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!

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}