Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$ Suppose that the function $f:[a,b]\rightarrow \mathbb{R}$ is continuous. For a natural number $k$, let $x_1,\cdots,x_k$ be points in $[a,b]$. Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$.

This can be proved by showing 
$\hspace{130pt}f(a)<\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}<f(b)$
and using the intermediate value theorem to say there exists a point $z\in [a,b]$ such that the statement is true.
I'm not sure how to show this though. Any suggestions?
 A: Pick $x_u, x_v$ among $x_1, \ldots, x_n$ such that
$$\begin{align}
f(x_u) = & \min( f(x_1), f(x_2), \ldots, f(x_n) )\\
f(x_v) = & \max( f(x_1), f(x_2), \ldots, f(x_n) )
\end{align}$$
and let $x_l = \min(x_u, x_v)$ and $x_r = \max(x_u, x_v)$. We have 
$$\begin{align} 
& f(x_u) \le f(x_1), f(x_2), \ldots, f(x_n) \le f(x_v)\\
\implies &
f(x_u) \le \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n} \le f(x_v)
\end{align}$$
Since $f(x)$ is continuous over $[a,b]$ and $x_u, x_v \in [a,b]$, $f(x)$ is continuous
over the closed interval $[x_l, x_r]$. By Intermediate value theorem, there
is a number $z \in [ x_l, x_r] \subseteq [a,b]$ such that
$$f(z) = \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}$$
A: Here is a hint: $$f(a)=\frac{f(a)+f(a)+\cdots+f(a)}{k}$$ (with $k$ terms in the numerator).
Edit: Two upvotes already, and yet the answer is wrong, as are the inequalities you are trying to prove! But replace $a$ in my hint with the $x_j$ for which $f(x_j)$ has the smallest value, and you're good to go. Similarly for $b$, but the largest value instead.
A: I'll leave here an alternative proof, It might be helpful to someone:

Lemma 1:
If $\{x_1,...,x_n\} \subset [a,b]$ then: $\exists x \in [a,b]:$
$$x = \frac{x_1 + ... + x_n}{n}$$
Proof:
Let $g: [a,b] \to \Bbb R$ such that:
$$g(x) = \sum_{k = 1}^n x_k - nx $$
note that $g$ is continuous. Now, for each $i \in \{1,...,n\}$, we know that $a \leq x_i \leq b$. So if we sum all $x_i$ we must get the following:
$$na \leq \sum_{k = 1}^n x_k \leq nb$$
So now we have:

*

*$g(a) = \sum_{k = 1}^n x_k - na \geq na - na = 0$


*$g(b) = \sum_{k = 1}^n x_k - nb \leq nb - nb = 0$.
So, as $g(a) \geq 0 \wedge g(b) \leq 0$ we conclude that: $\exists x \in [a,b]: g(x) = 0$.
$$\begin{align}
g(x ) &= 0
\\
\sum_{k = 1}^n x_k - nx &= 0
\\
x &= \frac{1}{n}  \sum_{k = 1}^n x_k
\end{align}$$
So we conclude that $\exists x \in [a,b]:$
$$x = \frac{x_1 + ... + x_n}{n}$$

Now we can easily prove the result.
Let's denote $ m:= \min\{f(x_1),...,f(x_n)\}$ and $M:= \max\{f(x_1),...,f(x_n)\}$.
And let $i,j \in \{1,...,n\}$ such that $f(x_i) = m$ and $f(x_j) = M$.
Now, we know that $\{f(x_1),...,f(x_n)\} \subset [m,M]$. So according to lemma 1: $\exists c \in [m,M]: $
$$c = \frac{f(x_1) + ... + f(x_n)}{n}$$
We have $c \in [m,M]$ which is the same as saying that $m \leq c \leq M$ and this means that $f(x_i) \leq c \leq f(x_j)$.
Because $f$ is continuous, using the intermediate value theorem, we know that $\exists z \in [a,b]: f(z) = c$, so we have:
$$f(z)= \frac{f(x_1) + ... + f(x_n)}{n}$$
q.e.d.
