<s>[INST] How to solve $\sqrt{2} + \sqrt{3} - \frac{m}{n} > \frac{1}{cn^4}, \forall n \ge 1$? If $\frac{m}{n}$ is an irreducible fraction such that $1 > \sqrt{2} + \sqrt{3} - \frac{m}{n} > 0$, then there exists a constant $c > 0$ so that $$\sqrt{2} + \sqrt{3} - \frac{m}{n} > \frac{1}{cn^4}, \forall n \ge 1$$
From the first inequation we can show that $4n > m \ $ and $\ m > 2n$.
I was also intuitively thinking that as $n$ gets bigger and bigger $m$ can approximate $(\sqrt{2} +\sqrt{3})n$ better and better so then $\sqrt{2} + \sqrt{3} - \frac{m}{n}$ tends to $0$.
 [/INST] It is an old result of Liouville :https://en.m.wikipedia.org/wiki/Diophantine_approximation
It can be applied because your number $\sqrt{2}+\sqrt{3}$ is algebraic of degree $4$.
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