TITLE: sum of series using mean value theorem
QUESTION [5 upvotes]: Let $f(x)$ be a function which is differentiable on $[0,1]$ with $f(0)=0$ and $f(1)=1$.
Show that for every $n\in \Bbb N$ there exists numbers $x_1,x_2,\ldots,x_n\in [0,1]$ such as
$$
 \sum_{k = 1}^n \frac{1}{f' (x_k)} = n
$$
I think the mean value theorem should be applied.
So there exists $x_1$ in $[0,1]$  such that $f ' (x_1) = \frac{f(1) - f(0)}{1-0} =1$  and there exists $x_2$ in $[0,x_1]$ such that $f ' (x_2) = \frac{f(x1) - f(0)}{x1-0} = \frac{f(x1)}{x1}$, so on and so forth for $x_3 ,x_4, \ldots,x_n$ and we have the sum $$1+\frac{x_1}{f(x1)} + \frac{x_2}{f(x2)} +\cdots+\frac{x_{n-1}}{f(x_{n-1})}$$  and from here I have no idea what to do .
I was wondering if anyone could be so kind to help ?

REPLY [5 votes]: By the intermediate value property of continuous functions, there are $0=x_0,x_1,\ldots,x_{n-1},x_n=1\in[0,1]$ such that:
$$ f(x_i) = \frac{i}{n} $$
for every $i\in[0,n]$, and we may assume WLOG $x_0<x_1<\ldots<x_n$. Moreover, there is some $\xi_i$ in the interval $(x_{i-1},x_i)$ such that:
$$ \frac{f(x_{i})-f(x_{i-1})}{x_{i}-x_{i-1}}=f'(\xi_i) $$
that is:
$$ \frac{1}{f'(\xi_i)} = n(x_i-x_{i-1}).$$
By summing those identities over $i\in[1,n]$ we get:

$$ \sum_{i=1}^{n}\frac{1}{f'(\xi_i)}= n(x_n-x_0) = \color{red}{n} $$

as wanted.