Difference between revisions of "1993 USAMO Problems/Problem 5"

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<center><math>\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}</math>.</center>
 
<center><math>\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}</math>.</center>
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==Solution==
  
 
== Resources ==
 
== Resources ==

Revision as of 20:55, 22 February 2012

Problem 4

Let $a_0, a_1, a_2,\cdots$ be a sequence of positive real numbers satisfying $a_{i-1}a_{i+1}\le a^2_i$ for $i = 1, 2, 3,\cdots$ . (Such a sequence is said to be log concave.) Show that for each $n > 1$,

$\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}$.

Solution

Resources

1993 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
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All USAMO Problems and Solutions
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