Difference between revisions of "1994 USAMO Problems/Problem 4"
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− | ==Problem | + | ==Problem == |
Let <math>\, a_1, a_2, a_3, \ldots \,</math> be a sequence of positive real numbers satisfying <math>\, \sum_{j = 1}^n a_j \geq \sqrt {n} \,</math> for all <math>\, n \geq 1</math>. Prove that, for all <math>\, n \geq 1, \,</math> | Let <math>\, a_1, a_2, a_3, \ldots \,</math> be a sequence of positive real numbers satisfying <math>\, \sum_{j = 1}^n a_j \geq \sqrt {n} \,</math> for all <math>\, n \geq 1</math>. Prove that, for all <math>\, n \geq 1, \,</math> | ||
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== Solution == | == Solution == | ||
− | + | Note that if we try to minimize <math>(a_j)^2</math>, we would try to make the <math>a_j</math> as equal as possible. However, by the condition given in the problem, this isn't possible, the <math>a_j</math>'s have to be an increasing sequence. Thinking of minimizing sequences, we realize that the optimal equation is <math>a_n = \sqrt{n} - \sqrt{n-1} = 1/(\sqrt{n} + \sqrt{n-1})</math>. Note that this is strictly greater than <math>1/(2\sqrt{n})</math>. So it is greater than the sum of <math>(1/\sqrt{4n})^2</math> over all n from 1 to <math>\infty</math>. So the sum we are looking to minimize is strictly greater than the sum of <math>1/4n</math> over all <math>n</math> from 1 to <math>\infty</math>, which is what we wanted to do. | |
− | <math> | ||
− | + | == See Also == | |
− | + | {{USAMO box|year=1994|num-b=3|num-a=5}} | |
− | + | [[Category:Olympiad Algebra Problems]] | |
+ | [[Category:Olympiad Inequality Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:30, 4 July 2013
Problem
Let be a sequence of positive real numbers satisfying for all . Prove that, for all
Solution
Note that if we try to minimize , we would try to make the as equal as possible. However, by the condition given in the problem, this isn't possible, the 's have to be an increasing sequence. Thinking of minimizing sequences, we realize that the optimal equation is . Note that this is strictly greater than . So it is greater than the sum of over all n from 1 to . So the sum we are looking to minimize is strictly greater than the sum of over all from 1 to , 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 |
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