TITLE: How to prove there exist distinct $a_{i}$ such $f'(a_{1})f'(a_{2})f'(a_{3})\cdots f'(a_{n})=1$
QUESTION [11 upvotes]: Let $f$ be a continuous map from $[0,1]$ to $R$ that is differentiable on $(0,1)$,with $f(0)=0,f(1)=1$, show that for each postive integer $n$ there exist distinct numbers $a_{1},a_{2},\cdots,a_{n}\in (0,1)$,such that
$$f'(a_{1})f'(a_{2})f'(a_{3})\cdots f'(a_{n})=1$$Thanks in advance.

REPLY [5 votes]: An inductive proof.
We prove a formally more general (in fact obviously equivalent) statement:
If $c>0$, $f:[0,c]\to\mathbb{R}$ is continous and differentiable in $(0,c)$ and $f(0)=0$, then there are distinct numbers $0<a_1<\ldots<a_n<c$ such that $f'(a_1)\cdots f'(a_n)=\left(\dfrac{f(c)}{c}\right)^n$.
The case $n=1$ follows from Lagrange's MVT.
Suppose that $n\ge2$, and the statement is true for $n-1$.
Apply Cauchy's MVT to the functions $(f(x))^n$ and $x^n$. By the MVT, there is some $\xi\in(0,c)$ such that 
$$
\dfrac{n\cdot (f(\xi))^{n-1} \cdot f'(\xi)}{n\cdot (\xi)^{n-1}}
= \dfrac{f^n(c)}{c^n}
$$
$$
\left(\dfrac{f(\xi)}{\xi}\right)^{n-1} \cdot f'(\xi) = \left(\dfrac{f(c)}{c}\right)^n. 
$$
Choose $a_n=\xi$; then we have
$$
\left(\dfrac{f(a_n)}{a_n}\right)^{n-1} \cdot f'(a_n) = \left(\dfrac{f(c)}{c}\right)^n. \tag1
$$
Now apply the induction hypothesis to the interval $[0,a_n]$: there are 
some $0<a_1<\ldots<a_{n-1}<a_n$ such that 
$$
f'(a_1)\cdots f'(a_{n-1}) = 
\left(\dfrac{f(a_n)}{a_n}\right)^{n-1}.
\tag2
$$
Plugging (2) in (1), the statement follows.