I am considering the following system:
$$0 < x < 1,$$ $$x = y + z - 1,$$ $$x \geq y \geq 0,$$ $$x \geq z \geq 0,$$
which I am pretty sure does not have any solutions. But I'm struggling to prove this formally. A first observation is of course that $1 < y + z < 2$, but I am not sure how to proceed.
On
Notice $x,y,z$ are all between $0$ and $1$.
Since you want $z$ to be smaller or equal to $x$ we write
$$z = x - v^2$$
for suitable $v$. Clearly $v$ is at most $1$.
You also want $y$ to be smaller or equal to $x$ so we also write
$$y = x - w^2 $$
for suitable $w$. Clearly $w$ is at most $1$.
And you also want $x = y + z - 1$
We rewrite that as
$$x = 2x - v^2 - w^2 - 1$$
So we get
$$0 = x - v^2 - w^2 - 1$$
and
$$1 = x - v^2 - w^2 $$
$$1 = z - w^2 = y - v^2 $$
Since $y,z$ is at most $1$ this implies that $z = y = 1$ and $w = v = 0$.
This in turn implies $x= y = z = 1$.
And that seems like the only solution.
But hold on , you actually wanted $x,y,z$ strictly smaller than $1$.
So there are no solutions.
With the latest edit, it is true that there are no solutions whatsoever. The reason is quite simple also: $$x-y=z-1\geq0$$ by the second line and third line. Meaning $$z\geq1$$ But $$1>x\geq z\geq1$$ gives a contradiction.