Track / Stage 6 / 38 of 400 #1038 of 1964

Problem 1038

National Olympiad, first round

Algebra Difficulty 6.0 Prove it

Given $x_{i} \in \mathbf{R}(i=1,2, \cdots, n, n \geqslant 2)$ , satisfying $\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}=0$ .
Prove: $\left|\sum_{i}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n}$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Proof: Let $\sum_{i=1}^{n} x_{i}=A+B, \sum_{i=1}^{n} \frac{x_{i}}{i}=a+b$ , where $A, a$ are the sums of positive terms, and $B, b$ are the sums of negative terms,
From the problem, we know that $A+B=0, A-B=1$ ,
Thus, $A=-B=\frac{1}{2}$ ,
Since $\frac{A}{n} \leqslant a \leqslant A, B \leqslant b \leqslant \frac{B}{n}$ , we have $B+\frac{A}{n} \leqslant a+b \leqslant A+\frac{B}{n}$ ,
That is, $\left(\frac{1}{2}-\frac{1}{2 n}\right) \leqslant \sum_{i=1}^{n} \frac{x_{i}}{i} \leqslant \frac{1}{2}-\frac{1}{2 n}$ ,
Therefore, $\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.