On the one hand, let $j_{0}\in\{0,\dots,d\}$ be an index such that $r_{j_{0}}$ is minimal. By taking $\kappa_{j}=0$ for $j\neq j_{0}$ and $\kappa_{j_{0}}=1$, one sees that

$$\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\rvert_{K}\cdot r_{j}\Big\}\leq r_{j_{0}}=\min_{j\in\{0,\dots,d\}}\{r_{j}\}.$$

On the other hand, by the ultrametricity of $\lvert\mathord{\cdot}\rvert_{K}$, if $\sum_{j}\kappa_{j}=1$, then there exist at least one $j_{1}\in\{0,\dots,d\}$ such that $\lvert\kappa_{j_{1}}\rvert\geq 1$, so

$$\begin{split}\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\rvert_{K}\cdot r_{j}\Big\}&\geq\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}\lvert\kappa_{j_{1}}\rvert_{K}\cdot r_{j_{1}}\\ &\geq\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}r_{j_{1}}=r_{j_{1}}\geq\min_{j\in\{0,\dots,d\}}\{r_{j}\}.\end{split}$$

Hence the two sides are equal. ∎