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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>   <cmath>\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).</cmath> 
− 
 
−  == Solution ==
 
−  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>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.
 
   
 == See Also ==   == See Also == 
 {{USAMO boxyear=1994numb=3numa=5}}   {{USAMO boxyear=1994numb=3numa=5}} 
Revision as of 22:40, 13 April 2011