Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 7.7 National olympiad, round 2 Prove it

6. Let $\left\{a_{1}, a_{2}, a_{3}, \cdots\right\}$ be an infinite sequence of positive numbers. Prove the inequality $\sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 4 \sum_{n=1}^{N} a_{n}^{2}$ for any positive integer $N$ . Here $\alpha_{n}$ is the average of $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ , i.e., $\alpha_{n}=$ $\frac{a_{1}+a_{2}+a_{3}+\cdots+a_{n}}{n} .(2005$ Korean Mathematical Olympiad Problem)

Solution

6. If we set $\frac{1}{c} \sum_{n=1}^{N} \alpha_{n}^{2} \leqslant \sum_{n=1}^{N} \alpha_{n} a_{n}$ , then we have $\sum_{n=1}^{N} \alpha_{n} a_{n} \leqslant c \sum_{n=1}^{N} a_{n}^{2}$ , and the problem can be transformed into handling by the Abel method:
$\begin{array}{l} \sum_{n=1}^{N} \alpha_{n} a_{n}=\sum_{n=1}^{N} \alpha_{n}\left[n \alpha_{n}-(n-1) \alpha_{n-1}\right]= \\ \sum_{n=1}^{N} n a_{n}^{2}-\sum_{n=1}^{N}(n-1) \alpha_{n} \alpha_{n-1} \geqslant \\ \sum_{n=1}^{N} n a_{n}^{2}-\frac{1}{2}\left[\sum_{n=1}^{N}(n-1) \alpha_{n}^{2}+\sum_{n=1}^{N}(n-1) \alpha_{n-1}^{2}\right]= \\ \frac{1}{2} \sum_{n=1}^{N} \alpha_{n}^{2}+\frac{1}{2} n \alpha_{n}^{2} \geqslant \frac{1}{2} \sum_{n=1}^{N} \alpha_{n}^{2} \end{array}$

By the Cauchy-Schwarz inequality, we get $\left(\sum_{n=1}^{N} \alpha_{n} a_{n}\right)^{2} \leqslant \sum_{n=1}^{N} \alpha_{n}^{2} \sum_{n=1}^{N} a_{n}^{2}$ , i.e.,
$\sum_{n=1}^{N} \alpha_{n} a_{n} \leqslant \sqrt{\sum_{n=1}^{N} \alpha_{n}^{2} \sum_{n=1}^{N} a_{n}^{2}}$

Therefore, $\sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 4 \sum_{n=1}^{N} a_{n}^{2}$ holds for any positive integer $N$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.