Track / Stage 5 / 193 of 400 #793 of 1964

Problem 793

AIME late

Algebra Difficulty 5.4 Prove it

Does there exist a sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of positive real numbers satisfying both of the following conditions:
(i) $\sum_{i=1}^{n} a_{i} \leq n^{2}$ , for every positive integer $n$ ;
(ii) $\sum_{i=1}^{n} \frac{1}{a_{i}} \leq 2008$ , for every positive integer $n$ ?

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

The answer is no.
It is enough to show that
if $\sum_{i=1}^{n} a_{i} \leq n^{2}$ for any $n$ , then $\sum_{i=2}^{2^{n}} \frac{1}{a_{i}}>\frac{n}{4}$ . (or any other precise estimate)
For this, we use that $\sum_{i=2^{k}+1}^{2^{k+1}} a_{i} \sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}} \geq 2^{2 k}$ for any $k \geq 0$ by the arithmetic-harmonic mean inequality.
Since $\sum_{i=2^{k}+1}^{2^{k+1}} a_{i} \leq 4 \cdot 2^{2k}$ and hence $\sum_{i=2}^{2^{n}} \frac{1}{a_{i}} > \sum_{k=0}^{n-1} \sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}} > \frac{n}{4}$ . (it can be stated in words)

## Remark: no points for using some inequality, that doesn't lead to solution

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