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 $n >m$.

We may assume without loss of generality that $m, n$ are both positive as this is equivalent to them both being negative and if only one of them is negative then this inequality is trivial.
We can then consider 2 cases:
Case 1: m< 2n
By your argument we need only show that $|\sqrt{3}n^2|+|mn|<5n^2$. This follows from $|\sqrt{3}n^2|+|mn|<|2n^2|+|2n^2|=4n^2<5n^2$.
Case 2: $m\geq 2n$
In this case we have: $|\sqrt{3} - \frac{m}{n}| > |\frac{9}{5} - 2| = \frac{1}{5} \geq \frac{1}{5n^2}$.
