Weird definition of norm's triangle inequality Cassels (Froehlich & Cassels, ANT) uses a rather unusual definition of triangle inequality when he defines a norm.
He states that a norm should satisfy the following (in lieu of $|a+b|\le|a|+|b|$):

There exists a constant $C$ such that $|1+\alpha|\le C$ whenever $|\alpha|\le 1$.

It is clear that the triangle inequality implies his property: if we assume that $|a|\le|b|$ we can set $\alpha:=a/b \Rightarrow |\alpha|\le 1$ and we get $|1+\alpha|\le 1+|\alpha| \le 2$.
But do we get the other implication? Is Cassels using an obsolete definition of norm?
 A: It depends on the value of $C$: 


*

*For $C\leq 2$ the norm satisfies the triangle inequality ($\star$)

*The norm is non-archimedean iff $C\leq 1$. The condition then implies the stronger non-Archimedean triangular inequality $|x+y| \leq \max(|x|,|y|)$.

*If you cannot choose $C\leq 2$, then there is $|x| \leq 1$ such that $|1+x|> 2 \geq |1|+|x|$, hence the triangular inequality does not hold.

*However, by defining $|x|' :=|x|^{\log_C 2}$ every norm with $C\geq 2$ is equivalent to a norm satisfying the triangle inequality.


Proof of ($\star$) taken from my lecture notes: Let $x,y \in k$ such that $|x| \geq |y|$. Then $|1+\frac{y}{x}|\leq 2 $ showing $ |x+y|\leq 2|x|.$ Inductively we show that for $x_1,\dots x_{2^r} \in K$ we have $$\left | \sum_{i=1}^{2^r}x_i \right | \leq 2^r \max |x_i|.$$ By choosing $2^{r-1}<n\leq 2^r$ we get in particular $| n\cdot 1_K| \leq 2n$ for any $n \in \Bbb N$ and conclude that $$\begin{eqnarray}|x+y|^n &=& \left |\sum_{i=0}^n {n \choose i}x^iy^{n-i} \right| \\ &\leq & 2(n+1)\max_{0 \leq i \leq n} \left |{n \choose i}x^iy^{n-i}\right | \\&\leq &4(n+1)(|x|+|y|)^n\end{eqnarray}.$$
The triangular inequality follows by taking $n$-th roots and the limit $n \to \infty$.
