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 $\sqrt{3}$.       
$\lceil \frac{n}{\sqrt{3}} \rceil > \frac{n^2}{\sqrt{3n^2-5}}$

A tight result is $$\lceil \frac{n}{\sqrt{3}}\rceil \ge \frac{n^2}{\sqrt{3n^2-6+\frac{12}{n^2+2}}}$$ 
This means that the desired inequality holds for $n \ge 4.$ The cases $n=2,3$ can be checked separately.
As noted elsewhere, for $m=\lceil \frac{n}{\sqrt{3}}\rceil$  we have $n^2 \le 3m^2-2$ with equality exactly for the cases  $n=1,5,19,71,265,989,\cdots$ A001834 with $\frac{n}{m} \lt \sqrt{3}$ a convergent to that square root. A bit of manipulation then gives the result above and the information that equality holds for exactly those values of $n.$
Details: Change the desired inequality to $$m^2 \ge \frac{n^4}{3n^2-k}$$ with $k$ to be determined. Since $n^2=3m^2-j$ with $j \ge 2,$ we have $m^2=\frac{n^2+j}{3}.$ This yields $$k \le \frac{3jn^2}{n^2+j} =3j-\frac{3j^2}{n^2+j}.$$ Since we do sometime have $j=2,$ The bound $k \le 6-\frac{12}{n^2+2}$ is sometime an equality. It is not hard to show that $\frac{3jn^2}{n^2+j}$ increases when $j \ge 2$ does. 
