Silk Road Mathematics Competition Xx

Let $[\sqrt{m}] = k$. Then
$$|\{\sqrt{m}\} - \frac{1}{2}| = \frac{1}{2}|2k + 1 - 2\sqrt{m}| = \frac{1}{2} \cdot \frac{|4k^2 + 4k + 1 - 4m|}{2k + 1 + 2\sqrt{m}}.$$
The numerator of the last fraction is an odd positive integer, and therefore at least $1$. Thus,
$$|\{\sqrt{m}\} - \frac{1}{2}| \ge \frac{1}{2} \cdot \frac{1}{2k + 1 + 2\sqrt{m}} \ge \frac{1}{2(4\sqrt{m} + 1)} > \frac{1}{8(\sqrt{m} + 1)},$$

Q.E.D.