, 2021

We will make use of the following lemma.

Lemma. Let $D_1, \dots, D_6$ be disjoint disks in the plane with radii $R_1, \dots, R_6$. Let $P_i$ be a point in $D_i$, and let $O$ be an arbitrary point. Then there exist indices $i$ and $j$ such that $OP_i \ge R_j$.

Proof. Let $O_i$ be the center of $D_i$. Consider six rays $OO_1, \dots, OO_6$ (if $O = O_i$, then the ray $OO_i$ may be assumed to have an arbitrary direction). These rays partition the plane into six angles (one of which may be non-convex) whose measures sum up to $360^\circ$; hence one of the angles, say $\angle O_iOO_j$, has measure at most $60^\circ$. Then $O_iO_j$ cannot be the unique largest side in (possibly degenerate) triangle $OO_iO_j$, so, without loss of generality, $OO_i \ge O_iO_j \ge R_i + R_j$. Therefore, $OP_i \ge OO_i - R_i \ge (R_i + R_j) - R_i = R_j$, as desired. $\square$

Now we prove the required inequality by induction on $n \ge 5$. The base case $n = 5$ is trivial. For the induction step, apply the Lemma to the six largest disks, in order to find indices $i$ and $j$ such that $1 \le i, j \le 6$ and $OP_i \ge R_j \ge R_6$. Removing $D_i$ from the configuration and applying the inductive hypothesis, we get
$$\sum_{k \ne i} OP_k \ge \sum_{\ell \ge 7} R_\ell.$$
Adding up this inequality with $OP_i \ge R_6$ we establish the induction step.