<s>[INST] Is $\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$ for all $n > 1$? An equivalent inequality for integers follows:
$$(3n^2-5)\left\lceil n/\sqrt{3} \right\rceil^2 > n^4.$$
This has been checked for n = 2 to 60000.  Perhaps there is some connection to the convergents to $\sqrt{3}$.       
$\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$
 [/INST] Equivalently, we want to know if
$\mathrm{sqceiling}(n^2/3) > n^4/(3n^2-5) = n^2/3 + 5/9 + 25/(27n^2) + ...$
where $\mathrm{sqceiling}()$ is the function taking a real to the next exact square.
This is equivalent to
$\mathrm{sqceiling}(n^2/3) - n^2/3 > 5/9 + 25/(27n^2) + ...$
but since $n^2/3$ is never a square for $n>0$, and $n^2 \equiv 0\ \textrm{or}\ 1 (\textrm{mod}\ 3)$, we have
$\mathrm{sqceiling}(n^2/3) - n^2/3 \ge 2/3$.
Clearly for large enough $n$
2/3 > 5/9 + 25/(27n^2) + ...
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