Problem
ALG-B2-M02-P021 Circular version of the game
There are \(2n\) nonnegative numbers with sum \(1\), \(n\ge2\). They must be arranged around a circle so that the maximum product of neighbouring numbers is as small as possible. Prove that for any numbers one can make the maximum at most \(\frac1{8(n-1)}\), and give a set for which this cannot be improved.
Hint 1. For the example, take \(0,\frac12\), and \(2n-2\) equal numbers.
Hint 2. For the bound, order the numbers and place large ones with gaps, then small ones in reverse order.
Lower bound: take \(0,\frac12,\frac1{4(n-1)},\ldots,\frac1{4(n-1)}\). The number \(\frac12\) has two neighbours; at least one of them is \(\frac1{4(n-1)}\), so the maximum product is at least \(\frac1{8(n-1)}\).
For the upper bound, order \(x_1\ge\cdots\ge x_{2n}\). Arrange the numbers so that all potentially large neighbouring products have the form \(x_kx_{2n-k}\), \(1\le k\le n-1\). As in the pairing problem, if \(S=x_1+\cdots+x_k\), then \[x_kx_{2n-k}\le\frac{S(1-S)}{k(2n-2k)}\le\frac1{4k(2n-2k)}\le\frac1{8(n-1)}.\] Hence the bound is attainable for all choices and is sharp.
A. Source analysis. Main objects: positive quantities, a product or a sum, choosing the right terms for AM-GM, and a strict equality case. The obvious first attempt usually applies AM-GM to the visible terms, but it misses the hidden structure. The hidden observation is to normalize, order, multiply, or substitute first so that the product of the chosen terms is controlled.
B. New problem. The statement is rewritten; only the method architecture is preserved.
C-D. Hints. The hint fields contain a gentle hint and a method-level hint.
E. Full solution. Given in the solution field.
F. Difficulty justification. Regional level 7: the strategy requires an arrangement construction and a global estimate through \(S(1-S)\).
G. Why this is not a one-step exercise. Even the correct product estimate is useless without an arrangement that reduces all adjacencies to controlled pairs.