How to prove that the distance of $n\sqrt{3}$ to an integer is larger than $\frac{1}{3n\sqrt{3}}$?

Question

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

Answer 1

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 $(*)$.