Nearest-point minimization outside a Euclidean ball (${\mathrm{argmin}}_{\Vert x\Vert \ge R}\Vert x-a{\Vert }^2=Ra/\Vert a\Vert (0<\Vert a\Vert <R)$)

Suppose $0<\Vert a\Vert <R$ in a Euclidean space. Among points outside or on the radius-$R$ ball centred at zero, the unique closest point to $a$ is $x_{\ast }=Ra/\Vert a\Vert$. Its squared distance is $(R-\Vert a\Vert {)}^2$.

Indeed, the reverse triangle inequality gives $\Vert x-a\Vert \ge \Vert x\Vert -\Vert a\Vert \ge R-\Vert a\Vert$. Equality requires $\Vert x\Vert =R$ and $x$ pointing in the same direction as $a$, giving the stated unique point. If $a=0$, every point on the sphere is a minimizer. If $\Vert a\Vert \ge R$, the point $a$ itself is feasible and the minimum distance is zero.

The exterior is not a convex set, so a direct geometric argument is useful: it certifies the global answer without treating every candidate satisfying the Karush-Kuhn-Tucker conditions as a minimum.