<s>[INST] If $\frac{n^2}{3} < xy \leq n^2 - 5$, then does it follow that $n > \max(x,y)$? My question here is pretty basic:

If $\frac{n^2}{3} < xy \leq n^2 - 5$, then does it follow that $n > \max(x,y)$?

Here, $n, x, y$ are positive integers.
MY ATTEMPT
Let $n, x, y$ be positive integers such that
$$\frac{n^2}{3} < xy \leq n^2 - 5.$$
Suppose to the contrary that we have
$$n \leq \max(x,y) = \frac{x+y+|x-y|}{2}.$$
We get the bounds
$$\frac{n^2}{3x} < y \leq \frac{n^2 - 5}{x}$$
which implies that
$$\frac{5 - n^2}{x} \leq -y < -\frac{n^2}{3x}.$$
It follows that
$$n \leq \frac{\bigg(x + \dfrac{n^2 - 5}{x}\bigg) + |x - \dfrac{n^2}{3x}|}{2}.$$
I then asked WolframAlpha to solve the last inequality for $x$ and then for $n=a$, but I am not sure I can interpret the outputs by myself, as they are a bit messy.
Can anybody help?  Thanks!
 [/INST] For every $n\ge 3$, a counter example is given by $x=n-2$ and $y=n+1$
indeed in that case we have
\begin{equation}
\frac{n^2}{3}< n^2 - n - 2 = x y \le n^2 - 5
\end{equation}
</s>