the polygon's area is $\pi_n\cdot r_n^2$ and its circumference equals $2\pi_n\cdot r_n$.
(For example take $n=3$, a triangle with side $a$.  Then $r_3=\frac{a/2}{\sqrt{3}}$ and $\pi_3=3\cdot\tan(\pi/3)=3\sqrt{3}$. We calculate its area to $\pi_3\cdot r_3^2=\frac{\sqrt{3}}{4}\cdot a^2$.)
Given $P$, the radius $r_n=\frac{P}{2\pi_n}$, hence the area is
$$\frac{P^2}{4}\cdot\frac{1}{n\tan(\pi/n)}.$$
Now $\displaystyle\frac{1}{n\tan(\pi/n)}$ is strictly increasing with limit $\pi$.
