TITLE: There exist $x_1, x_2, x_3$ such that $\frac{1}{f'(x_1)} + \frac{1}{f'(x_2)} + \frac{1}{f'(x_3)} = 3$
QUESTION [6 upvotes]: Let $f$ be a real-valued function defined in $[a, b] \subset \mathbb{R}$, with
$f(a) = a, f(b) = b$. Suppose that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Show that there exist three distinct points $x_1, x_2, x_3$ such that
$$\frac{1}{f'(x_1)} + \frac{1}{f'(x_2)} + \frac{1}{f'(x_3)} = 3$$
My hunch is to use the mean-value theorem or Rolle's theorem somehow. But these theorems only guarantee the existence of a certain point, and not a triple of points, so I am stuck.

REPLY [3 votes]: Let $z_1$ and $z_2$ with $z_1<z_2$ be such that 
$$f(z_1)=\frac{2a+b}{3}$$ and 
$$f(z_2)=\frac{a+2b}{3}$$
Then it is easy to see that the slopes formed by the segements 
$$(a,a), (z_1,f(z_1)), (z_2,f(z_2)),  (b,b)$$ sum in their reciprocals to $3$.
Namely,
$$\left(\frac{\frac{b-a}{3}}{z_1-a}\right)^{-1}+
\left(\frac{\frac{b-a}{3}}{z_2-z_1}\right)^{-1}+
\left(\frac{\frac{b-a}{3}}{b-z_2}\right)^{-1}=3
$$
This the gives the points $x_1 \in (a,z_1)$ $x_2 \in (z_1,z_2)$ $x_3 \in (z_2,b)$ by the mean value theorem.