Suppose that $f(x)'>0$ and that $f(a)<0$ while $f(b)>0$, so that $f(x)=0$ has a root $r$ in the interval $(a,b)$. Newton's method of finding $r$, starting at $c$ in $(a,b)$ is as follows: let $x_0=c$, and for $n \ge 1$, define: $$x_n=x_{n-1}-f(x_{n-1})/f'(x_{n-1}).$$ If we suppose in addition that $f''(x)>0$ and $x_0=c>r$ show that: $$f(x_1)>f(x_2)>...>0$$ and $$x_1>x_2>...>a$$ and deduce that $x_n \to r$.
The following is my attempt to solve this question. The statement required to prove is partly equivalent to: $x_{n+1}-x_n<0$ for all integers $n\ge0$. So I thought about using mathematical induction:
Step 1: prove this statement to be true for $n=0$
$$x_1-x_0=-\frac{f(x_0)}{f'(x_0)}$$
Given that $f(x_0)>0$ (as $x_0>r$) and $f'(x_0)>0$,
$$x_1-x_0<0$$ This statement is hence true for $n=0$.
Step 2: assume this statement to be true for $n=k$
that is, we assume that
$$x_{k+1}-x_k<0$$ and $$f(x_k)>0$$
Step 3: prove this statement to be true for $n=k+1$
$$x_{k+2}-x_{k+1}=-\frac{f(x_{k+1})}{f'(x_{k+1})}$$
And here is where I got stuck.
Clearly $f'(x_{k+1})>0$ but proving $f(x_{k+1})>0$ is quite difficult.
This is what I've tried:
Suppose $f'(x_{k+1})<0$ (i.e. $x_{k+1}<r$) then$$x_{k+2}>x_{k+1}$$ and $$x_{k+1}<x_{k}$$
and hence $$f(x_k)>f(x_{k+1})<f(x_{k+2})$$ $$f'(x_k)>f'(x_{k+1})<f'(x_{k+2})$$
From here onward, I have attempted different methods but have failed to obtain a contradiction.
And by sketching it on graph, you can see that this statement holds true.
What is a more common approach to this type of question?
The idea is the geometric observation that a strictly convex function lies strictly above its tangent lines everywhere except the point of tangency. One can prove this analytically by using the mean value theorem (or, for some overkill, Taylor's theorem).
Therefore if you have a tangent line function $T(x;x_0)=f(x_0)+f'(x_0)(x-x_0)$, and $f(x_0) \neq 0$, then whenever $T(y;x_0)=0$, you have $f(y)>0$. This means that in the Newton iteration, if $f(x_{n-1})>0$ then $f(x_n)>0$, so if $f(x_0)>0$ then $f(x_n)>0$ for all $n \geq 0$ by induction. (Note that if $f(x_{n-1})<0$ then actually $f(x_n)>0$ for all $n \geq 1$, which also follows by induction.) On the other hand, since $f'>0$, if $f(x_{n-1})>0$ then $x_n<x_{n-1}$ so you move monotonically toward the root.