We discuss the common three types of Pythagorean means (i.e. Arithmetic Mean, Geometric Mean and Harmonic Mean) in this post, with emphasis on the interpretation of the Harmonic Mean.
Definitions
The definitions of the three Pythagorean means are the following:
\[\begin{aligned} AM &= \frac{1}{N} {\sum_{n=1}^N x_n} \\ GM &= \sqrt[N]{\lvert \prod_{n=1}^N x_n \rvert} \\ HM &= \frac{N}{\sum_{n=1}^N \frac{1}{x_n}} \end{aligned}\]where \(AM, GM\) and \(HM\) are abbreviations of Arithmetic Mean, Geometric Mean and Harmonic Mean, respectively.
Interpretations

\(AM\) is the most intuitive mean.

\(GM\) is actually the median of the geometric series \((x_n)\).
Proof: Let \(x_n = c r^n\), then the product of the first \(N\) terms is:
\[P = \prod_{n=1}^N x_n = \prod_{n=1}^N cr^n = c^N r^{\sum_{n=1}^N n} = c^N r^{\frac{N(1+N)}{2}}\]Thus, the \(n\)th root of the product is \(c r^{\frac{1+N}{2}}\), i.e. the median \(\blacksquare.\)

The formula of \(HM\) does not reveal intuitive understanding. We present a sequence of examples to understand \(HM\) as follows:

Modify the formula as follows: \((\frac{\sum \frac{1}{x_n}}{N})^{1}\), we can interpret \(\frac{1}{x_n}\) as the amount of work we need to finish the \(n\)th task given the rate \(x_n\). Then \(\frac{\sum \frac{1}{x_n}}{N}\) can be deemed as the average amount of work to finish a task, the inverse is the average rate to finish a task.
Note: The arithmetic mean of rate \(\bar{x}_n\) has no meaning here because the rates are not of equal weights.

Moreover, if we take \(x_n\) as the speed of \(n\)th vehicle used for travelling, then \(\frac{1}{x_n}\) will be the time used for travelling a unit route with \(n\)th vehicle and the sum of all the reciprocals is the mean time. Thus \(HM\) will be the average speed of the total trip divided into \(N\) continuous parts with different vehicles.

If \(P\) gives the Precision (\(\frac{tp}{tp + fp}\)) and \(R\) gives the Recall (\(\frac{tp}{tp + fn}\)), then we can interpret \(P\) to be the correct rate of the machine and \(R\) to be the retrieval rate. The two “vehicles” here are Model and Data in the sense that the rate of getting a correct (model) prediction is \(x_1\) and the rate of retrieve a reference (from data) is \(x_2\).
Note: Both the correct prediction and the retrieved reference refer to a True Positive (\(tp\)), but from different aspects. The former considers the model and the latter considers the data.

\(HM(P, R) = \frac{2PR}{P + R}\) is the standard F1score. Why not arithmetic mean? The reason is simply that Positive and Truth may not coincide with each other. We are measuring the machine performance from two aspects: [1] To what extent can we trust the Positive? [2] To what the extent of is the Truth being treated properly? By default, there is no preference to these two aspects. But if we do prefer one to another, then we need a weighted version of means.

The general \(F_{\beta}\) score has the following formula:
\[F_{\beta} = \frac{1}{\frac{\beta^2 / (1 + \beta^2)}{R} + \frac{1 / (1 + \beta^2)}{P}} = \frac{(1 + \beta^2)PR}{\beta^2 P + R}\]So the total “distance” is now \(1+\beta^2\) instead of \(2\) (\(\beta^2\) for Data and \(1\) for Model). If \(\beta^2 > 1\), we need larger Recall to increase the score (it favours Recall) and for \(\beta^2 < 1\) the metric favours Precision. To obtain the AM, we need \(R = \beta^2 P\) such that
\[F_{\beta} = \frac{(1 + \beta^2)\beta^2P^2}{2 \beta^2 P} = \frac{(1 + \beta^2)P}{2} = \frac{P + R}{2}\]Note: \(R = \beta^2 P\) can be interpreted as follows: Data takes \(\beta^2\) distance so it needs \(\beta^2\) times of the speed to keep the same time as Model. \(2\) in the denominator indicates the same time intervals spent by both vehicles.

Geometric Interpretations
*Fig. 1: Geometric relationship between each of the Pythagorean mean and the half circle. 
Claim: \(A, G\) and \(H\) correspond to the three Pythagorean means defined above, respectively.
Proof: \(A\) is clearly \(\frac{a + b}{2}\), i.e. the radius \(r\) of the circle. Thus, \(z = r  b = \frac{ab}{2}\).
Since \(z = \frac{a  b}{2}\), \(H + x = \frac{a + b}{2}\), we have, by the Pythagorean Theorem, \(G = \sqrt{ab}\).
Since \(\frac{x}{z} = \frac{y}{G}\) (similar triangles), we have \(\frac{x}{(ab)/2} = \frac{y}{\sqrt{ab}}\), thus \(x = \frac{ (ab)y }{ 2\sqrt{ab} }\). Again, by the Pythagorean Theorem,
\[\begin{aligned} & \frac{(ab)^2y^2}{4ab} + y^2 = \frac{(ab)^2}{4} \\[10pt] \implies & (a+b)^2y^2 = ab(ab)^2 \\[10pt] \implies & y = \frac{\sqrt{ab}ab}{(a+b)} \\ \implies & H = \sqrt{G^2  y^2} = \sqrt{ab  ab\frac{(ab)^2}{(a+b)^2}} \\ =& \sqrt{ab(\frac{(a+b)^2  (ab)^2}{(a+b)^2})} \\ =& \sqrt{\frac{4a^2b^2}{(a+b)^2}} = \frac{2ab}{a + b} = \text{Harmonic Mean}. \end{aligned}\]\(\blacksquare\).