$x_n^2 > 2,$ how to show that $x_{n+1}^2 > 2$?

Question

$x_n^2 > 2,$ how to show that $x_{n+1}^2 > 2$? I have tried using induction on this but haven't been able to solve this for a while.

The sequence is defined as $x_1 = 2,$ $x_{n+1} = \frac{1}{2}(x_n + \frac{2}{x_n}).$ All I got by induction was that 2 > 1.5, which is not sufficient (I just squared and expanded the terms). How can I solve this problem?

Srivatsan · Accepted Answer · 2012-01-18 00:08:24Z

up vote 1 down vote accepted

This feels like homework, so I will give just a hint initially.

HINT: Use the AM-GM inequality. Alternatively, you can do: $$ \begin{align*} x_{n+1}^2 = \frac{1}{4} \left( x_n^2 + \frac{4}{x_n^2} + 4 \right) = \frac{1}{4} \left[ \left( x_n - \frac{2}{x_n} \right)^2 + 4 + 4 \right] \gt \cdots \end{align*} $$ (Complete the chain of (in)equalities.)

edited Jan 18 '12 at 0:08

answered Jan 18 '12 at 0:01

Srivatsan
15.6k33786

	I can't use that because the book I am using hasn't taught it to me or proven it anywhere previously. We didn't do any of am-gm in class either. This is a problem for pre-prelim/midterm practice. – tyler1 Jan 18 '12 at 0:03
	@tyler1 Well, what you are seeking to prove is -in some sense- equivalent to AM-GM for two numbers, so using it is the best route. But if you prefer, you could "cheat" by mimicking the proof of the inequality: see the addition to my answer. – Srivatsan Jan 18 '12 at 0:10

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$x_n^2 > 2,$ how to show that $x_{n+1}^2 > 2$?

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