How can the AM-GM inequality be used in this case if the given variables are real numbers?

Question

I was wondering - if we are given a question like this:

Prove that for any $a, b \in ℝ$ and $ε > 0$,
$ab \leq \frac{a^2}{2ε}+\frac{εb^2}{2}$.

Since $a, b \in ℝ$, how can we make use of $\sqrt{ab} \leq \frac{a+b}{2}$? Because as I understand, this inequality can not be used right away since $a, b$ can be negative numbers and this inequality only holds if $a,b \ge 0$.
I thought about using $ab \leq (\frac{a+b}{2})^2$, but if we let $a = \frac{a^2}{2ε}$ and $b = \frac{εb^2}{2}$, we will get $\frac{a^2b^2}{4} \leq (\frac{\frac{a^2}{2ε}+\frac{εb^2}{2}}{2})^2$ which isn't what we want to show. If we were allowed to square both sides, this $\sqrt{ab} \leq \frac{a+b}{2}$ inequality could be used, but as stated above $a,b$ might be negative so this step is illegal.

So, my question is: what can be done in this case?

Thanks in advance!

Answer 1

If you set $a' = \frac{a}{\sqrt{\epsilon}}$ and $b' = \sqrt{\epsilon}b$, you get $ab= a'b'$.

Hence, it is equivalent with showing

$$ab = a'b' \leq \frac{a'^2 +b'^2}{2} \Leftrightarrow (a'-b')^2\geq 0$$

Answer 2

We can establish $$\dfrac{x^2+y^2}2\ge xy$$ for all real $x,y$

Set $x^2=a^2/\epsilon,y^2=\epsilon b^2$

$x^2y^2=a^2b^2\implies xy=\pm ab$

Answer 3

So:

By AM-GM: $$ \frac{a^2}{2ε}+\frac{εb^2}{2}\geq2\sqrt{ \frac{a^2}{2ε}\cdot\frac{εb^2}{2}}=|ab|\geq ab.$$