Typically, a Euclidean domain is defined to be an integral domain which admits a function defined on all nonzero elements whose values on the set of all non-negative integers. Some author requires additional condition which says that for every nonzero elements $a$, $b$ the value of the function at $ab$ is no less than that at $a$.
One may use those properties to prove that every ED is a PID, and by the theorem which says that PID is UFD one gets that every ED is a UFD.
Sometimes, it is required to prove ED is UFD without the concept of PID. However, hard part appears to be that every elements in ED can be written as a finite product of irreducibles. For me, it is unavoidable to pass through PID argument. However, as most ED function satisfies, if we have that the value of the function at a nonzero element is strictly larger than that at a proper divisor (non-unit, not associated), then typical least value argument provides that in the domain every elements can be factored into irreducibles.
My question is that is there any ED which does not admit any ED function with the above stronger property.
Oops. There is no such ED because we can prove that if $f$ is a euclidean function of an ED then for every nonzero $a$ and a proper divisor $d$, we can prove that $f(d) < f(a)$.
Let $a=dq$. Let $q$ and $r$ be such that $d=ka+r$, $r=0$ or $f(r)<f(a)$. Because $d$ is a proper divisor, $r \neq 0$. Therefore $f(a)>f(r) = f(d(1-kq)) \ge f(d)$, since $k$ is not a unit so $1 \neq kq$.