Does this particular inequality hold after applying (term to term)different monotone functions? Suppose that $A, B, C, D$ are finite subsets of $\mathbb{N}\cup\{0\}$.
I have defined a similarity measure $sim_f$ for any monotonically strictly increasing positive function $f:\mathbb{N}\cup \{0\}\rightarrow \mathbb{R}^+_0$ as $$sim_f(A,B) := \frac{\sum\limits_{e\in A\cap B}f(e)}{\sum\limits_{e\in A\cup B}f(e)}$$
Having said that, I've been struggling for days to prove that for any two monotonically strictly increasing positive functions $f,g$ such that $f,g:\mathbb{N}\cup \{0\}\rightarrow \mathbb{R}^+_0$,
$$sim_f(A,B)\leq sim_f(C,D) \hspace{1cm}\text{implies}\hspace{1cm} sim_g(A,B)\leq sim_g(C,D) $$
I think this statement holds, because I can't find any counter example, but I can't prove it either. Is it true?

This statement turns out to be false and here is a counter example.
\begin{align}
    A &= \{10\} \\
    B &= \{1,2,3,4,5,6,7,8,9,10\} \\
    C &= \{1,2\} \\
    D &= \{2,3\} \\
    f(x) &= x \\
    g(x) &= \begin{cases}
        x & x < 10 \\
        90 + x & x \geq 10
    \end{cases}
\end{align}
A direct calculation shows that
\begin{align}
    \text{sim}_f(A,B) &= \frac{10}{55} = \frac{2}{11} \\
    \text{sim}_f(C,D) &= \frac{2}{6} = \frac{1}{3} \\
    \text{sim}_g(A,B) &= \frac{100}{145} > \frac{2}{3} \\
    \text{sim}_g(C,D) &= \frac{2}{6} = \frac{1}{3}
\end{align}
The trick is to find sets $A,B,C,D$ and a function $f$ with the following properties:

*

*$\text{sim}_f(A,B) < \text{sim}_f(C,D) < 1$

*The largest element $e^* \in A \cup B \cup C \cup D$ is in $A \cap B$ and $e^* \notin C \cup D$.

*Choose $g$ to match $f$ up to $e^* - 1$ and $g(x) = g(e^*) >> f(e^*)$ for all $x \geq e^*$.

with this set up, by taking $g(e^*)$ large enough you can drive $\text{sim}_f(A,B) \rightarrow 1$ while $\text{sim}_g(C,D) = \text{sim}_f(C,D)$ since $e^* \notin C \cup D$ and its the largest element amongst the four sets.
After that its just a bit of tinkering to find a concrete example.
