Is this property correct?
If yes, is there a better way to prove it?
If not, what would be an example of a ring that does not satisfy the condition?
Theorem. In a ring with non-linearly cyclically ordered additive group there is no such element $x$ that $[0, ax, a]$ for any positive element $a$.
Proof:
From the properties of Positive and negative elements of a cyclically ordered group:
Lemma 1.11. If $a$, $b$, and $c$ are positive elements of a cyclically ordered group such that $a + b + c = 0$, and $[0, a', a]$, $[0, b', b]$, $[0, c', c]$ for some elements $a'$, $b'$, $c'$, then $a' + b' + c' \ne 0$.
Lemma 1.12. If $a$, $b$, $c$, $d$ are positive elements of a cyclically ordered group such that $a + b + c + d = 0$, and $[0, a', a]$, $[0, b', b]$, $[0, c', c]$, $[0, d', d]$ for some elements $a'$, $b'$, $c'$, $d'$, then $a' + b' + c' + d' \ne 0$.
- Applying The rule of three steps for a cyclically ordered group to the additive group of a ring: for a positive element $a$ there is a combination of not more than three positive elements $b$, $c$, $d$: $a + b + c + d = 0$;
- Case $a + b = 0$ is not applicable since $b = -a$ is negative;
- Case $a + b + c = 0$:
- Assuming there is an element $x$: $[0, ax, a]$, $[0, bx, b]$, $[0, cx, c]$;
- Applying lemma 1.11: $a + b + c = 0 \implies ax + bx + cx \ne 0$;
- Applying distributivity: $ax + bx + cx = (a + b + c)x = 0$; contradiction;
- Case $a + b + c + d = 0$:
- Assuming there is an element $x$: $[0, ax, a]$, $[0, bx, b]$, $[0, cx, c]$, $[0, dx, d]$;
- Applying lemma 1.12: $a + b + c + d = 0 \implies ax + bx + cx + dx \ne 0$;
- Applying distributivity: $ax + bx + cx + dx = (a + b + c + d)x = 0$; contradiction.