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

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==Problem 4==
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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>
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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>
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== Solution ==
 
Since each <math>a_{i}</math> is positive, by Muirhead's inequality,  
 
Since each <math>a_{i}</math> is positive, by Muirhead's inequality,  
 
<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>
 
<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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<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>.
 
<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>.
 
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.
 
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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== See Also ==

Revision as of 11:45, 12 April 2011

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

Since each $a_{i}$ is positive, by Muirhead's inequality, $2(\sum a_{i}^2) \ge (\sum a)^2 \ge n$. Now we claim that $\frac{n}{2}> frac{1}{4}(1+...\frac{1}{n)}$

$n=1$, giving $1/2>1/4$ works, but we set the base case $n=2$, which gives $1>3/8$. Now assume that it works for $n$. By our assumption, now we must prove that, for $n+1$ case, $1/2>\frac{1}{n+1}$, which is clearly true for $n>1$. So we are done.

See Also

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