How prove this $|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\frac{1}{20n^3}$ Prove that 
  $$|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\dfrac{1}{20n^3}$$
let $t=\{n\sqrt{2}\}-\{n\sqrt{3}\}$ and $k=[n\sqrt{3}]-[n\sqrt{2}]$
then we have $$t=k-(\sqrt{3}-\sqrt{2})n=k-\sqrt{5-2\sqrt{6}}n\neq 0$$
so 
\begin{align*}
t&=\dfrac{(k-(\sqrt{3}-\sqrt{2})n)(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}{(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}\\
&=\dfrac{k^4-10k^2n^2+n^4}{(t-2\sqrt{2}n)(t+2(\sqrt{3}-\sqrt{2})n)(t+2\sqrt{3}n)}
\end{align*}
notice that
$$|t-2\sqrt{2}n|\le 2\sqrt{2}n+\dfrac{1}{20}\le(2\sqrt{2}+\dfrac{1}{20})n$$
$$|t+2(\sqrt{3}-\sqrt{2})n|\le2(\sqrt{3}-\sqrt{2})n+\dfrac{1}{20}\le(2\sqrt{3}-2\sqrt{2}+\dfrac{1}{20})n$$
$$|t+2\sqrt{3}n|\le2\sqrt{3}n+\dfrac{1}{20}\le(2\sqrt{3}+\dfrac{1}{20})n$$
so 
$$|t|\ge\dfrac{1}{(2\sqrt{2}+\dfrac{1}{20})(2\sqrt{3}+2\sqrt{2}-\dfrac{1}{20})(2\sqrt{3}+\dfrac{1}{20})n^3}>\dfrac{1}{7n^3}$$
so 
$$|t|\ge\min{\left(\dfrac{1}{20},\dfrac{1}{7n^3}\right)}\ge\dfrac{1}{20n^3}$$
\
This post https://math.stackexchange.com/questions/465419/how-prove-this-t-2-sqrt2n-le2-sqrt2n-frac120 is  not true?  so This methods is wrong, so How prove it? Thank you 

What I have done is not a full solution, but it may even lead to the solution of a more general problem. Also, using this method I can get only a lower bound. Here it is.
For any positive integer $n$, let $$\sqrt {3} = \frac {a_n} {n} + \varepsilon_n \qquad \text {and} \qquad \sqrt {2} = \frac {b_n} {n} + \delta_n,$$ where $a_n$ and $b_n$ are positive integers, $\varepsilon_n$ and $\delta_n$ are positive real numbers, and $\max (\varepsilon_n, \delta_n) < \frac {1} {n}$. Such a configuration exists, for, let $a_n = [\sqrt {3} n]$ and $b_n = [\sqrt {2} n]$. Since $$a_n < \sqrt {3} n < a_n + 1 \qquad \text {and} \qquad b_n < \sqrt {2} n < b_n + 1,$$ we have $0 < \varepsilon_n, \delta_n < \frac {1} {n}$.
Then, we have
$$\begin {eqnarray}
|\{n\sqrt{3}\}-\{n\sqrt{2}\}| & = & |\{a_n + n \varepsilon_n\}|
 - |\{b_n + n \delta_n\}| \nonumber \\ & = & |n \varepsilon_n - n \delta_n| \nonumber \\ & = & n |\varepsilon_n - \delta_n|.
\end{eqnarray}$$
Since we know very crudely that $\varepsilon_n = O \left (\frac {1} {n^2} \right)$ and $\delta_n = O \left (\frac {1} {n^2} \right)$, we have $$|\{n\sqrt{3}\}-\{n\sqrt{2}\}| = O \left (\frac {1} {n} \right),$$ that is, there exists an absolute constant $c$ such that $$|\{n\sqrt{3}\}-\{n\sqrt{2}\}| < \frac {c} {n}.$$
Note: This does not answer the question, it is rather related to it. The reason I put it here (rather than not) is that I feel it may be useful. If you guys feel otherwise, delete it. 
