<s>[INST] 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 $\dfrac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$? 
 [/INST] 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
 $$x_{n} - \sqrt{2} ≥0$$
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