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How to prove $\forall n \in \mathbb{N}^*$, the following inequation is correct. $$ \min\{n\sqrt{3}-\lfloor n\sqrt{3}\rfloor, \lfloor n\sqrt{3}\rfloor+1-n\sqrt{3}\} \geqslant \dfrac{1}{3n\sqrt{3}} $$

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    $\begingroup$ For every integers $n\geqslant1$ and $k\geqslant0$, $$(n\sqrt3-k)(n\sqrt3+k)=3n^2-k^2\ne0$$ hence $$|n\sqrt3-k|\geqslant\frac1{n\sqrt3+k}$$ If $k$ realizes the minimum distance between $n\sqrt3$ and an integer, then $$k<n\sqrt3+1$$ hence $$|n\sqrt3-k|>\frac1{2n\sqrt3+1}$$ which proves a stronger form of the desired result. $\endgroup$ – Did Mar 1 '17 at 12:58
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Let $m$ be the integer closest to $n\sqrt3$. Then we have $$ |(n\sqrt3-m)(n\sqrt3+m)|=|3n^2-m^2|\ge1, $$ because $3n^2-m^2$ is an integer, and cannot be equal to zero because $\sqrt3$ is irrational.

Consequently $$ |n\sqrt3-m|\ge\frac1{n\sqrt3+m}.\qquad(*) $$ Because $|n\sqrt3-m|<1/2$ we have $n\sqrt3-\dfrac12<m<n\sqrt3+\dfrac12$. Therefore $m<2n\sqrt3$, and the claim follows from $(*)$.

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  • $\begingroup$ Looks like you could use $2n\sqrt3+1$ as the denominator instead of $3n\sqrt3$? $\endgroup$ – Jyrki Lahtonen Mar 1 '17 at 13:00

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