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?:)

$x_1\geq\sqrt{2}$, $x_2\geq\sqrt{2}$. Now suppose $x_{n-1}\geq\sqrt{2}$, then by AM-GM we have $x_n=\frac{1}{2}(x_{n-1}+\frac{2}{x_{n-1}})\geq\frac{1}{2}(2\sqrt{x_{n-1}\frac{2}{x_{n-1}}})=\sqrt{2}$
