I'm studying the book of Elon Lages Lima, Análise Real volume 1. In the section of applications of derivatives, the author wants to estimate the error for calculating the nth root of $c>0$. Let $f(x)=x^n-c$, by using the Taylor's formula, there exists $d$ between $a=\sqrt[n]{c}$ and $x_{k}$, such that
\begin{equation} 0=f(a)=f\left(x_{k}\right)+f^{\prime}\left(x_{k}\right)\left(a-x_{k}\right)+\frac{f^{\prime \prime}(d)}{2}\left(a-x_{k}\right)^{2}. \end{equation} So,
\begin{equation} f^{\prime}\left(x_{k}\right) x_{k}-f\left(x_{k}\right)-f^{\prime}\left(x_{k}\right) a=\frac{f^{\prime \prime}(d)}{2}\left(x_{k}-a\right)^{2} . \end{equation}
Dividing by $f^{\prime}\left(x_{k}\right)$ we obtain: \begin{equation} x_{k}-\frac{f\left(x_{k}\right)}{f^{\prime}\left(x_{k}\right)}-a=\frac{f^{\prime \prime}(d)}{2 f^{\prime}\left(x_{k}\right)}\left(x_{k}-a\right)^{2} \end{equation} that is, \begin{equation} x_{k+1}-a=\frac{f^{\prime \prime}(d)}{2 f^{\prime}\left(x_{k}\right)}\left(x_{k}-a\right)^{2} . \end{equation}
Finally he concludes that \begin{equation} \frac{f^{\prime \prime}(d)}{2 f^{\prime}\left(x_{k}\right)}=(n-1) \frac{d^{n-2}}{2 x_{k}^{n-1}} \leq \frac{n-1}{2 x_{k}} \text {. } \end{equation}
My question is:
How to prove that \begin{equation} (n-1) \frac{d^{n-2}}{2 x_{k}^{n-1}} \leq \frac{n-1}{2 x_{k}}? \end{equation}
I think that we don't have control of the size of $d$, which depends o $k$.
$$ x_{k+1}-a=\frac{f^{\prime \prime}(d)}{2 f^{\prime}\left(x_{k}\right)}\left(x_{k}-a\right)^{2} . $$ shows that $x_k \ge a$ after the first iteration. And if $a \le x_k$ then $a \le d \le x_k$ and consequently $$ (n-1) \frac{d^{n-2}}{2 x_{k}^{n-1}} \leq (n-1) \frac{x_k^{n-2}}{2 x_{k}^{n-1}} = \frac{n-1}{2 x_{k}} $$