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Let $f:[0,1]\to \mathbb R$ be a differentiable function with $f(0)=0$, $f(1)=1$. Prove that for every positive integer $n$, there exist $n$ distinct numbers $x_1,x_2,\cdots,x_n\in(0,1)$ such that $$\frac{1}{n}\sum_{i=1}^n \frac{1}{f'(x_i)}=1.$$

My friend sent this problem, but I couldn't even work on the question.

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Let $0=y_0 <y_1 < y_2< \cdots <y_n=1$, so that $f(y_i) = i/n$ for $i=0, 1, \cdots, n$. Such $y_i$'s can be found whenever $f$ is continuous.

Then by mean value theorem, there are $x_i \in (y_{i-1}, y_i)$, $i=1, \cdots, n$ such that

$$ f'(x_i) = \frac{f(y_i) - f(y_{i-1})}{y_i - y_{i-1}}\Leftrightarrow \frac{1}{n} \frac{1}{f'(x_i)} = y_i - y_{i-1}$$

summing over $i$,

$$\frac{1}{n} \sum_{i=1}^n \frac{1}{f'(x_i)} = \sum_{i=1}^n (y_i - y_{i-1}) = 1-0 = 1.$$

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