TITLE: For f continuous on $[0,1]$, show that there exist points $\alpha_k$ such that $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $
QUESTION [10 upvotes]: Suppose that $f$ is continous on $[0,1]$ , differentiable on $(0,1)$ , and $f(0)=0$ and  $f(1)=1$.For every integer $n$ show  that there must exist $n$ distinct points $\alpha_1,\alpha_2\cdots,\alpha_n$ in that interval so that
   $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $

Is it theorem or i can show it with basic differentiation knowledge .Please give me hint .

REPLY [15 votes]: $f$ is continuous with $f(0) = 0$ and $f(1) = 1$. It follows from the
intermediate value theorem that there are numbers
$$
   0 = b_0 < b_1 < \dots < b_{n-1} < b_n = 1
$$
such that 
$$
  f(b_j) = \frac jn \text{ for } j= 0, \ldots, n \quad .
$$
Now use the fact that $f$ is differentiable and apply the
mean value theorem
to each interval $[b_{j-1}, b_j]$, $j = 1, \ldots n$.
It follows that there are numbers $\alpha_j \in (b_{j-1}, b_j)$ such that
$$
   f'(\alpha_j) = \frac{f(b_j) - f(b_{j-1})}{b_j - b_{j-1}} = \frac 1{n(b_j - b_{j-1})} \quad .
$$
Then
$$
\sum_{j=1}^n \frac 1{f'(\alpha_j)} = n \sum_{j=1}^n (b_j - b_{j-1}) = n(b_n - b_0) = n
\quad .
$$
The $\alpha_j$ are necessarily distinct because they are in distinct
intervals $(b_{j-1}, b_j)$.