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.