If the area of square A and rectangle B are equal what will be relation between their perimeters? I found by actual calculation that the perimeter of Square A will be less than the perimeter of Rectangle B, but how to formalize it mathematically?
 A: Given any rectangle with area $A$. Let $a, b$ be its sides and $p$ is its perimeter.   
We have $p = 2(a+b)$ and $A = ab$. By AM $\ge$ GM, we have
$$p = 2(a+b) = 4\left(\frac{a+b}{2}\right) \ge 4\sqrt{ab} = 4\sqrt{A}$$
This means the perimeter for any rectangle with area $A$ is bounded from below by $4\sqrt{A}$.
Since a square with side $\sqrt{A}$ has area $A$ and perimeter $4\sqrt{A}$,
above lower bound is achieved by a square. From this, we can conclude

Among all rectangle with a given area $A$, the square has the smallest perimeter $4\sqrt{A}$.

A: Let $M^2$ be the area of the square, with $M>0$. Then, the side length of the square is $M.$ Moreover, The rectangle has sides $M+s$ and $\frac{M^2}{M+s}$, where we can assume $s\ge0$. In fact:
$$Area(A) = M^2 = (M+s)\frac{M^2}{M+s} = Area(B).$$
Now, the perimetere of $A$ and $B$ are, respectively
$$p(A) = 4M \qquad\text{and}\qquad p(B)=2\left(M+s+\frac{M^2}{M+s}\right).$$
So now you need to prove that
$$4M \le 2\left(M+s+\frac{M^2}{M+s}\right),\quad\text{when }s\ge0.$$
