Show that $|\sin(n\pi\sqrt{5})| \ge \frac{c}{n}$ for some $c>0$ Consider the sequence $\{x_n\}$ defined by
$$
x_n = |\sin (n\pi\sqrt{5})|.
$$
Show that there exists a real number $c>0$ such that $x_n \ge \frac{c}{n}$ for all $n \ge 1$.

I have been given two hints:

*

*On $\left[0,\frac{\pi}{2}\right]$, one has $\sin{t} \ge t - \frac{t^3}{6}$. (This is an elementary argument in calculus.)

*For any integers $p$ and $q$ with $p^2+q^2 \ne 0$, one has
$$
|p\sqrt{5}-q| \ge \frac{1}{p\sqrt{5}+q}.
$$
(This corresponds to the fact that $\sqrt{5}$ is irrational.)


For this question, given fixed $n$, I will consider finding $m$ such that $|n\sqrt{5}-m|\le\frac{1}{2}$ (such $m$ always exists, because $\left[n\sqrt{5}-\frac{1}{2},n\sqrt{5}+\frac{1}{2}\right]$ is an interval of length $1$, at least one integer will fall in it), so that
$$
|\sin(n\pi\sqrt{5})|=|\sin(n\pi\sqrt{5}-m\pi)|=|\sin(n\sqrt{5}-m)\pi|=\sin|(n\sqrt{5}-m)\pi|.
$$
Since then $|(n\sqrt{5}-m)\pi| \in \left[0,\frac{\pi}{2}\right]$, we may use these two hints. But I think I am stuck here.
By the first hint,
$$
\begin{aligned}
\sin|(n\sqrt{5}-m)\pi| &\ge |(n\sqrt{5}-m)\pi| - \frac{|(n\sqrt{5}-m)\pi|^3}{6} \\
                       &\ge \frac{1}{(n\sqrt{5}+m)\pi} - \frac{|(n\sqrt{5}-m)\pi|^3}{6}
\end{aligned}
$$
But I do not think this inequality brings me any close to the desired conclusion, for the existence of $n^3$ term on the right hand side.
Alternatively, I also think about applying the second hint immediately:
$$
\begin{aligned}
\sin|(n\sqrt{5}-m)\pi| & \ge \sin \frac{\pi}{n\sqrt{5}+m} \\
                       & \ge \frac{\pi}{n\sqrt{5}+m} - \frac{\pi^3}{6(n\sqrt{5}+m)^3}
\end{aligned}
$$
This approach looks more promising but it is still not obvious that I will deduce the desired result.
Is it possible to deduce the desired result from one of my approaches (most likely the second)? Or maybe I should try a even finer approach? I hope further steps should not be a bunch of lengthy fractions. Thanks in advance!
 A: From your last result, we have
$$
\begin{align}
\sin|(n\sqrt{5}-m)\pi| & \ge \sin \frac{\pi}{n\sqrt{5}+m} \\
                       & \ge \frac{\pi}{n\sqrt{5}+m} - \frac{\pi^3}{6(n\sqrt{5}+m)^3} \\
&= \alpha\frac{\pi}{n\sqrt{5}+m} + \left((1-\alpha)\frac{\pi}{n\sqrt{5}+m}- \frac{\pi^3}{6(n\sqrt{5}+m)^3} \right)  \tag{1}
\end{align}
$$
We will choose $\alpha \in(0,1)$ such that
$$(1-\alpha)\frac{\pi}{n\sqrt{5}+m}- \frac{\pi^3}{6(n\sqrt{5}+m)^3} \ge 0 \tag{2}$$
We have
$$\begin{align}
(2) &\Longleftrightarrow  6(n\sqrt{5}+m)^2\ge\frac{\pi^2}{1-\alpha}\\  
&\Longleftrightarrow \alpha \le 1-\frac{1}{6}\left(\frac{\pi}{n\sqrt{5}+m} \right)^2\\  \tag{3}
\end{align}$$
From $|n\sqrt{5}-m|\le\frac{1}{2}$, we deduce that $m \ge n\sqrt{5} -\frac{1}{2}$, then the right hand side of $(3)$ is
$$
RHS(3) \ge  1-\frac{1}{6}\left(\frac{\pi}{2\sqrt{5}n-\frac{1}{2}} \right)^2  \ge 1-\frac{1}{6}\left(\frac{\pi}{2\sqrt{5}-\frac{1}{2}} \right)^2 >\frac{1}{4}
$$
We can choose $\alpha = \frac{1}{4}$. Return back to $(1)$, we have
$$
\sin|(n\sqrt{5}-m)\pi| > \frac{1}{4}\frac{\pi}{n\sqrt{5}+m} > \frac{1}{4}\frac{\pi}{2\sqrt{5}n + \frac{1}{2}} > \frac{1}{4}\frac{\pi}{3\sqrt{5}n} 
$$
So, we can choose $c = \frac{1}{4}\frac{\pi}{3\sqrt{5}}$ and for all $n$
$$\sin|(n\sqrt{5})\pi| \ge \frac{1}{4}\frac{\pi}{3\sqrt{5}} \frac{1}{n}$$
