[proofplan]
We prove both directions of the equivalence. The forward direction is a direct application of the strong triangle inequality. The converse reduces the general strong triangle inequality to the special case $|\alpha + 1| \leq 1$ for $|\alpha| \leq 1$ by dividing through by the larger of the two terms.
[/proofplan]
[step:Prove the forward direction: the strong triangle inequality implies $|\alpha| \leq 1 \Rightarrow |\alpha + 1| \leq 1$]
Assume $|\cdot|$ is a non-archimedean absolute value on $L$, i.e., it satisfies the strong triangle inequality: $|x + y| \leq \max(|x|, |y|)$ for all $x, y \in L$.
Let $\alpha \in L$ with $|\alpha| \leq 1$. Applying the strong triangle inequality with $x = \alpha$ and $y = 1$:
\begin{align*}
|\alpha + 1| \leq \max(|\alpha|, |1|) = \max(|\alpha|, 1) = 1,
\end{align*}
where $|1| = 1$ follows from multiplicativity ($|1| = |1 \cdot 1| = |1|^2$ and $|1| \neq 0$, so $|1| = 1$).
[/step]
[step:Prove the converse: the condition $|\alpha| \leq 1 \Rightarrow |\alpha + 1| \leq 1$ implies the strong triangle inequality]
Assume $|\cdot|: L \to \mathbb{R}_{\geq 0}$ satisfies non-degeneracy ($|x| = 0 \iff x = 0$), multiplicativity ($|xy| = |x||y|$), and the condition: $|\alpha| \leq 1$ implies $|\alpha + 1| \leq 1$ for all $\alpha \in L$.
We show $|x + y| \leq \max(|x|, |y|)$ for all $x, y \in L$.
**Case 1: $y = 0$.** Then $|x + y| = |x| = \max(|x|, 0) = \max(|x|, |y|)$.
**Case 2: $y \neq 0$ and $|x| \leq |y|$.** Since $y \neq 0$, $|y| > 0$ (by non-degeneracy), and we may form $x/y \in L$. By multiplicativity, $|x/y| = |x|/|y| \leq 1$.
Applying the hypothesis with $\alpha = x/y$:
\begin{align*}
\left|\frac{x}{y} + 1\right| \leq 1.
\end{align*}
Multiplying both sides by $|y|$ (using multiplicativity: $|x/y + 1| = |(x + y)/y| = |x + y|/|y|$):
\begin{align*}
|x + y| = |y| \cdot \left|\frac{x}{y} + 1\right| \leq |y| \cdot 1 = |y| = \max(|x|, |y|).
\end{align*}
**Case 3: $y \neq 0$ and $|x| > |y|$.** By symmetry, interchange $x$ and $y$: since $|y| < |x|$ and $x \neq 0$, apply Case 2 with the roles of $x$ and $y$ swapped to get $|y + x| \leq \max(|y|, |x|) = \max(|x|, |y|)$. Since addition is commutative, $|x + y| = |y + x| \leq \max(|x|, |y|)$.
In all cases, $|x + y| \leq \max(|x|, |y|)$.
[guided]
The converse is the more substantial direction. The strategy is: to prove the strong triangle inequality for arbitrary $x, y$, reduce to the case where one of $|x|, |y|$ equals $1$ by dividing by the larger term.
Given $x, y \in L$ with $y \neq 0$ and $|x| \leq |y|$, set $\alpha = x/y$. Then $|\alpha| = |x|/|y| \leq 1$ (by multiplicativity and the assumption $|x| \leq |y|$). The hypothesis gives $|\alpha + 1| \leq 1$, which translates to $|x/y + 1| \leq 1$, i.e., $|(x + y)/y| \leq 1$.
Multiplying by $|y|$ (using $|(x+y)/y| = |x+y|/|y|$, which follows from multiplicativity applied to $x + y = y \cdot ((x+y)/y)$), we obtain $|x + y| \leq |y| = \max(|x|, |y|)$.
Why does this characterisation work? The strong triangle inequality is really about what happens when you add $1$ to something small. The general case $|x + y| \leq \max(|x|, |y|)$ is just a rescaled version of $|\alpha + 1| \leq 1$ when $|\alpha| \leq 1$. The multiplicativity of $|\cdot|$ provides the rescaling.
This criterion is useful in practice because it is often easier to verify the condition on the valuation ring $\{x : |x| \leq 1\}$ than to check the strong triangle inequality for all pairs.
[/guided]
[/step]
