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 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>. | ||
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as desired. | as desired. | ||
+ | |||
+ | == Solution 2== | ||
+ | Assume without loss of generality that <math>a_1 \geq a_2 \geq \cdots \geq a_n</math>. | ||
+ | Using the Cauchy–Bunyakovsky–Schwarz inequality and the inequality given, <cmath>\begin{align*} | ||
+ | &\left(a_1+a_2 + ... +a_n + 3a_n -\frac{3a_1}{4}\right)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) \\ | ||
+ | =&\left(\frac{a_1}{4}+a_2 + ... +a_{n-1}+4a_n\right)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) \\ | ||
+ | \ge& \left(\frac{1}{2}+n-2+2\right)^2 \\ | ||
+ | =&\left(n+\frac{1}{2}\right)^2 \\ | ||
+ | \ge& (a_1+a_2 + ... +a_{n})\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right).\end{align*}</cmath> | ||
+ | (Note that <math>n-2 \ge 0</math> since <math>n \ge 2</math> as given!) | ||
+ | This implies that <math>3a_n -\frac{3a_1}{4} \ge 0 \iff 4a_n \ge a_1</math> as desired. | ||
+ | |||
+ | ~Deng Tianle, username: Leole | ||
== See Also == | == See Also == | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:50, 14 July 2024
Contents
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.
Solution 2
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.
~Deng Tianle, username: Leole
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.