Maths Olympiad Prep

Library / /17 of 54

Algebra Difficulty 5.7 AIME, harder Prove it China

Let a1,a2,…,an,b1,b2,…,bna_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n be non-negative numbers satisfying the following conditions simultaneously:
(1)∑i=1n(ai+bi)=1; (1) \quad \sum_{i=1}^{n} (a_i + b_i) = 1;
(2)∑i=1ni(ai−bi)=0; (2) \quad \sum_{i=1}^{n} i(a_i - b_i) = 0;
(3)∑i=1ni2(ai+bi)=10. (3) \quad \sum_{i=1}^{n} i^2 (a_i + b_i) = 10.
Prove that max⁡{ak,bk}≤1010+k2\max\{a_k, b_k\} \le \frac{10}{10+k^2} for all 1≤k≤n1 \le k \le n.

Solution

For any 1≤k≤n1 \le k \le n, it follows from the given conditions and Cauchy's Inequality that
(kak)2≤(∑i=1niai)2=(∑i=1nibi)2≤(∑i=1ni2bi)(∑i=1nbi)=(10−∑i=1ni2ai)(1−∑i=1nai)≤(10−k2ak)(1−ak)=10−(10+k2)ak+k2ak2. \begin{align*} (ka_k)^2 &\le \left(\sum_{i=1}^n i a_i\right)^2 = \left(\sum_{i=1}^n i b_i\right)^2 \\ &\le \left(\sum_{i=1}^n i^2 b_i\right) \left(\sum_{i=1}^n b_i\right) \\ &= (10 - \sum_{i=1}^n i^2 a_i) (1 - \sum_{i=1}^n a_i) \\ &\le (10 - k^2 a_k) (1 - a_k) \\ &= 10 - (10 + k^2)a_k + k^2 a_k^2. \end{align*}
It follows from that ak≤1010+k2a_k \le \frac{10}{10+k^2}. Similarly, bk≤1010+k2b_k \le \frac{10}{10+k^2}, and hence the result follows. □\square

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.