Selection and Training Session

Let $m=[n\sqrt{6}]$. Then $n\sqrt{6}>m$, or $6n^2-m^2>0$. Note that $6n^2-m^2\ne 1$ or $4(\mathrm{mod}3)$, $6n^2-m^2\ne 2(\mathrm{mod}4)$ (recall that $n$ is even), $6n^2-m^2\ne 3(\mathrm{mod}9)$, so $6n^2-m^2\ge 5$. Now we have $$\{n\sqrt{6}\}=n\sqrt{6}-m=\frac{6n^2-m^2}{n\sqrt{6}+m}\ge \frac{5}{n\sqrt{6}+m}>\frac{5}{2n\sqrt{6}}>\frac{1}{n},$$ thus a) is proved. In particular, $m<n\sqrt{6}-\frac{1}{n}$, so $$\{n\sqrt{6}\}\ge \frac{5}{n\sqrt{6}+m}>\frac{5}{2n\sqrt{6}-1/6}>\frac{1}{n-\frac{1}{5n}},$$ thus b) is proved.