$B _{\ell ^{2}} ^{+}$ with the norm $\Vert\vert \cdot \Vert\vert _{\sqrt{2}}$ don’t have normal structure

Consider the space $ \ell ^{2}$ with the standard norm \begin{align*} \Vert x \Vert _{2} = \left( \sum _{i =1} ^{\infty} x _{i} ^{2} \right) ^{1/2} \end{align*} and we define the equivalent norm \begin{align*} \Vert\vert x \Vert\vert _{\sqrt{2}}= \max\{ \Vert x \Vert _{2}, \sqrt{2}\Vert x \Vert _{\infty} \} \mbox{.} \end{align*}

Let’s define the positive part of the unit ball \begin{align*} B _{\ell ^{2}} ^{+} = \lbrace x \in \ell ^{2}: \; \Vert x \Vert _{2} \leqslant 1, \; x _{i} \geqslant 0 \rbrace \mbox{.} \end{align*} I want show that $ B _{\ell ^{2}} ^{+}$ with the norm $ \Vert\vert \cdot \Vert\vert _{\sqrt{2}}$ don’t have normal structure, and for show that, I should show that diam$ (B _{\ell ^{2}} ^{+})$ = $ r _{x}(B _{\ell ^{2}} ^{+})$ . I showed that diam$ (B _{\ell ^{2}} ^{+}) = 1$ , but I don’t know why $ r _{x}(B _{\ell ^{2}} ^{+}) =1$ . What element in $ B _{\ell ^{2}} ^{+}$ can take for prove that $ r _{x}(B _{\ell ^{2}} ^{+}) =1$ ?