TITLE: $ af'(\xi)+bf'(\eta)=0 $ for some $ 0<\xi<\eta<1 $
QUESTION [5 upvotes]: Suppose $ f(x) $ is a differentiable function defined on $ [0, 1] $, satisfying $ f(0)=f(1) $ and for every $ 0<a<b $, there exist $ 0<\xi<\eta<1 $ such that $ af'(\xi)+bf'(\eta)=0 $.

My attempt: WLOG, assume that $ f(0)=f(1)=0 $. If $ f'(x) $ has at least two zeros on $ (0, 1) $, then set $ f'(\xi)=f'(\eta)=0 $ and we are done. It leaves us to prove when $ f'(x) $ has exactly one zero on $ (0, 1) $. It should be clear that $ \xi $ and $ \eta $ will not be zeros of $ f'(x) $ on $ (0, 1) $ by our assumption. So it suffices to show that $$ \frac{f'(\xi)}{f'(\eta)}<-1 $$
for some $ \xi $ and $ \eta $. Then I am stuck......

EDIT:
It is pretty clear to see that $ \frac{f'(\xi)}{f'(\eta)}<0 $ for some $  0<\xi<\eta<1 $. Since $ f'(x) $ has only one zero on $ (0, 1) $, then WLOG, we assume that $ f(x)>0 $ when $ x\in (0, 1) $.

REPLY [3 votes]: Observe that by the mean value theorem, there exist $\displaystyle\xi\in \left(0,\frac{a}{a+b}\right)$ and $\displaystyle\eta\in\left(\frac{a}{a+b},1\right)$ such that
$$\begin{align*}
\frac{a}{a+b}f'(\xi)&=f\left(\frac{a}{a+b}\right)-f(0)\\&=-f(1)+f\left(\frac{a}{a+b}\right)\\&=-\frac{b}{a+b}f'(\eta).
\end{align*}$$ It follows $0<\xi<\eta<1$ and $\displaystyle af'(\xi)+bf'(\eta)=0$.  Note: In fact, we don't need to assume $a<b$.