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China Western Mathematical Olympiad

Algebra Difficulty 5.7 AIME, harder Prove it China

Let $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ be non-negative numbers satisfying the following conditions simultaneously:
$(1) \quad \sum_{i=1}^{n} (a_i + b_i) = 1;$
$(2) \quad \sum_{i=1}^{n} i(a_i - b_i) = 0;$
$(3) \quad \sum_{i=1}^{n} i^2 (a_i + b_i) = 10.$
Prove that $\max\{a_k, b_k\} \le \frac{10}{10+k^2}$ for all $1 \le k \le n$ .

Solution

For any $1 \le k \le n$ , it follows from the given conditions and Cauchy's Inequality that
$\begin{align*} (ka_k)^2 &\le \left(\sum_{i=1}^n i a_i\right)^2 = \left(\sum_{i=1}^n i b_i\right)^2 \\ &\le \left(\sum_{i=1}^n i^2 b_i\right) \left(\sum_{i=1}^n b_i\right) \\ &= (10 - \sum_{i=1}^n i^2 a_i) (1 - \sum_{i=1}^n a_i) \\ &\le (10 - k^2 a_k) (1 - a_k) \\ &= 10 - (10 + k^2)a_k + k^2 a_k^2. \end{align*}$
It follows from that $a_k \le \frac{10}{10+k^2}$ . Similarly, $b_k \le \frac{10}{10+k^2}$ , and hence the result follows. $\square$

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