# 1994 USAMO Problems/Problem 4

## Problem 4

Let $\, a_1, a_2, a_3, \ldots \,$ be a sequence of positive real numbers satisfying $\, \sum_{j = 1}^n a_j \geq \sqrt {n} \,$ for all $\, n \geq 1$. Prove that, for all $\, n \geq 1, \,$ $$\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).$$

## Solution

Note that if we try to minimize (a_j)^2, we would try to make the a_j as equal as possible. However, by the condition given in the problem, this isn't possible, the a_j's have to be an increasing sequence. Thinking of minimizing sequences, we realize that the optimal equation is a_n = \sqrt{n} - \sqrt{n-1} = 1/(\sqrt{n} + \sqrt{n-1}). Note that this is strictly greater than 1/(2\sqrt{n}). So it is greater than the sum of (1/\sqrt{4n})^2 over all n from 1 to infinity. So the sum we are looking to minimize is strictly greater than the sum of 1/4n over all n from 1 to infinity, which is what we wanted to do.

## See Also

 1994 USAMO (Problems • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 All USAMO Problems and Solutions
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