I apologize in advance, if this is a stupid question. I thought about this for some time and don't see any mistake in my reasoning, even after carefully typing up the proof here. I think this is true
Claim. For any non-zero $x,y\in\mathbb{Q}$ we have that $$ |x+y|_p\color{red}{=}\mathrm{max}(|x|_p,|y|_p). $$
Proof.
Let $x=\frac{a}{b}p^n$, $y=\frac{c}{d}p^m$, where $a,b,c,d,n,m\in\mathbb{Z}$; $a,b,c,d\neq 0$ and $a,b,c,d$ are not divisible by $p$; $\mathrm{gcd} (a,b)=1$, $\mathrm{gcd} (c,d)=1$.
We have $|x|_p=\frac{1}{p^n}$, $|x|_p=\frac{1}{p^m}$.
Let $n>m$. Then
$$\mathrm{max}(|x|_p,|y|_p)=\frac{1}{p^m}.\tag{1}$$
We have $$ |x+y|_p=\Big|\frac{a}{b}p^n+\frac{c}{d}p^m\Big|_p=|p^m|_p\cdot \Big|\frac{a}{b}p^{n-m}+\frac{c}{d}\Big|_p\\ =\frac{1}{p^m}\cdot \Big|\frac{adp^{n-m}+bc}{bd}\Big|_p. $$ Since $n-m>0$ and the nonzero integers $ad$, $bc$ are not divisible by $p$, we have that the integer $adp^{n-m}+bc$ is nonzero and is not divisible by $p$. Since the non zero integer $bd$ is not divisible by $p$ either, we have $$ \Big|\frac{adp^{n-m}+bc}{bd}\Big|_p=1. $$ Thus using $(1)$ $$ |x+y|_p=\mathrm{max}(|x|_p,|y|_p)\cdot 1=\mathrm{max}(|x|_p,|y|_p) $$ as claimed. $\Box$
In the book I'm reading it is proved that $$ |x+y|_p \color{red}{\leq} \mathrm{max}(|x|_p,|y|_p). $$
I feel like I'm loosing my sanity. Please, help me.
