Prove all elements $x_n \geq \sqrt{2}.$ Given $x_1 = 2,$ and $$x_{n+1} = \frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg),$$ show that for all $n \in \mathbb{N},$ $x_n \geq \sqrt{2}.$
I tried the following. Suppose, for contradiction, that $x_{n+1} < \sqrt{2}.$ Then, $$\frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg) < \sqrt{2},$$ and $$x_n + \frac{2}{x_n} < 2\sqrt{2}.$$ Given that $x_n \geq \sqrt{2},$ it follows that $$\sqrt{2} + \frac{2}{x_n} \leq x_n + \frac{2}{x_n} < 2\sqrt{2},$$ and $$\sqrt{2} + \frac{2}{x_n} < 2\sqrt{2}.$$ So, $$\frac{2}{x_n} < \sqrt{2},$$ and $$\sqrt{2} < x_n.$$ No contradiction. I've tried proving it directly, by squaring both sides and by leaving the expression as is, and I consistently find that the $x_n$ and the $x_n^{-1}$ are together problematic.
I also tried something that allowed me to arrive at the following: $$x_{n+1}x_n \geq 2.$$ But, I don't think that's helpful.
Help?:)

This is immediate from AM-GM: $\frac{1}{2}\bigg(x + \frac{2}{x} \bigg)$ is the arithmetic mean of $x$ and $2/x$, which is always greater than or equal to the geometric mean $\sqrt{x\cdot\frac{2}{x}}=\sqrt{2}$ (for $x>0$).
Alternatively, if you don't know AM-GM, you can reach the same conclusion easily with a bit of calculus.  Letting $f(x)=x + \frac{2}{x}$, we have $f'(x)=1-\frac{2}{x^2}$ which is negative for $0<x<\sqrt{2}$ and positive for $x>\sqrt{2}$.  It follows that $f(x)$ is minimized (on $(0,\infty)$) at $x=\sqrt{2}$, so $\frac{1}{2}f(x)\geq\frac{1}{2}f(\sqrt{2})=\sqrt{2}$ for all $x>0$.
