<s>[INST] If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$. 
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the following operation $\biggl|\dfrac{\sqrt{3}n-m}{n}\biggr|\cdot \biggl|\dfrac{\sqrt{3}n+m}{\sqrt{3}n+m}\biggr|$ to get $$\biggl|\dfrac{3n^{2}-m^{2}}{\sqrt{3}n^{2}+mn}\biggr|$$
Since $n,m \ne 0$, we have that $|3n^{2}-m^{2}| \ge 1$. Now for the denominator, we have $$ |\sqrt{3}n^{2}+mn| \le |\sqrt{3n^{2}}| + |mn| $$
Thus it follows that $$\dfrac{1}{|\sqrt{3}n^{2}+mn|} \ge \dfrac{1}{|\sqrt{3}n^{2}| + |mn|}$$
Would I have to work in cases where $m<n$, for example? Then we have $$|\sqrt{3}n^{2}| + |mn| < |\sqrt{3}n^{2}| + n^{2} < 3n^{2} + n^{2} < 5n^{2}$$ which gives us the desired result. Although, the same method doesn't work when $n >m$.
 [/INST] The original inequality is equivalent to
$$
(5 m n+1)^2\leq 75 n^4
$$
or
$$
75 n^4\leq (5 m n-1)^2
$$
This reduces to
$$
\left(n\leq -1\land \left(m\leq \frac{1}{5 n}-\sqrt{3} \sqrt{n^2}\lor m\geq -\sqrt{3} \sqrt{n^2}-\frac{1}{5 n}\right)\right)
$$ 
or
$$ 
\left(n\geq 1\land \left(m\leq \sqrt{3} \sqrt{n^2}-\frac{1}{5 n}\lor m\geq \sqrt{3} \sqrt{n^2}+\frac{1}{5 n}\right)\right)
$$
Since $|n| \ge 1$ implies $2/(5n) \le 1$, we are done.
</s>