Proof using Induction? Any thoughts on how to prove the following proposition
If $x_1>0$ and $a>0$ and $x_{ n+1 }=\frac { 1 }{ 2 } \left( x_{ n }+\frac { a^{ 2 } }{ x_{ n } }  \right) $ then $a\leq x_{n+1}\leq x_n$.
 A: hint
$$x_{n+1}-a=\frac {(x_n-a)^2 }{2x_n}$$
$$x_{n+1}-x_n=\frac {a^2-x_n^2}{2x_n} $$
A: First, you see that $x_n > 0$ for all $n$.
Then $x_{n+1}=\dfrac{a}{2}(\dfrac{x_n}{a}+\dfrac{a}{x_n}) \geq \dfrac{a}{2} 2 =a$
And $x_{n+1}-x_n=\dfrac{a^2-x_n^2}{2x_n}$
And you have to assume that $x_1 > a$ to make the induction works.
A: btw. $n\ge 2$
Please do point out any flaws in the argument
$\mathcal{H}$ = $\sqrt{i\cdot j}\leq \frac{1}{2}(i+j)$
Assume that $x_1>0$, $a>0$ and let $n$ be an arbitrary element of the set $\{2,3,4,5,6,...\}$ from $\mathcal{H}$ we know that $\sqrt{x_{n}\cdot\frac{a^{2}}{x_{n}}} = a\leq x_{n+1} = \frac{1}{2}(x_n+\frac{a^2}{x_n})$. Now assume for purpose of contradiction that $x_{n}<x_{n+1}$ which implies that $x_n<\frac{1}{2}(x_n+\frac{a^2}{x_n})$ examine now the following series of implications $x_n<\frac{1}{2}(x_n+\frac{a^2}{x_n})\implies 2x_{n}^{2}<x_n^{2}+a^{2} \implies x_{n}^{2}-a^{2}<0 \implies (x_n+a)(x_n-a)<0$ we know that $a$ is positive and since $x_1$ is positive it is not difficult to see that $x_n$ is also positive this implies that $0<x_n+a$ but then it must be that $x_n<a$ and we know from $\mathcal{H}$ that $a\leq x_n$ resulting in a contradiction we therefore conclude that $x_{n+1}\leq x_n$ this together with our previous deduction of $a\leq x_{n+1}$ implies that $a\leq x_n$ and thus $a\leq x_{n+1}\leq x_n$.
A: For all $n\in\mathbb N$ we have
$$x_{n+1}-a=\frac{x_n}{2}+\frac{a^2}{x_n}-a=\frac{(x_n-a)^2}{2x_n}\geq0,$$
which says that $x_n\geq a$ for all $n\geq2$.
In another hand, for all $n\geq2$ we have
$$x_{n+1}-x_n=\frac{1}{2}\left(\frac{a^2}{x_n}-x_n\right)=\frac{(a-x_n)(a+x_n)}{2x_n}\leq0,$$
which says that for all $n\geq2$ we have
$$a\leq x_{n+1}\leq x_{n}$$
and your statement is wrong if $x_1<a$.
