MathDB
Geometric mean of fractions is larger

Source: 2012 Indonesia Round 2 TST 2 Problem 3

March 4, 2012
inequalities unsolvedinequalities

Problem Statement

Let a1,a2,,an,b1,b2,,bna_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n be positive reals such that a1+b1=a2+b2=+an+bna_1 + b_1 = a_2 + b_2 = \ldots + a_n + b_n and a1a2anb1b2bnnn.\sqrt[n]{\dfrac{a_1a_2\ldots a_n}{b_1b_2\ldots b_n}} \ge n. Prove that a1a2anb1b2bnna1+a2++anb1+b2++bn.\sqrt[n]{\dfrac{a_1a_2\ldots a_n}{b_1b_2\ldots b_n}} \ge \dfrac{a_1+a_2+\ldots+a_n}{b_1+b_2+\ldots+b_n}.