$a_1=1,a_{n+1}=\frac{n}{a_n}+\frac{a_n}{n}$. 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.
 A: 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
$$\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}\le\sqrt{n+2},n\ge 4$$
$$\Longleftrightarrow n^3-n^2-2n+1<(n^2-n)\sqrt{n^2-4}$$
$$\Longleftrightarrow 2n^3-6n^2+4n-1=2n^2(n-3)+4n-1>0$$
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
$\tt{\large\it\mbox{Hint:}}\ \mbox{For}\ n > 1:$
$$
a_{n + 1} = {n \over a_{n}} + {a_{n} \over n}\quad\imp\quad a_{n}^{2} - \pars{na_{n + 1}}a_{n} + n^{2} = 0
$$
Since $a_{n} \in {\mathbb R}$, we'll have $\pars{na_{n + 1}}^{2} - 4n^{2} \geq 0\quad
\imp\quad a_{n + 1} \geq 2$ and
$$
a_{n} =
\bracks{\vphantom{\LARGE A^{A^{A}}}%
{a_{n + 1} \over 2}  - \root{\pars{a_{n + 1} \over 2}^{2} - 1}}\,n
=
{n \over a_{n + 1}/2  + \root{\pars{a_{n + 1}/2}^{2} - 1}} 
$$ 
$$
\mbox{With}\quad b_{n} \equiv {a_{n} \over 2}\,,\qquad b_{1} = {1 \over 2}\,,\quad
b_{n} = {n/2 \over b_{n + 1}  + \root{b_{n + 1}^{2} - 1}}\,;
\qquad b_{n} \geq 1\,,\quad n \geq 2
$$
We have to prove that $\floor{b_{n}^{2}} = n/4$ when $n \geq 4$. Notice that
$b_{n} \leq n/2$.
