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

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== Solution ==
 
== Solution ==
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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>\infinity</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>\infinity</math>, which is what we wanted to do.
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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.
  
 
== 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 14:32, 15 April 2012

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 $\infty$. 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.

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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