
Proof that a square has the smallest perimeter of any rectangle with a given area
PROVE THE FOLLOWING STATEMENT: 
If a nonsquare rectangle has the same area as that of a square, prove that its perimeter will be larger than that of the square. 
Usually, this statement is proved by contradiction this way:
We're asked to prove given A=BH and A=S², and B‡ H, that 2B + 2H > 4S.
Assume the negation, namely that given the conditions above, 2B + 2H ≤ 4S.
B + H ≤ 2S
(B + H)² ≤ 4S², which is legal because B, H, and S are all positive
(B + H)² ≤ 4BH, because A=BH=S²
B² + 2BH + H² < 4BH
B²  2BH + H² < 0
(B  H)² < 0, and since the square of a real number can't be negative, we know that
(B  H)² = 0, which can only happen if
B = H, but that contradicts the assumption, above.
In this web page, I'll prove this a different way. But first, some interesting (but relevant) number facts:
I bet you've noticed that the two numbers that add up to, say, 16, have the highest product when they're equal. 8 times 8 is 64. 7 times 9 is less: 63. Maybe you haven't noticed that (8+x) (8x) = 64  x^{2}. Try a few more examples, and you'll see it's true. This is important, because all rectangles with a perimeter of 32 have sides that are 8+x and 8x in length. So the area of any such rectangle is 64x^{2}, where x is the amount by which each side differs from 8. x^{2}, you know, can never be negative (as long as x is a real number), so 64x^{2} will always be less than or equal to 64. This, in essence, is the method of proving this question.
First, if you're not familiar with factoring the difference of two squares, convince yourself that (a+b) (ab) = a^{2}  b^{2}. You can use the "F.O.I.L." method  multiply the first terms, the outer terms, the inner terms, and the last terms, and add 'em all up, and you'll see the result. In this proof, we'll use this important fact  backwards! By backwards, I mean we start off with something that doesn't look at all like the difference of two squares, and basically manufacture a difference of two squares out of nothing. You'll see...
Now for the proof. Let B be base, H be height, and A=BH be area of the rectangle.
A = BH
A = BH/2 + BH/2
A = (B^{2})/4 + BH/2 + (H^{2})/4  (B^{2})/4 + BH/2  (H^{2})/4
A = ((B^{2})/4 + BH/2 + (H^{2})/4)  ((B^{2})/4  BH/2 + (H^{2})/4)
A = (B/2 + H/2)^{2 } (B/2  H/2)^{2}
Do you follow all those steps? If so, move on to the next paragraph. If not, here's a quick recap: All I'm saying here is that the product of any two numbers is equal to the square of their average minus the square of half their difference. In more detail, here are the steps I did: You can see BH is BH/2 + BH/2, right? Now I capriciously add  and then subtract  some terms (B^{2})/4 and (H^{2})/4. This is legal, because it's a net zero change. Then I gather up the terms using parentheses, making two perfect square trinomials that I factor, showing you the difference of two squares. Cute, eh?
Now, let's turn our attention to the square with the same area A and side S.
S^{2} = A
S^{2} = (B/2 + H/2)^{2 } (B/2  H/2)^{2}
S^{2} < (B/2 + H/2)^{2}
S < B/2 + H/2
4S < 2B + 2H
Did you follow those steps? If so, you'll see the question is proved. If not, here's a quick recap: The area of a square is S^{2}. The statement of the proof tells us this is the same as the area of the rectangle, expressed as (B/2 + H/2)^{2 } (B/2  H/2)^{2}. Since (as we modified the question) B is not equal to H, then (B/2  H/2)^{2} is greater than zero. So S^{2} < (B/2 + H/2)^{2}. Take the primary square root of both sides (this is legal because S, B, and H are all nonnegative), and multiply both sides by four. Now it's clear that the perimeter of the square (4S) is less than the perimeter of the samearea rectangle (2B + 2H).
The webmaster and author of this Math Help site is Graeme McRae.