For $a>0$ and $x_0\ge a$ prove that the sequence defined as $x_{n+1}=(x_n+a/x_n)/2$ decreases and converges to $\sqrt a$ The problem seems slightly wrong in it assumptions. Please, check this proof.

For $a>0$ and $x_0\ge a$ prove that the sequence defined as $x_{n+1}=(x_n+a/x_n)/2$ decreases and converges to $\sqrt a$.

Then we have that $a/x_0\le1$ and then
$$x_1=\frac{x_0}2+\frac{a}{x_0}\frac12\le\frac{x_0}2+\frac12$$
but then
$$x_1\le\frac{x_0}2+\frac12<x_0\iff x_0>1$$
The exercise seems to be incorrect here, to continue I assume that $x_0>1$. Now I assume the induction hypothesis, i.e. $x_{n+1}<x_n$, then I will check if it is true that $x_{n+2}<x_{n+1}$. We have
$$x_{n+2}=(x_{n+1}+a/x_{n+1})/2=\frac{(x_{n+1})^2+a}{2x_{n+1}}<x_{n+1}$$
then $$(x_{n+1})^2+a<2(x_{n+1})^2\iff\sqrt a<x_{n+1}$$
but the induction hypothesis $x_{n+1}<x_n$ implies too that $\sqrt a<x_n$, then
$$\sqrt a<x_{n+1}\implies 2\sqrt a<\frac{(x_n)^2+a}{x_n}\iff 0<x_n^2-2\sqrt ax_n+a$$
To solve the inequality of second order first we solve the equation, where we get that the unique solution is $x=\sqrt a$. Then I substitute $x_n=\epsilon\sqrt a$ with $\epsilon>0$ to study the behavior of the inequality
$$0<x_n^2-2\sqrt ax_n+a\iff 0<\epsilon^2-2\epsilon+\epsilon\iff 0<(\epsilon-1)^2$$
Then for $x_0>1$ the sequence decreases.
Because the above we can see that $\sqrt a$ is a lower bound of the sequence, then (and because is decreasing) is a convergent sequence. Because the sequence converges then
$$\lim x_{n+1}=\lim \frac{(x_n)^2+a}{2x_n}\implies x=\frac{x^2+a}{2x}\iff x=\sqrt a$$

Trying to see what happen if $0<x_0\le 1$ I observe that is not monotone. What I know is that if $x_n<\sqrt a$ then $x_{n+1}\ge\sqrt a$.
The sequence seems oscillate around $\sqrt a$. I know that converges to $\sqrt a$ because the above sequence is a way to compute square roots, but I dont know how show it with these tools.
 A: Let $f(x)=\frac{1}{2}(x+\frac{a}{x})$
$f$ has two fixed points
$x=\sqrt{a}$ and $x=-\sqrt{a}$
those are possible limits of the sequence $(x_n)$ such that
$x_0\geq a>0$ and
$x_{n+1}=f(x_n)$.
by induction we can easily prove that
$\forall n\in \mathbb N  \; x_n>0$
thus if $(x_n)$ converges, the limit will be $\sqrt{a}$ as $f$ is continuous at $(0,+\infty)$.
on the other hand,
$f'(x)=\frac{1}{2}\frac{x^2-a}{x^2}$.
$f(x)-x=\frac{1}{2}\frac{a-x^2}{x}$.
now let us discuss three cases.
FIRST
$x_0=\sqrt{a}$.
in this case, the sequence is constant and convergent.
SECOND
$x_0>\sqrt{a}$
by induction we prove that in this case
$\forall n \;  x_n \in (\sqrt{a},+\infty)$
as $f$ is increasing.
thus
$f(x_0)-x_0<0 \implies x_1<x_0$
and also by induction we get
$x_{n+1}<x_n$
finally $(x_n)$ decreases and $\sqrt{a}<x_n \implies $(x_n)$ converges to $\sqrt{a}$.
THIRD CASE
$x_0<\sqrt{a}$
this case is little more difficult since $f$ is decreasing and i think that your hypothesis should be $x_0\geq\sqrt{a}$.
A: The sequence being decreasing means $x_{n+1}<x_n$, for all $n$. This becomes
$$
x_n+\frac{a}{x_n}<2x_n
$$
that is, $x_n^2>a$, because it is obvious that $x_n>0$ for all $n$. I believe that the statement should have $x_0^2>a$. For instance, if $a=1/4$ and $x_0=1/3$, we have $x_0^2<a$, which implies $x_1>x_0$.
So, I'll go with $x_0^2>a$ as the hypothesis, which is the base step in the induction proof.
Suppose $x_n^2>a$. Then $x_{n+1}^2-a$ becomes
$$
x_{n+1}^2-a=\frac{1}{4}\left(x_n+\frac{a}{x_n}\right)^{\!2}-a=
\frac{1}{4}\left(x_n^2+2a+\frac{a^2}{x_n^2}-4a\right)=
\frac{1}{4}\left(x_n-\frac{a}{x_n}\right)^{\!2}>0
$$

If the hypothesis is $x_0>a$, then the sequence need not be decreasing, as shown before. Of course, if we assume $a>1$, then $x_0>a$ implies $x_0^2>a$, because from $x_0>a>1$ we get
$$
x_0^2>ax_0>x_0>a
$$
So the statement should have either $x_0^2>a$ or $a>1$. In the case $x_0>a$ and $0<a\le1$ we may indeed obtain an oscillating sequence.
