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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== See Also == | == See Also == | ||
{{USAMO box|year=1994|num-b=3|num-a=5}} | {{USAMO box|year=1994|num-b=3|num-a=5}} | ||
| + | [[Category:Olympiad Inequality Problems]] | ||
Revision as of 10:53, 17 September 2012
Problem
Let 
be a sequence of positive real numbers satisfying 
for all 
. Prove that, for all 
![\[\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).\]](http://latex.artofproblemsolving.com/8/2/9/8293056937b2c93fb3ccfde0a9b743d24fd9660d.png)
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 | ||
