Given a suitable function $f$, often of class $C^2$, in the section of quadratic convergence for Newton's iterative method mentioned in WikipediA I found the following estimate $$ \displaystyle \left|{\varepsilon _{n+1}}\right|={\frac {\left|f''(\xi _{n})\right|}{2\left|f'(x_{n})\right|}} |\varepsilon _{n}|^2. \tag{6} $$ where $\alpha$ is a zero of $f$ over the interval $I$ and $$\varepsilon _{n+1} = \alpha - x_{n+1}.$$ Then, it says that the above estimate (6) can be expressed in the following way: $$ \displaystyle \left|{\varepsilon _{n+1}}\right|\leq M |\varepsilon _{n}|^2 $$ where $M$ is the supremum of the variable coefficient of $\varepsilon_n^2$, that is: $$ \displaystyle M=\sup _{x\in I}{\frac {1}{2}}\left|{\frac {f''(x)}{f'(x)}}\right|. $$ Given this new estimate, it is now clear that the order of convergence is at least quadratic if $x_0$ is chosen suitable and $$M |\varepsilon_0| < 1.$$ My question is how to prove that the above new inequality is sufficient? The apparent difficulty is to handle the two points $\xi_n$ and $x_n$ in (6).
In various references, the following inequality $$\varepsilon_{n+1} \leq \frac 12 \frac{\sup_I|f''|}{\inf_I |f'|} |\varepsilon_n|^2$$ often appears. Having this, it is now quite obvious to get the the order of convergence by seeing (6).
