How to prove that for $a_{n+1}=\frac{a_n}{n} + \frac{n}{a_n}$ , we have $\lfloor a_n^2 \rfloor = n$?

We will prove the following result instead.

$$n+{2\over n}<{a_n}^2<n+1$$

First we check it's true for $n=4$.

Now the induction step, first notice that $x>y>1$ implies $x+{1\over x}>y+{1\over y} \equiv 1>{1\over xy}$

Therefore for RHS $${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{n+{2\over n}\over n^2}+{n^2 \over n+{2\over n}}+2=2+{n^4+n^2+4+{4\over n^2}\over n^3+2n}$$$$= 2+n-{n^2-4-{4\over n^2}\over n^3+2n}<n+2$$

For LHS,

$${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2>{n+1\over n^2}+{n^2\over n+1}+2={n^4+n^2+2n+1\over n^3+n^2}+2$$ $$=2+(n-1)+{2n^2+2n+1\over n^3+n^2}>n+1+{2\over n}>n+1+{2\over n+1}$$