How to solve a maximisation problem using the AM-GM inequality?

Question

The sum of the squares of two positive numbers is $a$. Prove that their product is the maximum possible when the two numbers are equal.

I solved this problem initially by using differentiation, however, when I looked at alternative methods the AM-GM inequality came up.

My question is whether someone could explain to me how to use that inequality to solve this type of problem (I have very little knowledge about using this inequality). Many thanks.

Answer 1

We have that $x^2+y^2=a$, and are looking to maximize $xy$. We know that by AM-GM $$xy\leq\frac{(x+y)^2}4$$

But, note that when $x=y$, $xy=\frac{(x+y)^2}4$, showing that it is maximal.

Answer 2

We have

$$x^2+y^2=a$$

and by AM-GM inequality

$$\frac{x^2+y^2}2\ge \sqrt{x^2y^2} \implies xy\le \frac a 2$$

and equality holds if and only if $x=y=\frac{\sqrt a}4$.