Exercise 3.2.15 in Aluffi's Algebra: Chapter 0 goes as follows:
Given a Euclidean Domain $R$ with valuation $v$, show that there exists a Euclidean valuation $\overline{v}$ on $R$ such that $\overline{v}(ab) \geq \overline{v}(b)$ for all $a, b \in R$.
He then goes on to give a hint as to what the valuation should be, namely, $\overline{v}(a)$ is the minimum of $v(ab)$ across all $b \in R$. I've shown that this valuation satisfies the required property, but I'm having a little trouble showing that $\overline{v}$ is in fact Euclidean. Aluffi suggests writing arbitrary $a, b$ as $a = qb + r$, with $v(r) < v(b)$, assume that $\overline{v}(r) \geq \overline{v}(b)$, and reach a contradiction from this, but I haven't been able to do so. I feel like if I knew what form the contradiction should take I should be able to reach it, but more extensive or alternative solutions are more than welcome.
Relatively old question, but since it was never answered and I wanted to verify my own answer here it goes (for whoever stumbles upon it):
Define everything as Aluffi does in his hint. Let $k \in R$ be the element that minimizes $\tilde{v}(b)$; i.e. $\tilde{v}(b) = v(bk)$. Since $R$ is a Euclidean Domain, then for $a$ and $bk$ there are $q_1, r_1 \in R$ such that: $$a=bkq_1 + r_1$$ with $v(r_1) < v(bk)$ (since $a \nmid b$).
Now, by the definition we gave of $k$, $v(r_1) < v(bk) \leq v(b)$, so $v(r_1) < v(b)$. This implies we can define $kq_1 = q_2$ and write: $$a=bq_2 + r_1$$ with $v(r_1) < v(b)$. Since we know that $v(r)$ is minimal, then $v(r) \leq v(r_1) < v(bk)$. Which, by the definition of $\tilde{v}$, implies $\tilde{v}(r) < \tilde{v}(b)$; contradicting our assumption. Hence, $\tilde{v}$ defines a Euclidean valuation on $R$.