Difference between revisions of "1993 USAMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
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{{USAMO box|year=1993|num-b=4|after=Last Problem}} | {{USAMO box|year=1993|num-b=4|after=Last Problem}} | ||
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks] | ||
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+ | [[Category:Olympiad Inequality Problems]] |
Revision as of 10:53, 17 September 2012
Problem 5
Let be a sequence of positive real numbers satisfying
for
. (Such a sequence is said to be log concave.) Show that for
each
,

Solution
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See Also
1993 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |