# Prove there exist $0\le c_1<c_2<\dots<c_n\le1$ where $\sum_{k=1}^n\frac1{f'(c_k)}=n$ [duplicate]

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.

## marked as duplicate by Martin R, Lord Shark the Unknown, Henrik, kingW3, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 15 '17 at 12:59

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• 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. – u1571372 Nov 11 '16 at 0:46
• @u1571372 Thanks. I don't speak Spanish. Is the Sean's answer the same one? – Kay K. Nov 11 '16 at 1:23
• @u1571372 It looks so. Thanks for your help! – Kay K. Nov 11 '16 at 1:37

## 1 Answer

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.