<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] The inequality is true for $n\geq 2$. Let $m:=\lceil n/\sqrt{3}\rceil$, then
$$ m^2-\frac{n^2}{3} = \frac{3m^2-n^2}{3} \geq \frac{2}{3}, $$
because $3m^2-n^2>0$ and $3m^2-n^2\not\equiv 1\pmod{3}$. Hence the inequality follows from
$$ \frac{2}{3}>\frac{n^4}{3n^2-5}-\frac{n^2}{3}=\frac{5n^2}{3(3n^2-5)}. $$
The latter is equivalent to $n^2>10$, hence we have a proof for $n\geq 4$. The remaining cases $n=2$ and $n=3$ can be checked by hand.
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