Question: Is there a solution for the following system?
(if so I'm interested in an explicit solution)

$x_1>x_2, \ \ x_1>x_3, \ \ x_1>x_4,$
$ x_2>x_5, \ \ x_2>x_6, \ \ x_3>x_5, \ \ x_3>x_7, \ \ x_4>x_6, \ \ x_4>x_7,$
$x_5>1, \ \ x_6>1, \ \ x_7>1,$
$x_2 x_3 \le x_1 x_5, \ \ x_2 x_4 \le x_1 x_6, \ \ x_3 x_4 \le x_1 x_7$
$x_5 x_6 \le x_2, \ \ x_5 x_7 \le x_3, \ \ x_6 x_7 \le x_4$
$x_1-x_2-x_3-x_4+x_5+x_6+x_7-1 \le 0$

Motivation: Let $B_n$ be the rank $n$ boolean lattice (i.e. the subset lattice of $\{1,2, \cdots , n\}$).
A labeling $f: B_n \to \mathbb{R}_+$ is called allowed if it satisfies:

  • $f(\emptyset) = 1$
  • $a \subset b \Rightarrow f(a) < f(b)$
  • $ f(a)f(b) \le f(a \cup b)f(a \cap b)$, $\forall a,b \in B_n$

Let $\varphi(f) := (-1)^n\sum_{a \in B_n} (-1)^{|a|} f(a)$ be the Euler totient of the labeling $f$.

Remark: For $n=1$, we see directly that $\varphi(f)>0$, and for $n=2$ it is proved in this answer.

Our system admits a solution iff there is an allowed labeling of $B_3$ with a negative Euler totient.
To understand that take $f(\{1,2,3\})=x_1$, $ f(\{1,2\})=x_2 $, $f(\{1,3\})=x_3$, $f(\{2,3\})=x_4$, $f(\{1\})=x_5$, $f(\{2\})=x_6$, $f(\{3\})=x_7$ and $f(\emptyset)=1=x_8$.

enter image description here $ \ \ \ \ \ \ \ \ \ \ \ \ $ enter image description here

Remark: If $f$ maps to $\mathbb{N}$ and $x_1 < 100$ then $\varphi(f) > 0$ (checked by SAGE).
If moreover $a \subset b \Rightarrow f(a) \mid f(b)$, then we get the same for $x_1 < 10^5$.
If $ f(a)f(b) = f(a \cup b)f(a \cap b)$ then $\varphi(f) = (x_5-1)(x_6-1)(x_7-1)>0$.

Bonus question: What's the lowest $\varphi(f)$ for $f$ an allowed labeling of $B_3$?

Lemma: $\varphi(f)>-1$.
Proof: First $(x_1-x_2)(x_1-x_3)(x_1-x_4)>0$,
so $x_1^3 + x_1(x_2x_3+x_2x_4+x_3x_4) > x_1^2(x_2+x_3+x_4)+x_2x_3x_4$,
then $x_1 + \frac{x_2x_3}{x_1}+\frac{x_2x_4}{x_1}+\frac{x_3x_4}{x_1} > x_2+x_3+x_4+\frac{x_2x_3x_4}{x_1^2}$,
but $x_5 \ge \frac{x_2x_3}{x_1}, x_6 \ge \frac{x_2x_4}{x_1}, x_7 \ge \frac{x_3x_4}{x_1}$.
It follows that $\varphi(f) > \frac{x_2x_3x_4}{x_1^2}-1>-1$. $\square$

The following shows that $\frac{x_2x_3x_4}{x_1^2} \le 1$, so that we can't get a better lower bound for $\varphi(f)$ by this method.

Lemma: $x_2x_3x_4 \le x_1^2$.
Proof: First $(x_2x_3)(x_2x_4)(x_3x_4) \le (x_1x_5)(x_1x_6)(x_1x_7)$,
so $(x_2x_3x_4)^2 \le x_1^3x_5x_6x_7$,
Then, $(x_2x_3x_4)^4 \le x_1^6(x_5x_6)(x_5x_7)(x_6x_7) \le x_1^6(x_2x_3x_4)$.
It follows that $(x_2x_3x_4)^3 \le x_1^6$, so $x_2x_3x_4 \le x_1^2$. $\square$


Yes, take $0< \epsilon \le 1/100$ and $2 \le x \le 3$, then the following labeling is a solution:

enter image description here

by assumption on $\epsilon$ and $x$ we can check that:

  • $1< 1+\epsilon < \sqrt{x} < x$,
  • $x(1+\epsilon) \ge (\sqrt{x})^2$ and $\sqrt{x} > (1+ \epsilon)^2$,
  • $\varphi(f) = x-3\sqrt{x} + 3(1+ \epsilon)-1 = x+2-3\sqrt{x}+3\epsilon < 0 $.

Moreover $x+2-3\sqrt{x}$ has a minimum for $x=9/4$ where $\varphi(f) = -1/4+3 \epsilon$.

About the bonus question: Is it true that $\varphi(f) > -1/4$, for any allowed labeling $f$ of $B_3$?


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