# Name of proof that area of square>area of rectangle of the same perimeter

What is the proof called for the fact that the area of a square is always greater than the area of a non-square rectangle of the same perimeter?

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How about this: "Proof of the fact that area of a square>area of a rectangle of the same perimeter" – user31029 Jun 22 '12 at 2:16
This is a consequence what is commonly called the arithmetic-geometric mean inequality. – user17794 Jun 22 '12 at 2:23
This is a consequence of $(a+b)(a-b)=a^2-b^2$. – Jonas Meyer Jun 22 '12 at 2:55
Facts of this kind are sometimes called isoperimetric inequalities, of which the best-known is the fact that the area of a circle is always greater than that of any other region with the same perimeter. – Nate Eldredge Jun 22 '12 at 3:49
Using Lagrange multipliers is one of the approaches you can take. See this link [problem #5]: math.ualberta.ca/~thillen/math209/s3.pdf – math-visitor Jun 22 '12 at 4:08

I don't know what the name of the proof is (if there is a standard known name), but as stated in one of the comments above it is a consequence of the Arithmetic-Geometric (AM-GM) Mean Inequality which states that for non-negative real numbers $x_i$ the following inequality holds: $$\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\ge \sqrt[n]{x_1 x_2\cdots x_n}$$ With equality if, and only if $x_1=x_2=\cdots=x_n$
So suppose we fix the perimeter of a rectangle as $2(a+b)$ with side lengths $a$ and $b$. Hence, a square with the same perimeter has a side length of $\displaystyle\frac{a+b}{2}$. So by applying the AM-GM inequality with $n=2$ it follows that $\displaystyle \left(\frac{a+b}{2}\right)^2 \ge ab$. So we see that the area of the rectangle doesn't exceed the area of the square, and the areas are equal if and only if $a=b$. So in other words, for a fixed perimeter the maximum area of a rectangle is when it is in fact a square.
Alternatively, one can apply this to a fixed area and deduce that the perimeter of a rectangle is always greater than or equal to the perimeter of a square. This time suppose that $x$ and $y$ are the lengths of a rectangle with a fixed area $xy$. Hence, a square with the same area has side-lengths $\sqrt{ab}$. So, by AM-GM we obtain $2(x+y)\ge 4\sqrt{xy}$. Now we see that the for a fixed area $xy$, the rectangle with the smallest perimeter is a square.
This can be easily extended and generalized into $n$ dimensions and we can say that an $n$-cube has the smallest perimeter among all $n$-dimensional boxes with the same volume and that an $n$-cube has the largest volume among all $n$-dimensional boxes with the same perimeter.