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

$x_n^2 > 2,$ 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?

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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.)

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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

The formula for $x_{n+1}$ is the Newton's algorithm next estimate of $\sqrt{2}$. if the previous estimate were $x_n$. Picture a parabola $y = x^2-2$; the algorithm tries to find the zero to the right of the origin.

If guess $x_n$ is to the right of the zero (which is at $\sqrt{2}, 0)$) then guess $x_{n+1}$ is at the intersection of the X axis with the line tangent to $y = x^2-2$ at $(x_n,x_n^2-2)$.

(Try it, the slope is $2x_n$, the value is $x_n^2-2$, so the intercept is at $x_n - \frac{x_n^2-2}{2x_n} = \frac{1}{2}\left( x_n + \frac{2}{x_n} \right)$ ).

But since the second derivative is always positive (+2 in fact), the parabola will lie above the tangent everywhere but at the tangent point. So the X intercept will be to the right of $\sqrt{2}$. So $x_{n+1}^2 > 2$.

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