Difference between revisions of "2009 USAMO Problems/Problem 4"
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Prove that <math>\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)</math>. | Prove that <math>\text{max}(a_1, a_2, ... ,a_n) \le 4 \text{min}(a_1, a_2, ... , a_n)</math>. | ||
− | == Solution == | + | == Solution 1== |
Assume without loss of generality that <math>a_1 \geq a_2 \geq \cdots \geq a_n</math>. Now we seek to prove that <math>a_1 \le 4a_n</math>. | Assume without loss of generality that <math>a_1 \geq a_2 \geq \cdots \geq a_n</math>. Now we seek to prove that <math>a_1 \le 4a_n</math>. | ||
Revision as of 00:48, 14 July 2024
Problem
For let
,
, ...,
be positive real numbers such that
![$(a_1+a_2+ ... +a_n)\left( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \right) \le \left(n+ {1 \over 2} \right) ^2$](http://latex.artofproblemsolving.com/5/9/6/5963cac8c96d5f27b431ca50755a0e1b60ed69c1.png)
Prove that .
Solution 1
Assume without loss of generality that . Now we seek to prove that
.
By the Cauchy-Schwarz Inequality,
Since
, clearly
, dividing yields:
as desired.
Alternative Solution (by Deng Tianle, username: Leole)
Assume without loss of generality that .
Using the Cauchy–Bunyakovsky–Schwarz inequality and the inequality given,
(Note that
since
as given!)
This implies that
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.