Let us consider functions of the form $$ f(x) = x^{2} - a, $$ now the task is to find a root of f.
I was asked to prove that for every initial guess $ x_{0} > 0 $ it is true that $ x_{1} \geq x_{2} \geq ... \geq \sqrt a $ and that the limit $ \lim_{x->\infty} x_{n} $ is convergent to $ \sqrt a $. Having seen a geometrical explanation of this approach I do see the two statements clearly must hold, though I do not know how to rigorously justify them. Any ideas?
Suppose $x_n>\sqrt{a}$. The tangent line to $x^2-a$ at $x_n$ lies below the curve (why?), so it hits the $x$ axis to the right of $\sqrt{a}$ (why?). Where it hits the $x$ axis is $x_{n+1}$ (why?), so $x_{n+1}>\sqrt{a}$. Thus once $x_{n_0}>\sqrt{a}$ for some $n_0$ then $x_n>\sqrt{a}$ for all $n \geq n_0$ (by induction).
This finishes the case $x_0>\sqrt{a}$ (in which case the desired inequalities can be extended to $n=0$ as well). What happens in the case $0<x_0<\sqrt{a}$? Is $x_1>\sqrt{a}$ or not?
Note$$x_n-x_{n+1}=\frac{f(x_n)}{f^\prime(x_n)}=\frac{x_n^2-a}{2x_n},$$so the claimed chain of inequalities holds iff $x_n$ aren't underestimates of $\sqrt{a}$. If $x_n\ge\sqrt{a}$,$$x_{n+1}-\sqrt{a}=\frac{x_n^2+a-2\sqrt{a}x_n}{2x_n}=\frac{(x_n-\sqrt{a})^2}{2x_n}.$$So regardless of $x_0$, $x_1\ge\sqrt{a}$, after which $x_1\ge x_2\ge\sqrt{a}$ etc.
In short, this is Darboux theorem.
$x_0$ being the starting point of Newton method for solving $f(x)=0$, there will not be any overshoot of the solution if $$f(x_0)\times f''(x_0) >0$$
Applied to your case, it then requires that $$2(x_0^2-a) >0\implies x_0 > \sqrt a$$
