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 $n >m$.

Using a technique similar to Liouville's theorem, $\sqrt{3}$ is a root of $P_2(x)=x^2-3$. Then, for any $\frac{m}{n}$ we have an $\varepsilon$ in between $\sqrt{3}$ and $\frac{m}{n}$ such that (this is MVT)
$$\left|P_2\left(\frac{m}{n}\right)\right|=
\left|P_2(\sqrt{3})-P_2\left(\frac{m}{n}\right)\right|=
|P_2'(\varepsilon)|\cdot \left|\sqrt{3}-\frac{m}{n}\right|$$
or
$$\left|\sqrt{3}-\frac{m}{n}\right|=
\left|\frac{m^2-3n^2}{2\varepsilon \cdot n^2}\right|\geq
\frac{1}{2\left|\varepsilon\right| \cdot n^2}\tag{1}$$

Now, if $\frac{m}{n}<\varepsilon<\sqrt{3}$ then $(1)$ becomes $\frac{1}{2\left|\varepsilon\right| \cdot n^2}>\frac{1}{2\sqrt{3}n^2}>\frac{1}{5n^2}$ and we are done.

If $\sqrt{3}<\varepsilon<\frac{m}{n}<\frac{5}{2}$ then $2\varepsilon<5$ and $(1)$ becomes $\frac{1}{2\left|\varepsilon\right| \cdot n^2}>\frac{1}{5n^2}$. So, we are done.

If $\sqrt{3}<\frac{5}{2}<\varepsilon<\frac{m}{n}$ then 
$$\left|\frac{m}{n}-\sqrt{3}\right|>
\left|\frac{5}{2}-\sqrt{3}\right|=
\frac{\frac{25}{4}-3}{\frac{5}{2}+\sqrt{3}}=
 \frac{13}{10+ 4\sqrt{3}}>
\frac{1}{5\cdot 1^2}\geq
\frac{1}{5\cdot n^2}$$
for all $n\geq1$.
