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 
 A: If the statement is false, then there exists a $n \in \mathbb{Z}_{+}$ such that
$$\left|\{n\sqrt{3}\} - \{n\sqrt{2}\}\right| \le \frac{1}{20n^3}\tag{*1}$$
This in turn implies existences of $m \in \mathbb{Z}$ and 
$\delta \in [ -\frac{1}{20n^3}, \frac{1}{20n^3} ]$ such that:
$$n (\sqrt{3} - \sqrt{2}) = m + \delta$$
Consider the polynomial $f(x) = x^4 - 10x^2 + 1$, it has the factorization:
$$f(x) = (x - \sqrt{3}-\sqrt{2})(x - \sqrt{3} + \sqrt{2})(x + \sqrt{3} - \sqrt{2})( x + \sqrt{3} + \sqrt{2})$$
We have:


*

*$f(\frac{m+\delta}{n}) = f(\sqrt{3} - \sqrt{2}) = 0$.

*$f(\frac{m}{n}) \ne 0$ because the roots of $f$ are all irrational.

*$n^4 f(\frac{m}{n}) = m^4 - 10 m^2 n^2 + n^4 \in \mathbb{Z}$.


(2) and (3) together implies $$n^4 \left|f(\frac{m}{n})\right| \ge 1\quad\iff\quad \left|f(\frac{m}{n})\right| \ge \frac{1}{n^4}$$
On the other hand, (1) and Mean value theorem implies existence of $\xi \in [0,1]$ such that
$$f(\frac{m}{n}) = f'(\frac{m+\xi\delta}{n}) \frac{\delta}{n}$$
Notice 
$$|\delta| \le \frac{1}{20n^3} \quad\implies\quad |\frac{\xi\delta}{n}| \le \frac{1}{20}$$
A plot of $f'(x)$ tells us $|f'(x)| \le 7.2$ whenever $\left|x - (\sqrt{3}-\sqrt{2})\right| \le \frac{1}{20}$. From this, we obtain a contradiction:
$$\frac{1}{n^4} \le \left|f(\frac{m}{n})\right| = \left| f'(\frac{m+\xi\delta}{n}) \frac{\delta}{n}\right| \le 7.2\left|\frac{\delta}{n}\right| \le \frac{7.2}{20n^4}$$
As a result, the original assumption $(*1)$ is false and we have
$$\left|\{n\sqrt{3}\} - \{n\sqrt{2}\}\right| > \frac{1}{20n^3}$$
In fact, we can improve the constant in above inequality form $20$ to something around $8.4033154$.
A: 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. 
