A correction. The problem should be $|\{\sqrt{m}\}-1/2|$ where $\{t\}=t-\lfloor t \rfloor$ is the fractional part of $t$. Let $k\triangleq \lfloor \sqrt{m}\rfloor$. That is, $k^2\le m<(k+1)^2$. It then boils down verifying \[ |2\sqrt{m}-2k-1|>\frac{1}{4(\sqrt{m}+1)}\Leftrightarrow \frac{|4m-(2k+1)^2|}{2\sqrt{m}+2k+1}>\frac{1}{4(\sqrt{m}+1)}. \]Note that $|4m-(2k+1)^2|\in\mathbb{N}$, and in particular is positive. Now, if this term is at least two, then we immediately have the conclusion as one would then have \[ \frac{|4m-(2k+1)^2|}{2\sqrt{m}+2k+1}>\frac{1}{4(\sqrt{m}+1)}\ge \frac{2}{2\sqrt{m}+2k+1}\ge \frac{2}{4\sqrt{m}+1}>\frac{1}{4\sqrt{m}+4} \]for trivial reasons. Next, suppose $|4m-(2k+1)^2| = 1$. Then $4m-(2k+1)^2 = -1$, that is $m=k^2+k$. In this case, it suffices to show \[ \frac{1}{2\sqrt{k^2+k}+2k+1}>\frac{1}{4\sqrt{k^2+k}+1}\Leftrightarrow 2\sqrt{k^2+k}>2k, \]which is clearly the case.

Hopeooooo wrote: For every positive integer $m$ prove the inquility $|{\sqrt m} - \frac{1}{2}| \geq \frac{1}{8(\sqrt m+1)} $ You sure? since $m \ge 1$, then \[ \sqrt{m} - \frac{1}{2} \ge \frac{1}{2} > \frac{1}{16} \ge \frac{1}{8(\sqrt{m} + 1)} \]

Hopeooooo wrote: For every positive integer $m$ prove the inquility $|\{\sqrt{m}\} - \frac{1}{2}| \geq \frac{1}{8(\sqrt m+1)} $ Sorry..... The problem was edited

Hopeooooo wrote: For every positive integer $m$ prove the inquility $|\{\sqrt{m}\} - \frac{1}{2}| \geq \frac{1}{8(\sqrt m+1)} $ (The integer part $[x]$ of the number $x$ is the largest integer not exceeding $x$. The fractional part of the number $x$ is a number $\{x\}$ such that $[x]+\{x\}=x$.)

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a solution in Russian here