For $x\in \mathbb{R}$, we define $\{x\}=x-\lfloor x\rfloor$.
Prove that for all $n\in \mathbb{N}^*$, $\displaystyle{\{n\sqrt{2}\}>\frac{1}{2n\sqrt{2}}}$.
I really don't see how to start. The fact that $\sqrt{2}\notin \mathbb{Q}$ is surely important but I don't know how to use it.
If $p=\lfloor n\sqrt{2}\rfloor $, we have:
$\{n\sqrt{2}\}= n\sqrt{2}-p=\dfrac{2n^2-p^2}{n\sqrt{2}+p}\geqslant \dfrac{1}{n\sqrt{2}+p} >\dfrac{1}{2n\sqrt{2}} $