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

Question

Let $(a_n)_{n\ge 1}$ be the sequence defined as the following : $$a_1=1 ,\ a_{n+1}=\dfrac{a_n}{n} + \dfrac{n}{a_n} ,\ n\ge1$$ Show that for every $n\ge4,\ \lfloor a_n^2 \rfloor = n$.
My approach to this problem was trying induction and using the function $f_n(x)=\dfrac{x}{n} + \dfrac{n}{x}$ :
Proving the base case for $n= 4$ and then by the inductive hypothesis $\lfloor a_n^2 \rfloor = n$ implies that $$\sqrt{n} \le a_n \lt \sqrt{n+1}$$
We then apply $f_n$ knowing that it is decreasing in that interval following it up with the floor function and some polishing, all leads to this inequality : $$n+1\le \lfloor a_{n+1}^2 \rfloor \le n+2$$
So I can't exactly get $n+1$ since $n+2$ is a possibility, this problem is a product of the fact that if $a\lt b$ then $\lfloor a \rfloor \le \lfloor b \rfloor$.

Any insights would be greatly appreciated! I wonder if my result is correct because it seems like the only way.

Answer 1

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,

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