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

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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 box|year=1994|num-b=3|num-a=5}}
 
{{USAMO box|year=1994|num-b=3|num-a=5}}

Revision as of 22:40, 13 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 $\, n \geq 1, \,$

\[\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).\]

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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