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A real valued function $f(x)$is differentiable on $[0,1]$, where $f(0)=0$ and $f(1)=1$.

Prove that for any integer $n$ there exist $0\le c_1<c_2<\dots<c_n\le1$ where $$\sum_{k=1}^n\frac1{f'(c_k)}=n$$

This is a question a student asked me (I'm a teacher). I have no clue on how to prove it except that I guess it may use the mean-value theorem and the intermediate value theorem. Thanks in advance.

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marked as duplicate by Martin R, Lord Shark the Unknown, Henrik, kingW3, Namaste calculus Oct 15 '17 at 12:59

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    $\begingroup$ This is from the OIMU 2012 olympiad, problem 2. If you speak spanish, you can find the oficial solution here: oimu.eventos.cimat.mx/sites/oimu/files/…. If you don't speak spanish, let me know and I can try to explain the solution to you. $\endgroup$ – u1571372 Nov 11 '16 at 0:46
  • $\begingroup$ @u1571372 Thanks. I don't speak Spanish. Is the Sean's answer the same one? $\endgroup$ – Kay K. Nov 11 '16 at 1:23
  • $\begingroup$ @u1571372 It looks so. Thanks for your help! $\endgroup$ – Kay K. Nov 11 '16 at 1:37
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Use the intermediate value theorem to deduce the existence of an increasing sequence $(y_k)_{k=0}^{n}$ with $f(y_k)=k/n$, where $y_0=0,y_n=1$.

Application of the mean value theorem on the interval $I_k=[y_{k-1},y_k]$ yields the existence of an $x_k\in I_k$ with $f'(x_k)(y_{k}-y_{k-1})=\frac{1}{n}$.

Divide by $f'(x_k)$ and sum from $k=1$ to $k=n$ to complete the proof.

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