Track / Stage 7 / 132 of 300 #1532 of 1964

Problem 1532

National olympiad second round; IMO P1/P4

Algebra Difficulty 7.3 Prove it

10. (1) Let $\left\{b_{n}\right\}$ be a sequence of positive integers, and for all $n \geqslant 1$ we have $b_{n+1}^{2} \geqslant \frac{b_{1}^{2}}{1^{3}}+\frac{b_{2}^{2}}{2^{3}}+\cdots+\frac{b_{n}^{2}}{n^{3}}$ . Prove: There exists a positive integer $k$ , such that $\sum_{n=1}^{k} \frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}}>\frac{1993}{1000}$ . (34th IMO, Turkey)
(2) Let $x_{1}, x_{2}, \cdots, x_{2001}$ satisfy $x_{i}^{2} \geqslant \frac{x_{1}^{2}}{1^{3}}+\frac{x_{2}^{2}}{2^{3}}+\cdots+\frac{x_{i-1}^{2}}{(i-1)^{3}}, 2 \leqslant i \leqslant 2001$ , prove: $\sum_{i=2}^{2001} \frac{x_{i}}{x_{1}+x_{2}+\cdots+x_{i-1}}>1999$ . (2001 Yugoslav Mathematical Olympiad)

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

10. (1) By the Cauchy inequality, we have
$\left(1^{3}+2^{3}+\cdots+n^{3}\right) b_{n+1}^{2} \geqslant\left(1^{3}+2^{3}+\cdots+n^{3}\right)$
$\left(\frac{b_{1}^{2}}{1^{3}}+\frac{b_{2}^{2}}{2^{3}}+\cdots+\frac{b_{n}^{2}}{n^{3}}\right)=\left(b_{1}+b_{2}+\cdots+b_{n}\right)^{2}$

Since $1^{3}+2^{3}+\cdots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$ , we have
$\frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}} \geqslant \frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}$

Therefore,
$\sum_{n=1}^{k} \frac{b_{n+1}}{b_{1}+b_{2}+\cdots+b_{n}} \geqslant \sum_{n=1}^{k}\left(\frac{2}{n}-\frac{2}{n+1}\right)=2\left(1-\frac{1}{k+1}\right)$

Thus, by taking $k=999$ , we get
$2\left(1-\frac{1}{1000}\right)=\frac{1998}{1000}>\frac{1993}{1000}$
(2) By taking $k=2001$ .

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