<s>[INST] $a_1=1,a_{n+1}=\frac{n}{a_n}+\frac{a_n}{n}$. Prove that for $n\ge4$, $\lfloor{a_n^2}\rfloor=n$ Define a sequence $\left\lbrace a_{n}\right\rbrace$ by
$\displaystyle{a_{1} = 1\,,\ a_{n + 1} = {n \over a_n} + {a_n \over n}.\quad}$ Prove that for $n \geq 4,\,\,\left\lfloor a_{n}^{2}\right\rfloor=n$

The substitution $b_{n} = a_{n}^{2}$ might be helpful, but I still haven't proved the assertion yet.
 [/INST] First,we  use Mathematical induction have following inequality

$$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$

we easy prove this function $$f(x)=\dfrac{x}{n}+\dfrac{n}{x} $$ is decreasing on $(0,n)$
becasue $$f'(x)=-\dfrac{n}{x^2}+\dfrac{1}{n}\le 0$$
since $a_{1}=1,a_{2}=2,a_{3}=2$,so
$$\sqrt{3}\le a_{3}\le\dfrac{3}{\sqrt{2}}$$
Assmue that
$$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$
then
$$a_{n+1}=f(a_{n})\ge f(\dfrac{n}{\sqrt{n-1}}=\dfrac{n}{\sqrt{n-1}}>\sqrt{n+1}$$
$$a_{n+1}=f(a_{n})\le f(\sqrt{n})=\dfrac{n+1}{\sqrt{n}}$$
In fact,we can prove 

$$a_{n}\le \sqrt{n+1},n\ge 4$$
  since
  $$a_{n+1}=f(a_{n})\ge f(\dfrac{n}{\sqrt{n-1}})=\dfrac{n}{\sqrt{n-1}},n\ge 3$$
  so
  $$a_{n}\ge\dfrac{n-1}{\sqrt{n-2}},n\ge4$$
  and note
  $$a_{n+1}=f(a_{n})\le f\left(\dfrac{n-1}{\sqrt{n-2}}\right)=\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}$$

it suffces prove that
$$\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}\le\sqrt{n+2},n\ge 4$$
$$\Longleftrightarrow n^3-n^2-2n+1<(n^2-n)\sqrt{n^2-4}$$
$$\Longleftrightarrow 2n^3-6n^2+4n-1=2n^2(n-3)+4n-1>0$$
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