Given

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

Show that

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

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I got it down to:

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

but not sure what to do now...