Difference between revisions of "1994 USAMO Problems/Problem 4"

Latest revision as of 12:30, 4 July 2013

Problem

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 $\infty$ . So the sum we are looking to minimize is strictly greater than the sum of $1/4n$ over all $n$ from 1 to $\infty$ , which is what we wanted to do.

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 [[Category:Olympiad Algebra Problems]]
 [[Category:Olympiad Inequality Problems]]
+{{MAA Notice}}

1994 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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Difference between revisions of "1994 USAMO Problems/Problem 4"

Latest revision as of 12:30, 4 July 2013

Problem

Solution

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