I've come across the following claim in these notes (pg 87, proof of 5.2.35).
Claim: If $K$ is a field, with $|\cdot|_1$ and $|\cdot|_2$ two inequivalent valuations (positive definite, multiplicative, satisfies triangle inequality) on $K$, then there exist elements $\alpha$ and $\beta$ such that $$|\alpha|_1 < 1, \quad |\alpha|_2 \ge 1,$$ and $$|\beta|_2 < 1,\quad |\beta|_1 \ge 1.$$
The notes claim that this follows from Prop 5.2.5, the relevant part of which appears to be that $|\cdot |_1$ and $|\cdot |_2$ are equivalent if and only if $$|\alpha|_1 < 1 \iff |\alpha|_2<1. $$ Then the inequivalence of the valuations tells us that there is $\alpha$ such that either $|\alpha|_1 < 1$ and $|\alpha|_2\ge 1$, or $|\alpha|_2 < 1$ and $|\alpha|_1 \ge 1$. Thus we can find either $\alpha$ or $\beta$, but its not clear that we can find both.
Being able to find both $\alpha$ and $\beta$ appears to be equivalent to the claim that if $|\alpha|_1 < 1 \implies |\alpha|_2 < 1$, then the valuations are equivalent.
Thus I'm attempting to either
- prove $|\alpha|_1 < 1 \implies |\alpha|_2 < 1$ tells us that the valuations are equivalent,
- or disprove (1) and prove directly that if the valuations are inequivalent, then there exists $\alpha$ with $|\alpha|_1 < 1$ and $|\alpha|_2 > 1$, which is what the notes are using the claim for.
Progress on proof/disproof:
If $|\alpha|_2 < 1$, then $|1/\alpha|_2 > 1$, so $|1/\alpha|_1 \ge 1$. Thus $|\alpha|_1 \le 1$.
However, I can't seem to prove that we can't have $|\alpha|_1 = 1$.
For non archimedean valuations, I tried thinking about this in terms of valuation rings. Let $\newcommand\calO{\mathcal{O}}\calO_i = \{ a\in K : |a|_i \le 1 \}.$ Then $\calO_2 \subseteq \calO_1$. Let $\newcommand{\mm}{\mathfrak{m}}\mm_i$ be the maximal ideals of $\calO_i$. We know $\mm_1 \subseteq \mm_2$. So suppose $x\in \calO_1\setminus \calO_2$. Then $|x|_1 = 1$ and $|x|_2 > 1$. Then $y=1/x$ has $|y|_1 = 1$ and $|y|_2 < 1$, so $y\in \mm_2\setminus \mm_1$. Hence if the rings are not equal, $\mm_2$ is a proper ideal.
This seems to be completely possible if $\newcommand\pp{\mathfrak{p}}\pp$ is a prime ideal inside the maximal ideal of a valuation ring, then localizing at $\pp$ should produce a valuation ring, and the associated valuations should be inequivalent, yet satisfy the given property. However, I'm not very familiar with valuation rings, and I don't have a concrete example of how this would work.
Related questions
There appears to be this related question, which appears to suggest that the answer is affirmative, rather than negative as my prior paragraph is leading me to believe.
The crucial issue, as in why it's not helping me, is that I can't figure out how the three implications in the question body were derived.