Proof-check: Proving that a rectangle of area $A$ necessarily has minimal perimeter when it is a square of side $|\sqrt A|$ Found many questions going the other way, but couldn't any like this. If I missed one, sorry.
Just want to confirm if this is totally valid.
Let the sides of the rectangle be $(|\sqrt A| +\alpha)$ and ($|\sqrt A|-\beta)$ with $\alpha,\beta\geq 0$
The perimeter is,
$$2(|\sqrt A|+\alpha)+2(|\sqrt A| - \beta)=4|\sqrt A| +2(\alpha-\beta)$$
we have that $$4|\sqrt A|+2(\alpha-\beta)\geq4|\sqrt A| \iff \alpha \geq \beta$$
($4|\sqrt A|$ is the perimeter of the square)
So this comes down to proving that $\alpha\geq\beta$
We take the area: $$(|\sqrt A|+\alpha)(|\sqrt A|-\beta)=A\implies \alpha|\sqrt A|-\beta|\sqrt A|-\alpha\beta=0$$
$$\to\beta=\frac{\alpha|\sqrt A|}{|\sqrt A| +\alpha}$$
and so we show $$\alpha\geq\frac{\alpha|\sqrt A|}{|\sqrt A|+\alpha}$$
$|\sqrt A|+\alpha \geq 0$ since both components are, hence:
$$\alpha|\sqrt{A}|+\alpha^2\geq\alpha|\sqrt A|\to\alpha^2\geq 0$$
 which is trivial.
So $\alpha\geq \beta$ is a necessity, hence the minimal perimeter is $4|\sqrt A|$
 A: Another way you could have proved this is with calculus. Suppose your rectangle has area $A$ and sides of length $x$ and $y$. We wish to minimize the perimeter $$P=2x+2y$$ subject to the constraint $$A=xy.$$
First we use the constraint to solve for a variable: $$y=\frac{A}{x}.$$ So that we have $$P=2x+\frac{2A}{x}.$$ Now we minimize the perimeter by taking its derivative and setting it equal to zero.
$$P'(x)=2-\frac{2A}{x^2}=0$$
$$\implies x=\sqrt{A}$$
We can check that this is indeed a minimum using the second derivative test. 
$$P''(x)=\frac{4A}{x^3}$$
$$\implies P''(\sqrt{A})=\frac{4}{\sqrt{A}}>0$$
So this is indeed a minimum. Now plugging our value for $x$ back into our equation for $y$, we have
$$y=\frac{A}{\sqrt{A}}=\sqrt{A}.$$
So the rectangle with a given area and minimal perimeter is a square with side length $\sqrt{A}$.
On a side note, there is a book called "When Least is Best" by Nahin, which solves lots of classic calculus problems without the use of calculus if this is something you are interested in.
