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1994 USAMO Problems/Problem 4: Difference between revisions

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Created page with '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+..…'
 
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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>


<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, which simplifies down to <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.

Revision as of 23:41, 11 April 2011

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.

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