I want to see in which situation the following inequality holds
$\underbrace{\frac{\prod_{d \in D_1} d^{-\alpha_1}}{f(\alpha_2, D_2)^{|D_2|} \cdot f(\alpha_3, D_3)^{|D_3|}}}_{A} < \underbrace{\frac{\prod_{d \in D_2} d^{-\alpha_2} \cdot \prod_{d \in D_3} d^{-\alpha_3} }{f(\alpha_1, D_1)^{|D_1|}}}_{B}$
,where $\alpha_i >2$.
$ f(\alpha, D) = \left[\sum_{k = \min(D)}^{\max(D)} k^{-\alpha} \right]^{-1}$
I would like to obtain some rules of the form (just for illustration):
$\alpha_1 < \alpha_3$ $\wedge$ $ \alpha_2 > \alpha_1$ $\wedge$ $|D_1| < |D_2|$ $\rightarrow$ $A<B$
Let me give you some details about the (multi-)sets of integers $D_i$. The set $D_1$ is the overall set which we want to "cover", with $D_2$ and $D_3$. Therefore, one special case is $D_1 = D_2 \cup D_3$, where $|D_1| = |D_2|+|D_3|$. However, $D_2$ can also contain only integer fractions of some of the values of $D_1$, while $D_3$ contains the remaining fractions of the values, and the remaining values to cover $D_1$.
Example:
$D_1 =\{1,1,2,3,3,2,3\}$ the set for covering
$D_2 =\{1,1,1,3,.,1,2\}$ "." just for visualization
$D_3 =\{.,.,1,.,3,1,1\}$
As we can see, $D_1$ is divided into two sets $D_2$ and $D_3$..
I tried to bound $A<1$ for some parameters and tried to verify if $B>1$. This didnt work. Generally, I have no idea how to approach such a problem. Moreover, the values of $D_1$ are sampled of a certain distribution, and I dont know how to include this.