The additive $p-$ adic valuation of $\mathbb{Q}_p$:
$$w_p: \left\{\begin{matrix} \mathbb{Q}_p \rightarrow \mathbb{Z} \cup \{\infty\}\\ p^m u \mapsto m\\ 0 \mapsto \infty \end{matrix}\right.$$
$$\forall x,y \in \mathbb{Q}: w_p(x+y) \geq \min \{ w_p(x), w_p(y)\}$$
If $w_p(x) \neq w_p(y)$, then the equality stands.
This is the proof, according to my notes:
$$x=p^m u_1 | u_1 \in \mathbb{Z}_p^*$$
$$y=p^n u^2 | u_2 \in \mathbb{Z}_p^*$$
$$m,n \in \mathbb{N}$$
Without loss of generality, we suppose that $m \leq n$.
$$x+y=p^m(u_1+p^{n-m}u_2)$$
$$w_p(x+y) \geq m$$
If $n>m$, then $u_1+p^{n-m}u_2 \in \mathbb{Z}_p^*$
In this case:
$$w_p(x+y)=m=\min \{ w_p(x), w_p(y) \}$$
If $n=m$, $\displaystyle{ w_p(x+y) \geq n=m=\min \{ w_p(x), w_p(y)\} }$
Could you explain me why it is : $w_p(x+y) \geq m$ and not $w_p(x+y)=m$ ?
Hint: e.g. $3^2|(2+3^0\cdot7)$. So when $m= n$...
Expanding a bit more based on the comment below, when $m=n$, you have $x+y = p^m(u_1+u_2)$. Now you could have the case where $w_p(u_1+u_2) > 0$ even though $w_p(u_i)=0$, as in the example above. For another e.g. note $w_2(2+2) = 2 \ge w_p(2)=1$.