An inequality that holds for every $\epsilon >0 $ 
For every $\epsilon > 0$ and $a,b \in \mathbb{R}$, we have
$$ |ab| \leq \epsilon a^2 + \frac{b^2}{4 \epsilon} $$

attempt:
I was thinking on using AM-GM inequality:
$$ \epsilon a^2 + \frac{b^2}{4 \epsilon} = \frac{ \frac{b^2}{2 \epsilon} + 2 \epsilon a^2 }{2} \geq \sqrt{ \frac{b^2}{2 \epsilon} (2 \epsilon a^2) } = \sqrt{a^2b^2} = |ab|$$
Is this corect? What other nice results can we derive from the inequality? thanks

Yes, it is correct. An equivalent way would be to use $2|xy|\leq x^2+y^2$ for $x=a\sqrt{\epsilon}$ and $y=\frac{b}{2\sqrt{\epsilon}}$
