I don't understand one step of proving of the Babylonian Method for $a=2,x_1=1$, and
$x_{n+1} =\frac{1}{2}( x_n + \frac{a}{x_n}). \quad (n=1,2,\ldots)$
$x_{n+1} - \sqrt{2} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - \sqrt{2} = \frac{(x_n^2-2\sqrt{2} x_n+2)}{2x_n} = \frac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$
For all values of $x_n>0$. Conclude $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$.
After using $x_n \geq \sqrt{2}$ we find
$x_{n+1} - x_n = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - x_n = \frac{x_n^2+2-2x_n^2}{2x_n} =\frac{2-x_n^2}{2x_n} \leq 0,$
which means the sequence $(x_n)$ is monotonically decreasing. Now we know $(x_n)$ is bounded below by $\sqrt{2}$ and $(x_n)$ is monotonically decreasing so $(x_n)$ has a limit.
My problem is I don't understand how could we say $x_n \ge \sqrt{2}$ by using $\dfrac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$?