Show that the area of a square is greater than the area of a rectangle. It is a common problem that : Show that the square is a rectangle of maximum area for a given perimeter and a rectangle of minimum perimeter for a given area.
It is an easy problem when we use calculus. It can done by using maxima &amp; minima but how we shall show it using inequality. I can do it using calculus by how I show it using basic inequality. Somebody help me .
One problem is just the dual of the other one, so it is enough to prove one of them to get them both for (almost) free. Assume that we have a fixed perimeter $(2p)$ and we want to maximize the area. Our problem can be written as
$$ \max_{\substack{a+b=p\\ a,b\geq 0}} ab = \large{?} $$
and we may exploit the AM-GM inequality, stating that if $a,b\geq 0$, 
$$ \sqrt{ab}\leq\frac{a+b}{2} $$
with a strict inequality if $a\neq b$. It follows that the maximum area ($p^2/4$) is achieved by $a=b$, i.e. by a square.
Assuming the same perimeter: Cauchy–Schwarz inequality