How does $(x_n - \sqrt{2})^2/(2x_n ) \geq 0$ mean $x_n \geq \sqrt{2}$ when $x_1=1$? 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 $\dfrac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$? 
 A: you have concluded that for $x_n > 0$,$$x_{n+1} - \sqrt{2} = {{((x_n−\sqrt{2})^2)}\over{(2x_n)}}≥0. $$ This means $$x_{n+1} - \sqrt{2} ≥ 0$$ Now since n is arbitrary, any value of n will satisfy this equation , put $n = n -1$ in this equation, you will get
 $$x_{n} - \sqrt{2} ≥0$$
A: There seems to be confusion on what you think they claim, and what they actually claim. The step you are talking about is as follows.

Conclude that $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$.

What is claimed here is that if $x_n \geq \sqrt{2}$, then $x_{n+1} \geq \sqrt{2}$. We know this holds, since, as shown in the previous step, 
$$x_{n+1} - \sqrt{2} = \frac{(x_n - \sqrt{2})^2}{2 x_n} \geq 0. \quad \quad (\text{if } x_n > 0)$$
So a stronger statement actually holds: $x_n > 0 \Rightarrow x_{n+1} \geq \sqrt{2}$. Since $x_n \geq \sqrt{2}$ implies $x_n \geq 0$, this then also implies $x_{n+1} \geq \sqrt{2}$.
However, this does not mean that $x_n \geq \sqrt{2}$ for any $n$! This only means that if $x_{n_0} \geq \sqrt{2}$ for some $n_0 \in \mathbb{N}$, then we can use induction to show that $x_n \geq \sqrt{2}$ for all $n \geq n_0$. 
So probably you would now also like to find such an $n_0$. Since $x_n > 0 \Rightarrow x_{n+1} \geq \sqrt{2}$ as shown above, and $x_1 = 1 > 0$, it follows that $x_2 \geq \sqrt{2}$. And indeed, $x_2 = 2 \geq \sqrt{2}$. So then, by induction we get $x_n \geq \sqrt{2}$ for all $n \geq 2$. 
