When can we be sure that the maximum of the harmonic mean of two functions is where we'd expect it to be? If we have two functions, P(x) and R(x), where P is monotonically increasing and R is monotonically decreasing, and their curves cross, and we take the harmonic mean: 
$$M(x) = \frac{2P(x)R(x)}{P(x)+R(x)}$$
in what circumstances can we expect the maximum of $M(x)$ to be at the point where $P(x) = R(x)$?
For example, is it enough for $P(x)$ and $R(x)$ to be continuous?
Here's an example of the data that prompted the question, and where F-measure is another name, used in information retrieval, for the harmonic mean.

 A: Note this is an answer to a previous version of the question, where $M=(P+R)/2$. Had it all typed up and then the question changed... See "Heh-heh" below for an answer to the current version.
So suppose $M=(P+R)/2$.
If there is a point $x$ where the two curves cross and if $P\ge0$ and $R\ge0$ then $M(x)$ is actually within a factor of $2$ of being the minimum of $M$. (Hence $M(x)$ is close to the maximum if $P,R\le 0$.)
Indeed, if $t>x$ then $$M(t)\ge P(t)/2\ge P(x)/2=(P(x)+R(x))/4=M(x)/2;$$similarly if $t<x$.
Heh-heh It follows that if $P,R>0$ and $H$ is the harmonic mean then $H(x)$ is within a factor of $4$ of the maximum of $H$. Because the result above shows that $1/P(x)+1/R(x)$ is within a factor of $2$ of the minimum of $1/P+1/R$, and $H=2/(1/P+1/R)$.
This can actually be useful. My favorite example:


If $f,f'\in L^2(\mathbb R)$ then $\hat f\in L^1(\mathbb R)$, and in fact $||\hat f||_1\le c(||f||_2||f'||_2)^{1/2}.$


Proof: For $A>0$ we have $$\int_{|t|<A}|\hat f(t)|\le(2A)^{1/2}||\hat f||_2=(2A)^{1/2}||f||_2.$$And since $||f'||_2^2=\int t^2|\hat f(t)|^2$ we get $$\int_{|t|>A}|\hat f(t)|\le\left(\int_{|t|>A}1/t^2\right)^{1/2}||f'||_2=(2/A)^{1/2}||f'||_2.$$Now set $A=||f'||_2/||f||_2$ and the result follows, with $c=2^{3/2}$.
(Of course we're free to set $A=||f'||_2/||f||_2$ even if we don't know that result about the minimum being essentially at the point where the two curves cross; that result reassures us that the value of $c$ we got is not totally stupid: we cannot get a much better $c$ from any other value of $A$.)
A: Differentiating wrt $M$ you find
$$M'(x) = 2\frac{(P'(x)R(x)+P(x)R'(x))(P(x)+R(x)) - P(x)R(x)(P'(x)+R'(x))}{(P(x)+R(x))^2}$$
Now assume $x^*$ is a point where $P(x^*) = R(x^*) = l$. Then 
$$M'(x^*) = 2 \frac{(P'(x^*)l+lR'(x^*))2l - l^2(P'(x^*)+R'(x^*))}{4l^2} =$$
$$= \frac 12 (2(P'(x^*)+R'(x^*))-P'(x^*)+R'(x^*)) = \frac 12 (P'(x^*)+R'(x^*)) \neq 0$$
So in general the maximum of $M$ is not in a point where $P(x^*) = R(x^*)$. 

You can see though that when additionally $P'(x^*)+R'(x^*) = 0$, then $M'(x^*)$ is a stationary point, which means that it can be a maximum (or a minimum - for that you need to investigate further). 
In your picture, when the two lines cross, one is increasing more or less linearly, one is decreasing more or less linearly. This means that in your picture $P'(x^*)+R'(x^*) \approx 0$ which is why the maximum appears to be the intersection of the lines.
