I have found an assertion in a scientific book (Hinds, Aerosol Technology, 2nd Edition, 1998, p. 83-84) that claims:
Given the general form [here for grouped data] for the diameter of an average property proportional to $d^p$:
$$ \bar{d}_p = \left(\frac{1}{N}\sum\limits_{i}n_i d_i^p\right)^{\frac{1}{p}} $$
For the same distribution [of diameter], the higher the moment [$p$], the larger the moment average [$\bar{d}_p$].
This assertion is left without any other argumentation. The author provides and example where:
$$ \bar{d} < d_\bar{s} < d_\bar{m} $$
The author makes the assumption that particles are spheres of equal density. Which is equivalent - according to our notation:
$$ \bar{d}_1 < \bar{d}_2 < \bar{d}_3 $$
All diameters are strictly positive real quantities. Is there a general proof for such assertion?
After digging into internet I found this inequality that seems to answer to my question. Lyapunov's inequality states:
$$ \left( \mathrm{E\left[|x|^s\right]} \right)^{1/s} \leq \left( \mathrm{E\left[|x|^t\right]} \right)^{1/t} \, , \quad 0 < s < t$$
This result relies on Jensen's inequality.