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

 
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Since each <math>a_{i}</math> is positive, by Muirhead's inequality,
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==Problem ==
<math>2(\sum a_{i}^2) \ge (\sum a)^2 \ge n</math>. Now we claim that <math> \frac{n}{2}> frac{1}{4}(1+...\frac{1}{n)}</math>
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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>
  
<math>n=1</math>, giving <math>1/2>1/4</math> works, but we set the base case <math>n=2</math>, which gives <math>1>3/8</math>. Now assume that it works for <math>n</math>.
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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>
By our assumption, now we must prove that, for <math>n+1</math> case, <math>1/2>\frac{1}{n+1}</math>, which is clearly true for <math>n>1</math>. So we are done.
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== 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>\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.
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== See Also ==
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{{USAMO box|year=1994|num-b=3|num-a=5}}
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[[Category:Olympiad Algebra Problems]]
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[[Category:Olympiad Inequality Problems]]
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{{MAA Notice}}

Latest revision as of 13:30, 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.

See Also

1994 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions

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