Say $f(r,\theta, \phi)$ is a function defined on the unit sphere, with $f \in [0, 1]$ for every point in the sphere. The average of this function over the unit sphere, $\overline{f}$, can be written as:
$$\overline{f} = \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} f(r,\theta, \phi) r^2 \sin{\theta}dr d\theta d\phi.$$
The average of the same function over the great circle whose parallel vector establishes an angle of $\theta$ with the $z$ axis, $\overline{f}_{\text{circ}_{\theta}}$, can be written as:
$$\overline{f}_{\text{circ}_{\theta}} = \int_{0}^{2\pi}\int_{0}^{1} f(r,\theta, \phi) rdr d\phi.$$
My question is: does the following hold?
$$\max_{\theta \in [0,\pi]} \overline{f}_{\text{circ}_{\theta}} \geq \overline{f}.$$
Or, in words: is the average of the function over the great circle which maximizes it lower-bounded by the average over the whole sphere?
It seems intuitive to me that this should hold, by an argument of the type: the average of the function over the sphere is something like the average of the averages over the different circles that make up the sphere, so the average over the best circle should not be smaller than the overall average. However, I really don't know how to go about formalizing this. Some googling hasn't given me an answer, either because I'm not using the right words or because this isn't a particularly interesting problem.
I don't think it's true. Let $$f(r,\theta,\phi)=\frac{r}{2\pi}$$ Then $$\bar f=2\int_0^1r^3dr=\frac12$$ and $$\bar f_{circ_\theta}=\int_0^1r^2dr=\frac13$$