How to show that a particular sequence is always above a certain value? I am trying to solve the following question but I am a little confused on how to solve it. The question is the following:
$x_{n+1} = \frac{x_{n}}{2} + \frac{a}{2x_{n}}, x_{0} = r >0, r \in Q, a > 0$.
I need to show that $x^{2}_{n} > a  \forall n \ge 1$ . I am a little confused on how to do this, since what I essentially did was express $x^{2}_{n} = \frac{1}{4}x^{2}_{n-1} + a^{2}\frac{1}{4x_{n-1}^{2}} + \frac{a}{2}$.
But I have no clue what to do from here on forth.
Like this shows that the $x_{n}^{2}$ is definitely at least greater than a/2. But how can I show that it is greater than a?
Also, I was trying to show that $x_{n} > x_{n+1}  \forall n \ge 1$. I am so confused on how to do this as well. I tried to prove this using induction but really could not get to an end result. If I split up the $x_{n+1}$ term I get that each term individually is less than $x_{n}$. But how can I cummulatively show that it is less than $x_{n}$?
Finally, I was asked to conclude using the two solutions above that $x_{n} -> L$ where $L^{2} = a$
I have no idea how to even get started on this part.
Any answer would be greatly appreciated.
Thanks!
 A: Let $f : (0, +\infty) \rightarrow \mathbb{R}$ defined for all $x > 0$ by
$$f(x)= \frac{x}{2} + \frac{a}{2x}$$
$f$ is differentiable on $(0, +\infty)$ and for all $x > 0$,
$$f'(x)= \frac{1}{2} - \frac{a}{2x^2}$$
so $f'(x) \geq 0$ iff $x \geq \sqrt{a}$. You deduce that $f$ attains its minimum at $x=\sqrt{a}$, so for all $y > 0$,
$$f(y) \geq f(\sqrt{a})= \frac{\sqrt{a}}{2} + \frac{a}{2\sqrt{a}} = \sqrt{a}$$
In particular, for all $n \geq 1$, $x_n = f(x_{n-1}) \geq \sqrt{a}$, so you deduce that
$$x_n^2 \geq a$$
A: If $x_0^2 = a$ then $x_n^2 = a$ for all $n$, i.e. the sequence is constant.
Otherwise
$$
 x_{n+1}^2 - a = \frac{1}{4}x^{2}_{n} + a^{2}\frac{1}{4x_{n}^{2}} - \frac{a}{2}
= \left( \frac{x_n}{2}- \frac{a}{2x_n}\right)^2
= \frac{(x_n^2 -a)^2}{4x_n^2}
$$
shows that $x_n^2 > a$ for all $n \ge 1$. Then
$$
x_{n+1} - x_n = -\frac{x_{n}}{2} + \frac{a}{2x_{n}} = - \frac{x_n^2-a}{2x_n} < 0
$$
shows that $x_{n+1} < x_n$ for $n \ge 1$.
So you have a decreasing sequence which is bounded below, and that is convergent by the monotone convergence theorem, i.e. $L = \lim_{n\to \infty} x_n$ exists.
Finally use the recursion formula to show that $L$ satisfies
$$
 L = \frac L2 + \frac{a}{2L}
$$
and conclude that $L^2=a$.
