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\sum_{i=1}^n ai=\Pi_{i=1}^n a_i, 0<a_i \le b_i=>\sum_{i=1}^n bi \le\Pi_{i=1}^n

Source: Polish second round 1994 p2

January 19, 2020
SumProductinequalitiesalgebra

Problem Statement

Let a1,...,ana_1,...,a_n be positive real numbers such that ∑i=1nai=∏i=1nai\sum_{i=1}^n a_i =\prod_{i=1}^n a_i , and let b1,...,bnb_1,...,b_n be positive real numbers such that ai≤bia_i \le b_i for all ii. Prove that ∑i=1nbi≤∏i=1nbi\sum_{i=1}^n b_i \le\prod_{i=1}^n b_i
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