Show that among all rectangles with area A the square has the minimum perimeter?

Suppose sqrt(A) = B ie the square with sides B has an area of A and its perimeter is 4*B.

Now consider a rectangle with sides C and D whose area is A.

So C*D = A = B*B so that D = B*B/C

Perimeter of the rectangle = 2*(C+D) = 2*C + 2*D = 2*C +2*B*B/C

Now consider (C-B)2 which, because it is a square, is always >= 0

ie C*C + B*B - 2*B*C >= 0

ie C*C + B*B >= 2*B*C

Multiply both sides by 2/C (which is >0 so the inequality remains the same)

2*C + 2*B*B/C >= 4*B

But, as shown above, the left hand side is perimeter of the rectangle, while the right hand side is the perimeter of the square.