Let $m,x,y \in (0,1)$. Suppose $|x-y|, 1-|x-y|, x,y,1-y,1-x \ge m$. Solve for $x$ and $y$ in terms of $m$.
What I tried:
$|x-y| \ge m$ and $1-|x-y| \ge m$ tell me $m \le |X-Y| \le 1-m$
$x,y,1-y,1-x \ge m$ tell me $m \le x \le 1-m$ and $m \le y \le 1-m$
After some long computation I got these. Are these correct?
- $x \in (m,1-2m)$, $y \in (x+m,1-m)$, $m \in (0,\frac13)$
- $x \in (2m,1-m)$, $y \in (m, x-m)$, $m \in (0,\frac13)$
(Of course you could also write with $y$ as the independent variable, but whatever.)
A geometric interpretation helps:
a. $x, 1-x \ge m$ constitutes a vertical stripe: $m < x < 1-m$
b. $y, 1-y \ge m$ constitutes a horizontal stripe: $m < y < 1-m$
c. $|x-y|, 1-|x-y| \ge m$ constitutes two 45 degree rising diagonal stripes:
for $y<x$: $m < x-y < 1-m$ ; for $y>x$: $-m > x-y > -(1-m)$.
From a. and b. it is clear that no solutions exist for $m > 0.5$.
Now the intersection of a. and b. constitute a square, and the intersection of this square with each of the diagonal stripes may constitute a triangle. The "may" results from the question whether the intersection of this square with each of the diagonal stripes exists at all. This will be the case if the upper left corner of the square meets the lower bound of the stripe, i.e. if the point $m, 1-m$ is on the line $-m = x-y$. We have that for $-m = m - (1-m)$ or $m = 1/3$.
So if $m\le 1/3$, we can again use the lower bound of the stripe to compute the bounds for $x$: We have that $x$ can grow up to the point where that line meets the upper bound of the square, i.e. $y = 1-m$, and hence $-m = x-y = x - (1-m)$ gives $x = 1 - 2m$.
So we can summarize:
For $m > 1/3$, the inequalities allow no solution.
For $1/3 \ge m \ge 0$, the regions allowed by the inequalities are the two triangles (mirrored on the diagonal):
$m\le x \le 1 - 2m$ ; $ m+x \le y \le 1-m$
and $m\le y \le 1 - 2m$ ; $ m+y \le x \le 1-m$ .
So yes this is what you have got.
Here is a picture which illustrates that for $m=0.2$: