$x_n^2 > 2,$ 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?
 A: 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.)
A: 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 $x_{n+1}^2 > 2$. 
