Means defined by tangent parallel to secant This question made me aware of the logarithmic mean, which I hadn’t encountered before. I found this derivation quite interesting: The logarithmic mean
$$
L(x,y)=\frac{y-x}{\log y-\log x}
$$
of two distinct positive numbers $x$ and $y$ is the value of $\xi$ at which the tangent to $\log \xi$ is parallel to the secant through $(x,\log x)$ and $(y,\log y)$. I hadn’t realized that in this way, any function can be used to define a mean by solving
$$
f'(\xi)=\frac{f(y)-f(x)}{y-x}
$$
for $\xi$. For instance, for $f(\xi)=\xi^2$ we recover the arithmetic mean:
$$
2\xi=\frac{y^2-x^2}{y-x}
\\
\rightarrow \xi=\frac{x+y}2\;.
$$
Some other obvious candidates yield:
\begin{array}{c|c}
f(\xi)&\xi\\\hline
\xi^2&\frac{x+y}2\\
\sqrt\xi&\left(\frac{\sqrt x+\sqrt y}2\right)^2\\
\xi^n&\left(\frac1n\sum_{k=0}^{n-1}x^ky^{n-1-k}\right)^\frac1{n-1}\\
\mathrm e^\xi&\log\frac{\mathrm e^y-\mathrm e^x}{y-x}
\end{array}
That last one is a major disappointment – I was expecting it to yield something nice, perhaps the geometric mean. I haven’t been able to find functions that yield the arithmetic or harmonic mean, or any other generalized mean except for $p=1$ with $f(\xi)=\xi^2$ and $p=\frac12$ with $f(\xi)=\xi^\frac12$ as above. The functional equations that would need to be satisfied, e.g.
$$
f'\left(\sqrt{xy}\right)=\frac{f(y)-f(x)}{y-x}
$$
for the geometric mean, seem intractable to me.
Wikipedia shows how to compute the Pythagorean means using the logarithmic mean:
\begin{eqnarray}
\frac{L\left(x^2, y^2\right)}{L(x,y)}&=&\frac{x+y}2\;,\\
\sqrt{\frac{L\left(x,y\right)}{L\left(\frac1x,\frac1y\right)}}&=&\sqrt{xy}\;,\\
\frac{L\left(\frac1x,\frac1y\right)}{L\left(\frac1{x^2},\frac1{y^2}\right)}&=&\frac2{\frac1x+\frac1y}\;,
\end{eqnarray}
but this doesn’t provide any indication how these means might be recovered with the secand method.
My questions are:


*

*Which function $f(\xi)$ would yield the geometric mean, the harmonic mean or any other generalized mean?

*Which functions $f(\xi)$ yield other means that don’t fit into the generalized means scheme but might be of interest?

*Any other insights you have about the logarithmic mean and the secant method to derive it would also be welcome.

 A: 
... any function can be used to define a mean by solving
  $$
f'(\xi)=\frac{f(y)-f(x)}{y-x}
$$

In order for the mean to be well-defined, the equation must have a unique solution $\xi$. This is for example satisfied if $f$ is strictly convex or strictly concave, so that the derivative is strictly monotone.
Strict convexity/concavity also guarantees that the mean defined by $f$ is strictly increasing in both arguments. 

Which function $f(\xi)$ would yield the geometric mean?

If we restrict the domain to strictly positive real numbers then $f(\xi) = 1/\xi$, $f'(\xi) = -1/\xi^2$ gives the geometric mean:
$$
 \frac{f(y)-f(x)}{y-x} = - \frac{1}{xy} = f'(\sqrt{xy}) \, .
$$

... the harmonic mean ...

There is no strictly convex or strictly concave, differentiable function $f:(0, \infty) \to \Bbb R$ which yields the harmonic mean.
Assume that such a function $f$ exists. Without loss of generality assume that $f$ is strictly convex, otherwise consider $(-f)$ instead. 
So we have
$$
 f'\left( \frac{2xy}{x+y}\right) = \frac{f(y)-f(x)}{y-x}
$$
for $0 < x < y$. For $x=1$ and $y > 3$ we get
$$
 f'(2) \ge f'\left( \frac{2y}{1+y}\right) = \frac{f(y)-f(1)}{y-1} \ge \frac{f(3) + f'(3)(y-3)-f(1)}{y-1} 
$$
and for $y \to \infty$ it follows that $f'(2) \ge f'(3)$, in contradiction to $f$ being strictly  convex. 

The same reasoning can be applied to show that there is no strictly convex or strictly concave, differentiable function $f:(0, \infty) \to \Bbb R$ which yields the generalized mean
$$
 M_p(x, y) = \left( \frac{x^p+y^p}{2}\right)^{1/p}
$$
with any $p < 0$, or any mean which remains bounded for fixed $x$ and $y \to \infty$. 
