Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1

Theorem
Let $R$ be a division ring.

Let $\norm{\,\cdot\,}_1: R \to \R_{\ge 0}$ be a norm on $R$.

Let $\norm{\,\cdot\,}_1$ satisfy:
 * $\forall x \in R, x \ne 0_R:\norm{x}_1 \ge 1$

Then:
 * $\norm{\,\cdot\,}_1$ is the trivial norm.

Proof
We prove the contrapositive.

Let $\norm{\,\cdot\,}_1$ be a nontrivial norm.

Then there exists $y \in R: \norm{y}_1 \neq 0$, and $\norm{y}_1 \neq 1$.

By Real Numbers form Totally Ordered Field either $\norm {y}_1 \lt 1$ or $\norm {y}_1 \gt 1$.

Suppose $\norm {y}_1 \gt 1$.

By Norm axiom (N1) (Positive Definiteness then $y \ne 0_R$

By Inverse of Norm then:
 * $\norm {y^{-1}}_1 = \dfrac 1 {\norm{y}_1 } \lt 1$

So either $\norm {y}_1 \lt 1$ or $\norm {y^{-1}}_1 \lt 1$

So:
 * $\exists x \in R, x \ne 0_R:\norm{x}_1 \lt 1$

The theorem now follows by the Rule of Transposition.