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sum \frac{b_{n+1}}{b_1+b_2+...+b_n} > \frac{2013}{1013}

Source: Singapore Senior Math Olympiad 2013 2nd Round p3 SMO

March 25, 2020
Sequenceinequalitiesalgebra

Problem Statement

Let b1,b2,...b_1,b_2,... be a sequence of positive real numbers such that for each n≥1 n\ge 1, bn+12≥b1213+b2223+...+bn2n3b_{n+1}^2 \ge \frac{b_1^2}{1^3}+\frac{b_2^2}{2^3}+...+\frac{b_n^2}{n^3} Show that there is a positive integer MM such that ∑n=1Mbn+1b1+b2+...+bn>20131013\sum_{n=1}^M \frac{b_{n+1}}{b_1+b_2+...+b_n} > \frac{2013}{1013}
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