Problem
ALG-B2-M01-P012 Three versus two or four
Given \(12\) distinct positive numbers. Prove that one can choose three numbers whose product is greater than the product of two other chosen numbers, or three numbers whose product is greater than the product of four other chosen numbers.
Hint 1. Order the numbers decreasingly.
Hint 2. Split according to whether the sixth number is at least \(1\).
Let \(a_1>a_2>\cdots>a_{12}>0\). Then \(a_1a_2a_3>a_4a_5a_6\). If \(a_6\ge1\), then \(a_1a_2a_3>a_4a_5a_6\ge a_4a_5\), so three numbers beat two. If \(a_6<1\), then \(a_7<1\), and \(a_1a_2a_3>a_4a_5a_6>a_4a_5a_6a_7\), so three numbers beat four.
A. Source analysis. Main objects: inequalities, order, an extremal element, or an invariant. The obvious first move usually gives only a local estimate. The hidden observation is to choose the right nondecreasing quantity, or to add/multiply inequalities only after signs are controlled. The needed step is an ordering, an invariant, a product transformation, or a boundary case.
F. Difficulty justification. Regional level 6: the method is simple, but the boundary \(1\) must be chosen correctly.
G. Why this is not a one-step exercise. Without ordering, it is unclear which five to seven numbers should be compared.