Difference between revisions of "2009 USAMO Problems/Problem 4"
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For <math>n \ge 2</math> let <math>a_1</math>, <math>a_2</math>, ..., <math>a_n</math> be positive real numbers such that | For <math>n \ge 2</math> let <math>a_1</math>, <math>a_2</math>, ..., <math>a_n</math> be positive real numbers such that | ||
<center><math> (a_1+a_2+ ... +a_n)\left( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \right) \le \left(n+ {1 \over 2} \right) ^2 </math></center> | <center><math> (a_1+a_2+ ... +a_n)\left( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \right) \le \left(n+ {1 \over 2} \right) ^2 </math></center> | ||
− | Prove that | + | Prove that <math>\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)</math>. |
== Solution == | == Solution == |
Revision as of 14:22, 27 February 2016
Problem
For let
,
, ...,
be positive real numbers such that

Prove that .
Solution
Assume without loss of generality that . Now we seek to prove that
.
By the Cauchy-Schwarz Inequality,
Since
, clearly
, dividing yields:
as desired.
See Also
2009 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.