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 $n >m$.

One aims to give an optimal lower bound as follows:
Theorem. $|\sqrt{3}-\frac mn |\geq \frac 1{(2+\sqrt{3})n^2}$
Proof. Clearly the lower bound is achieved when $n=1,m=2.$ As in other proofs, one may assume without loss of generality that $m,n$ are positive and $n\geq 2$. Observe first that $$\frac 53<\sqrt{3}<\frac 74.$$ We divide into two cases.
Case 1. $\frac m n<\frac 7 4.$ Then $$|\sqrt{3}-\frac m n|=\frac{|3n^2-m^2|}{n^2(\sqrt{3}+\frac m n)}$$ $$>\frac 1{n^2(\frac 7 4+\frac 7 4)}=\frac 1{(3.5)n^2}>\frac 1{(2+\sqrt{3})n^2}.$$
Case 2. $\frac m n\geq \frac 7 4.$ Here one considers three subcases: $n=2,n=3$ and $n\geq 4$. If $n=2,$ then $\frac m n\geq \frac 7 4\Rightarrow m\geq 4,$ so $\frac m n\geq 2$ and hence $$|\sqrt{3}-\frac m n|=\frac m n-\sqrt{3}\geq 2-\sqrt{3}\geq \frac 1{(2+\sqrt{3})n^2}.$$ If $n=3$, then $\frac m n\geq \frac 7 4\Rightarrow m\geq \frac{21}4,$ so $m\geq 6$ and $\frac m n\geq 2$ and the result follows in exactly the same way as the above subcase. It remains to check the subcase when $n\geq 4$, but then $$ |\sqrt{3}-\frac m n|=\frac m n-\sqrt{3}\geq \frac 7 4-\sqrt{3}>\frac 1{(2+\sqrt{3})\cdot 4^2}\geq \frac 1{(2+\sqrt{3})n^2},$$ where one used $$\frac 7 4-\sqrt{3}>\frac 1{(2+\sqrt{3})\cdot 4^2}$$ $$\Leftrightarrow \frac 7 4>\sqrt{3},$$ which is true. 
Combining all cases, the result is proven.
Edit. (More details) $$\frac 7 4-\sqrt{3}>\frac 1{(2+\sqrt{3})\cdot 4^2}$$ $$\Leftrightarrow \frac {7-4\sqrt{3}}4>\frac 1{(2+\sqrt{3})\cdot 4^2}$$ $$\Leftrightarrow 7-4\sqrt{3}>\frac 1{(2+\sqrt{3})\cdot 4}$$
$$\Leftrightarrow (7-4\sqrt{3})(2+\sqrt{3})>\frac 1 4$$
$$\Leftrightarrow 2-\sqrt{3}>\frac 1 4$$
$$\Leftrightarrow \frac 7 4>\sqrt{3},$$ which is true. 
