= Solution
Two <absolute values on a field> are <equivalent absolute values> when they induce the same <topology>. For nontrivial <absolute values on a field> this is equivalent to
$$
|x|_2=|x|_1^c\qquad(x\in K)
$$
for a fixed real number $c>0$. The <trivial absolute value> is equivalent only to itself.
Restriction of a <Non-Archimedean absolute value> plainly preserves the <ultrametric inequality>. Conversely, if the restriction is non-Archimedean, every integer, interpreted in either <field>, has <absolute value on a field> at most one. In particular every binomial <coefficient> has <absolute value on a field> at most one. The <triangle inequality> applied to the binomial expansion gives, for $x,y\in L$ and every positive integer $N$,
$$
|x+y|^N
\leq\sum_{j=0}^N\left|\binom Nj x^j y^{N-j}\right|
\leq (N+1)\max(|x|,|y|)^N.
$$
Raise both sides to the power $1/N$ and let $N\to\infty$. Since $(N+1)^{1/N}\to1$, this yields
$$
\boxed{|x+y|\leq\max(|x|,|y|).}
$$
This is the <bounded-integer criterion for a non-Archimedean absolute value>. The proof works for arbitrary <field extensions>, not just algebraic ones, and in positive characteristic as well.
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