Are the given sum of abolute value functions increasing? We have the set $\{0,\cdots,2^K-1\}$ in binary representation, $\{00,01,10,11\}$ for $K=2$. Then each $0$ is replaced by $b_k$ and each $1$ is replaced by $1-b_k$, This gives the set $$B=\{b_1b_2,b_1(1-b_2),(1-b_1)b_2,(1-b_1)(1-b_2)\}$$ The same process is done in reverse order for the $a_k$s i.e., by replacing each $0$ by $1-a_k$ and each $1$ by $a_k$. The result is then $$A=\{(1-a_1)(1-a_2),(1-a_1)a_2,a_1(1-a_2),a_1a_2\}$$
Then, we find the absolute difference sum of these two sets: $$f(a_1,b_1,a_2,b_2)=|b_1b_2-(1-a_1)(1-a_2)|+|b_1(1-b_2)-(1-a_1)a_2|+|(1-b_1)b_2-a_1(1-a_2)|+|(1-b_1)(1-b_2)-a_1a_2|$$
Here, each $(a_k,b_k)$ satisfies $1>a_K>\cdots>a_1>0$ and $0<b_K<\cdots<b_1<1$ and $a_k+b_k<1$.

Given any set of pairs $(a_k,b_k)$ for $k\in\{1,\ldots,K\}$, the function $f$ is non-increasing in every $b_k$.

Example: $(a_1,b_1)=(0.1,0.2)$ and $(a_2,b_2)=(0.2,0.1)$. Then if we take $b_1$ as variable we expect that $f(0.1,b_1,0.2,0.1)$ is decreasing for $0<b_1<0.9$. I verified that this is true in Mathematica. If we take $b_2$ as variable we expect that $f(0.1,0.2,0.2,b_2)$ is decreasing for $0<b_2<0.8$. This is also true according to Mathematica.
I guess this should be general for any example. I dont have any idea how to prove it, however.
 A: First, let us replace $a_i$ with $1-a_i$ to simplify notation. Note that we now have $b_i < a_i$. We denote $b_i^0 = b_i$ and $b_i^1 = 1-b_i$. Let $f(\vec{a}, \vec{b}) = \sum_{v\in\{0, 1\}^n}|\prod_{i=1}^nb_i^{v_i} - \prod_{i=1}^na_i^{v_i}|$. We wish to show that $\frac{\partial f}{\partial{b_i}} \leq 0$ for $i\in [n]$.
By symmetry it suffices to show that $\frac{\partial f}{\partial{b_1}} \leq 0$. The trick is to group the terms of $f$ in pairs, with each pair consisting of the $v$'s with $v_0$ either $0$ or $1$, and the remaining $v_i$'s fixed. For $w\in \{0, 1\}^{n-1}$, let $f_w = |b_1\prod_{i=2}^nb_i^{w_{i-1}} - a_1\prod_{i=2}^na_i^{w_{i-1}}| + |(1-b_1)\prod_{i=2}^nb_i^{w_{i-1}} - (1-a_1)\prod_{i=2}^na_i^{w_{i-1}}|$. Since $f = \sum_w f_w$, it suffices to show that for each $w$, that $\frac{\partial f_w}{\partial{b_1}} \leq 0$.
Fix $w$ and note that $f_w$ consists of two terms. For $\frac{\partial f_w}{\partial{b_i}}$ to be positive, we must have that the first term without the absolute value sign is positive and the second is negative. But suppose $b_1\prod_{i=2}^nb_i^{w_{i-1}} > a_1\prod_{i=2}^na_i^{w_{i-1}}$. Then since $(1-b_1)/b_1 > (1-a_1)/a_1$, we can multiply to get $(1-b_1)\prod_{i=2}^nb_i^{w_{i-1}} > (1-a_1)\prod_{i=2}^na_i^{w_{i-1}}$. Thus each $\frac{\partial f_w}{\partial{b_i}} \leq 0$ and we are done.
