Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\left\lceil{\frac{1}{n^2}}\right \rceil = 5 - 4$$
Therefore, for $n \in \mathbb R$, where $n \neq 0$
$$4 < \left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil \le 5$$
$$4 < \left\lceil{4}+\frac{1}{n^2}\right \rceil \le 5$$
$$0 < \left\lceil\frac{1}{n^2}\right \rceil \le 1$$
I feel kind of without purpose once I hit point. Should I be scrapping the inequality and solving for $n$ instead?

For all $ n \in \mathbb{N}$, we know that $0 < \frac{1}{n^2} \leq 1$, hence $\left\lceil \frac{1}{n^2}\right\rceil=1.$
$$\left\lceil  \frac{4n^2+1}{n^2} \right\rceil=\left\lceil 4 +  \frac{1}{n^2} \right\rceil= 4 + \left\lceil  \frac{1}{n^2} \right\rceil=4+1=5$$
