Belarus — 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 $4 \pmod{3}$, $6n^2 - m^2 \ne 2 \pmod{4}$ (recall that $n$ is even), $6n^2 - m^2 \neq 3 \pmod{9}$, so $6n^2 - m^2 \geq 5$. Now we have
$$\{n\sqrt{6}\} = n\sqrt{6} - m = \frac{6n^2 - m^2}{n\sqrt{6} + m} \geq \frac{5}{n\sqrt{6} + m} > \frac{5}{2n\sqrt{6}} > \frac{1}{n},$$
thus a) is proved. In particular, $m < n\sqrt{6} - \frac{1}{n}$, so
$$\{n\sqrt{6}\} \geq \frac{5}{n\sqrt{6} + m} > \frac{5}{2n\sqrt{6} - 1/6} > \frac{1}{n - \frac{1}{5n}},$$
thus b) is proved.