Show that $\sum_{i=1}^{n}x_if'(x_i)=f'(\xi)$ Let $f:(a,b)\to\mathbb{R}$ be differentiable and let $x_1,\dots,x_n\in (a,b)$. Suppose $x_i>0$ for all $i$ and $\sum_{i=1}^{n}x_i=1$. Show that there is $\xi\in(a,b)$ s.t.
$$\sum_{i=1}^{n}x_if'(x_i)=f'(\xi).$$
Clearly by mean value theorem there is $\xi\in(a,b)$ s.t.
$$f'(\xi)=\frac{f(b)-f(a)}{b-a}$$. Moreover since $\sum_{i=1}^{n}x_i=1$, it follows that
$$\sum_{i=1}^{n}x_if'(\xi)=f'(\xi)$$
so it suffices to show that
$$\sum_{i=1}^{n}x_i(f'(x_i)-f'(\xi))=0$$
Thus I am trying to construct a function which can give me the desired form. I have some trouble here.
 A: Here is one way to prove this, which relies on the following slightly nontrivial fact:
Fact (Darboux's Theorem$^{[1]}$): If a function is differentiable on an interval $(a,b)$, then the image of the derivative is an interval; i.e. the Intermediate Value Theorem holds on any closed subinterval.
This fact can be proved by noting that the slopes of chords of a graph of a continuous function form an interval, and then applying the Mean Value Theorem.
Assuming this, your work can be finished as follows.
Set $$g(x)=\sum_{i=1}^{n}x_i(f'(x_i)-f'(x))=(\sum_{i=1}^{n}x_if'(x_i))-f'(x).$$
Then the image of $g(x)$ is an interval by the fact above.
Let $m$ be the minimum of the $x_i$ and $M$ be the maximum of the $x_i$. If $m=M$, then all the $x_i$ are equal and we may take $\xi=x_i$. Otherwise,
$$g(m)=\sum_{i=1}^{n}x_i(f'(x_i)-f'(m))>0, \qquad g(M)=\sum_{i=1}^{n}x_i(f'(x_i)-f'(M))<0.$$
Applying the fact to the inteval $[m,M]\subset(a,b)$, we see that there must be a $\xi\in [m,M]$ such that $g(\xi)=0$; i.e., 
$$\sum_{i=1}^{n}x_if'(x_i)=f'(\xi)$$ as required.
A: Since $x_{i} > 0$ and $\sum x_{i} = 1$ it is clear that $\sum x_{i} f'(x_{i}) = \{\sum x_{i} f'(x_{i})\}/\{\sum x_{i}\}$ is a sort of generalized mean of the numbers $f'(x_{i})$ (this fact is easy to understand if the $x_{i}$ are rational, but not difficult to generalize for irrational $x_{i}$). Hence the quantity $A = \sum x_{i} f'(x_{i})$ lies between the least and greatest of the $f'(x_{i})$'s and hence by intermediate value theorem for derivatives we must have some $\xi \in (a, b)$ for which $f'(\xi) = A = \sum x_{i} f'(x_{i})$.
