Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it United States

Problem:

Let SS be a finite set of positive real numbers. If SS's average is at most 11 but its product is at least 0.90.9, show that any three elements of SS can form the sides of a triangle.

Solution

Solution:

Assume otherwise for the sake of contradiction, i.e. that there exist x,y,zSx, y, z \in S so that x+yzx + y \leq z. For fixed zz, the product xyx y is then maximized when x=yx = y. Then, if x,y,zx, y, z have average aa, their product is at most that when x+y=zx + y = z, which happens at x=3a/4x = 3a/4, y=3a/4y = 3a/4, z=3a/2z = 3a/2, for a product of 27a3/3227a^3/32.

Now consider replacing all of x,y,zx, y, z with aa in SS. The product multiplies by at least 32/2732/27 for a product of at least 0.932/27>10.9 \cdot 32/27 > 1, while the average is unchanged, a contradiction to AM-GM.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.