Does the HM follow the same pattern as AM and GM of the particular MVT application/theorem? This seems like something which should be well known, but I don't know how to find it so I'm asking it here.

Suppose $f:[a,b]\to \mathbb{R}$ (with $f(a)\neq f(b)$) is continuous at $[a,b]$ and differentiable at $(a, b)$ and let $n\in\mathbb{N}^+$.
One can prove the following results having to do with each of AM and GM

Arithmetic Mean: There exist numbers $a < \xi_1 <\cdots < \xi_n < b$ such that
$$\frac{1}{n}\sum_{i=1}^n f'(\xi_i) = \frac{f(b)-f(a)}{b-a}$$
Geometric Mean: There exist numbers $a < \xi_1 <\cdots < \xi_n < b$ such that
$$\prod_{i=1}^n f'(\xi_i)= \left(\frac{f(b)-f(a)}{b-a}\right)^n$$
or, if $f'$ is positive,
$$\left(\prod_{i=1}^n f'(\xi_i)\right)^{1/n}= \frac{f(b)-f(a)}{b-a}$$

It only makes sense to ask whether the following holds
Harmonic Mean Claim: There exist numbers $a < \xi_1 <\cdots < \xi_n < b$ such that
$$\frac{n}{\sum_{i=0}^n \frac{1}{f'(\xi_i)}}= \frac{f(b)-f(a)}{b-a}$$
(Suppose that perhaps $f'(x_0)\neq 0$ for all $x_0\in(a,b)$)
However I have absolutely no idea where to start. For the other two looking at the $n=2$ case helped, however I can't seem to think of anything here.
 A: Here's a proof of the IVT version of this claim, for a very general class of means that includes AM, GM, and HM:
Theorem. Let $g \colon [c,d] \to \mathbb R$ be a continuous function with $g(c) \ne g(d)$, and let $y$ lie between $g(c)$ and $g(d)$. Let $M$ be a continuous "mean" whose only real mean-like property we care about is that $M(x,x,\dots,x) = x$ for all $x$. Then we can choose $c < \xi_1 < \xi_2 < \dots < \xi_n < d$ such that $M(g(\xi_1), \dots, g(\xi_n)) = y$.
Proof. For a parameter $t \in [0,1]$, define $$\xi_i(t) = (1 - t^{n+1-i}) c + t^{n+1-i} d.$$ This has three important properties:

*

*When $t = 0$, $\xi_1(t) = \xi_2(t) = \dots = \xi_n(t) = c$.

*When $0 < t < 1$, $c < \xi_1(t) < \xi_2(t) < \dots < \xi_n(t) < d$.

*When $t = 1$, $\xi_1(t) = \xi_2(t) = \dots = \xi_n(t) = d$.

So the function $m(t) := M(g(\xi_1(t)), g(\xi_2(t)), \dots, g(\xi_n(t)))$ is continuous with $m(0) = g(c)$ and $m(1) = g(d)$.
By the intermediate value theorem, there is some $t$ between $0$ and $1$ such that $m(t) = y$, and then $ \xi_1(t) < \xi_2(t) < \dots < \xi_n(t)$ is the solution we wanted.

If $f'$ is continuous, then we can prove the claim we want by taking $g = f'$.
First, we need to find values $c,d$ such that $a < c < d < b$ and $\frac{f(b)-f(a)}{b-a}$ lies strictly between $f'(c)$ and $f'(d)$. This is possible provided $f$ is not linear. (Or, if $f$ is linear, then any $\xi_1, \dots, \xi_n$ will work.)
In that case, there is some $x \in (a,b)$ such that $\frac{f(x)-f(a)}{x-a} \ne \frac{f(b)-f(a)}{b-a}$. Without loss of generality, $\frac{f(x)-f(a)}{x-a} < \frac{f(b)-f(a)}{b-a}$; then we must have $\frac{f(b)-f(x)}{b-x} > \frac{f(b)-f(a)}{b-a}$ to compensate. Pick $c \in (a,x)$ so that $f'(c) = \frac{f(x)-f(a)}{x-a}$ and pick $d \in (x,b)$ so that $f'(d) = \frac{f(b)-f(x)}{b-x}$, both by the mean value theorem. Now apply the theorem proved earlier with $g = f'$ on the interval $[c,d]$ with $y = \frac{f(b)-f(a)}{b-a}$.
