Double posted https://artofproblemsolving.com/community/c6h2625916p22698473

Denote the center of disk $i$ as $O_i$, for all $i = 1, 2, \dots, n$. Claim 01. (IGO 2017/2) Given any six non-intersecting disk, each having radius at least $r$. Then the radius of the circle intersecting all of the six disks is at least $r$. Proof. Consider $OO_i$ for each $i$. By Pigeon Hole Principle, there exists two circles with center $O_i, O_j$ such that $\angle O_i O O_j \le 60^{\circ}$. Suppose that the circle intersecting all of the six disks has radius $r_0 < r$. By Law of Cosine on $\triangle OO_i O_j$, we have \[ O_iO_j^2 = OO_i^2 + OO_j^2 - 2 \cdot OO_i \cdot OO_j \cdot \cos O_i OO_j \le OO_i^2 + OO_j^2 - OO_i \cdot OO_j \]WLOG $OO_i \ge OO_j$, then we have \[ OO_i \cdot OO_j \ge OO_j^2 \ \text{and} \ O_i O_j > R_i + R_j \ge R_i + r > R_i + r_0 \ge OO_i \]from which this forces $O_i O_j^2 \le OO_i^2 + (OO_j^2 - OO_i \cdot OO_j) \le OO_i^2 < O_i O_j^2 $, which is a contradiction. Now, we just prove the inequality by induction on $n \ge 6$, the base case is trivial. To do this, applying the lemma to the six largest disks, and we will have find index $i$ and $j$ such that $OP_i \ge R_j \ge R_6$. Delete this disk and apply the induction hypothesis, and we get \[ OP_1 + OP_2 + \dots + OP_n \ge R_6 + R_7 + \dots + R_n \]

Proposed by Mohammad Ali Abam and Morteza Saghafian from Iran

Dadgarnia wrote: Proposed by Mohammad Ali Abam and Morteza Saghafian from Iran

I contend \(OP_i\ge R_6\) for some \(i\le6\). This suffices by induction. Drop the condition \(R_1\ge R_2\ge\cdots\ge R_6\), and instead number the disks \(\mathcal D_1\), \ldots, \(\mathcal D_6\) counterclockwise with respect to \(O\). Let the center of \(\mathcal D_i\) be \(O_i\). Since \(\angle O_1OO_2+\cdots+\angle O_6OO_1=360^\circ\), for some index \(i\) we have \(\angle O_iOO_{i+1}\le60^\circ\). Without loss of generality \(OO_i\ge OO_{i+1}\). Evidently \(\angle O_iOO_{i+1}\) is not the largest angle in \(\triangle O_iOO_{i+1}\), i.e.\ \(\overline{O_iO_{i+1}}\) is not the longest side, so \[OO_i\ge O_iO_{i+1}\ge R_i+R_{i+1}.\]This allows us to conclude \[OP_i\ge OO_i-R_i\ge R_{i+1}\ge\min\{R_1,\ldots,R_6\}.\]

Let $C_i$ denote the center of $D_i$. First, we can delete some disks $D_i$ with $i\geq 6$ and $\overline{OC_i}\geq 2R_i$, if the inequality holds in this case, then it holds for all cases. Let $A_i=\{j|\overline{OC_j}>R_i+R_j,\ j<i\}$. If there is 6 disks $D_{j_1}, D_{j_2}, \dots, D_{j_6},\ j_1, j_2, j_3, j_4, j_5, j_6<i$ such that $\{j_1, j_2, j_3, j_4, j_5, j_6\}\cap A_i=\emptyset$, then (WLOG suppose that $D_{j_1}, D_{j_2}, \dots, D_{j_6}$ is clockwise order to $O$) there exists two adjacent disk $D_{j_k}, D_{j_{k+1\mod 6}}$ (WLOG suppose that is $D_{j_1}$ and $D_{j_2}$) such that $\angle C_{j_1}OC_{j_2}\leq 60^o$. But $\overline{C_{j_1}C_{j_2}}>R_{j_1}+R_{j_2}>\max(R_{j_1}+R_i, R_{j_2}+R_i)>\max(\overline{OC_{j_1}}, \overline{OC_{j_2}})$, which is a contradiction. $\therefore |A_i|\geq i-5$ From $i=6$ to $n$, we can choose one element $j_i$ from $A_i$ such that it hasn't be chosen. As we know $OP_{j_i}\geq R_i$, therefore $\overline{OP_1}+\overline{OP_2}+\cdots+\overline{OP_n}\geq\overline{OP_{j_6}}+\overline{OP_{j_7}}+\cdots+\overline{OP_{j_n}}\geq R_6+R_7+\cdots+R_n$.

We proceed with induction. $n\le 5$ is clearly true. Let $O_i$ be the center of $D_i$. Then, there exist $1\le i<j\le 6$ such that $\angle O_iOO_j\le 60^\circ$. Thus, $O_iO_j$ is not the longest side of that triangle. Without loss of generality, \[OP_i+R_i\ge OO_i\ge O_iO_j\ge R_i+R_j\]so $OP_i\ge R_j\ge R_6$. Remove $D_i$ and the result is still true. Therefore, we can add in $OP_i$ to the left side and $R_6$ to the right and it is still true.