Definition of absolute value Usually absolute value $v$ on a field $K$ is defined as a map $x\to |x|_v$ such that $\forall x,y\in K$ one has
$$\begin{align}
&|x|_v=0\iff x=0\\
&|xy|_v=|x|_v|y|_v\\
&|x+y|_v\leq |x|_v+|y|_v
\end{align}$$
If one has $|x+y|_v\leq \max(|x|_v,|y|_v)$ which is stronger than the triangle inequality, the absolute value is ultra metric or non Archimedean.
Now Marc Hindry in his “arithmétique” suggests a broader definition replacing the third inequality by 
$$\exists C_v\gt 0,\,\,|x+y|_v\leq C_ v\max(|x|_v,|y|_v)$$
This definition covers the ultra metric case with $C_v=1$ and if the absolute value verifies the triangle inequality, it verifies the above with $C_v=2$.
How to prove the converse i.e a map such that $|x+y|_v\leq 2\max(|x|_v,|y|_v)$ verifies the triangle inequality.
 A: I tried to solve it my self for hours but I realized that the problem is far from trivial. So after searching in some books I found the proof and it is not straightforward. 
See chapter 3, 18.6-18.9 of the book: Topological Fields, S. Warner, Volume 157 of North-Holland Mathematics Studies, Elsevier, 1989
A: To simplify the notations we will write $v(x)=|x|_v$. the definition of absolute value rewrites as follows
$$\begin{align}
&v(x)0\iff x=0\tag{*}\label{*}\\
&v(xy)=v(x)v(y)\tag{**}\label{**}\\
&v(x+y)\leq2\max(v(x),v(y))\tag{***}\label{***}
\end{align}$$
We will first prove by induction on $r$ that
$$\forall\left(x_i\right)_{1\leq i \leq 2^r},\,v\left(\sum_i x_i\right)\leq 2^r\max_{1\leq i\leq 2^r}\left(v(x_i)\right)\tag{1}\label{1}$$
With $\ref{1}$ in hand we will prove
$$\forall n\in\Bbb{N},\,v(n\cdot 1)\leq 2.n\tag{2}\label{2}$$
Indeed consider $n$ and $r$ such that $2^{r-1}\leq n\lt 2^r$.
Now apply $\ref{1}$ to the $(x_i)_{1\leq i\leq 2^r}$ such that $x_i=1$ for $1\leq i\leq n$ and $x_i=0$ for $n\lt i\leq 2^r$ and you get the result.
$\ref{2}$ combined with the inequality $\ref{1}$ will lead to the triangle inequality and complete the proof.
Indeed assume $\ref{1}$ and $\ref{2}$ and take $n=2^r-1$. One has
$$\begin{align}
v(1+x)^n&\stackrel{\ref{**}}=v\left((1+x)^n\right)=v\left(\sum_{k=0}^n{n\choose k}x^k\right)\\
&\stackrel{\ref{1}}\leq 2^r\max_k\left(v\left({n\choose k}x^k\right)\right)\\
&\stackrel{\ref{**}}=2^r\max_k\left(v\left({n\choose k}\cdot 1\right)v\left(x^k\right)\right)\\
&\stackrel{\ref2}\leq 2\cdot2^r\max_k\left({n\choose k}v(x)^k\right)\\
&\leq 2^{r+1}\sum_{k=0}^n{n\choose k}v(x)^k=2^{r+1}\left(1+v(x)\right)^n
\end{align}$$
Now take the $n^{th}$ root and let $r$ go to infinity, one gets
$$v(1+x)\leq\lim_{r\to\infty}2^{r+1\over 2^r-1}\left(1+v(x)\right)=1+v(x)$$
We can now conclude.
$$v(x+y)=v(x(1+yx^{-1})=v(x)v(1+yx^{-1})\leq v(x)(1+v(yx^{-1})=v(x)+v(y)$$
