1. ## Sequence inequality

Given

$x_1=1,\quad x_{n+1} = \frac{1}{2}\left( x_n + \frac{2}{x_n}\right),\quad n = 1, 2, 3, \dots$

Show that

$x_{n+1}^2 -2 \ge 0$

-------

I got it down to:

$x_{n+1}^2 -2=\frac{x_n^2}{4}+\frac{1}{x_n^2}-1$

but not sure what to do now...

2. $x_{n+1}^2=\frac{1}{4}\left(x_n^2+4+\frac{4}{x_n^2} \right)$

$=2+\frac{1}{4}\left(x_n^2-4+\frac{4}{x_n^2}\right)$

$=2+\frac{1}{4}\left(x_n-\frac{2}{x_n}\right)^2>2.$

3. Originally Posted by scorpion007
Given

$x_1=1,\quad x_{n+1} = \frac{1}{2}\left( x_n + \frac{2}{x_n}\right),\quad n = 1, 2, 3, \dots$

Show that

$x_{n+1}^2 -2 \ge 0$

-------

I got it down to:

$x_{n+1}^2 -2=\frac{x_n^2}{4}+\frac{1}{x_n^2}-1$

but not sure what to do now...
Induction.

So our base case is obviously $n=1$ and $x_2^2=\frac{1}{4}\left(1+\frac{2}{1}\right)^2=\fra c{9}{4}\geqslant2$. Now assume that $x_{n}^2\geqslant 2$ then $x_{n+1}^2=\frac{1}{4}\left(x_n+\frac{2}{x_n}\right )^2=\frac{1}{4}\left(x_n^2+4+\frac{1}{x_n^2}\right )\geqslant\frac{1}{4}\left(4+4+\frac{1}{4}\right)= 2+\frac{1}{16}\geqslant 2$

4. Thanks.

I follow everything except this bit:
$
\frac{1}{4}\left(x_n^2+4+\frac{1}{x_n^2}\right)\ge qslant\frac{1}{4}\left(4+4+\frac{1}{4}\right)$

shouldn't it be

$
\frac{1}{4}\left(x_n^2+4+\frac{1}{x_n^2}\right)\ge qslant\frac{1}{4}\left({\color{red}2}+4+\frac{1}{\ color{red}2}\right)$

since we have that
$
x_{n}^2\geqslant 2
$

?

5. related question.

For the same sequence,

Show that

$x_{n+1}-x_n\le 0$ for all $n \ge 2$

I did the following:

Base case: n=2

$x_3 -x_2=\frac{17}{12}-\frac{3}{2}=-\frac{1}{12}\le 0$

Induction:
If $x_{n+1}-x_n \le 0$ then

$x_{n+2}-x_{n+1} = \frac{1}{2}\left(x_{n+1} + \frac{2}{x_{n+1}}\right) - \frac{1}{2}\left(x_{n} + \frac{2}{x_{n}}\right)$

$= \frac{1}{2}\left(\underbrace{x_{n+1}-x_n}_{\le 0} +\frac{2}{x_{n+1}} - \frac{2}{x_{n}}\right)$

$\le \frac{1}{2}\left(\frac{2}{x_{n+1}} - \frac{2}{x_{n}}\right)
= \frac{1}{x_{n+1}} - \frac{1}{x_{n}}$

but now what? I need to show that that expressions is <= 0.