Difference between revisions of "1994 USAMO Problems/Problem 4"
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<cmath>\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).</cmath> | <cmath>\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).</cmath> | ||
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+ | == 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 == | == See Also == | ||
{{USAMO box|year=1994|num-b=3|num-a=5}} | {{USAMO box|year=1994|num-b=3|num-a=5}} |
Revision as of 23:50, 23 February 2012
Problem 4
Let be a sequence of positive real numbers satisfying for all . Prove that, for all
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 by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |