Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$

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 $$f(z) = \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}$$

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.

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.