The formula for $x_{n+1}$ is the Newton's algorithm next estimate of $\sqrt{2}$. if the previous estimate were $x_n$. Picture a parabola $y = x^2-2$; the algorithm tries to find the zero to the right of the origin.
If guess $x_n$ is to the right of the zero (which is at $\sqrt{2}, 0)$) then guess $x_{n+1}$ is at the intersection of the X axis with the line tangent to $y = x^2-2$ at $(x_n,x_n^2-2)$.
(Try it, the slope is $2x_n$, the value is $x_n^2-2$, so the intercept is at $x_n - \frac{x_n^2-2}{2x_n} = \frac{1}{2}\left( x_n + \frac{2}{x_n} \right)$ ).
But since the second derivative is always positive (+2 in fact), the parabola will lie above the tangent everywhere but at the tangent point. So the X intercept will be to the right of $\sqrt{2}$. So $x_{n+1}^2 > 2$.