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2009 USAMO Problems/Problem 4: Difference between revisions

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== Problem ==
== Problem ==
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  
<cmath> (a_1+a_2+ ... +a_n)\big( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \big) \le \big(n+ {1 \over 2} \big) ^2 </cmath>
<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 max <math>(a_1, a_2, ... ,a_n)</math> <math> \le </math> 4 min <math>(a_1, a_2, ... , a_n)</math>.
Prove that max <math>(a_1, a_2, ... ,a_n) \le 4 \text{min}\, (a_1, a_2, ... , a_n)</math>.


== Solution ==
Assume without loss of generality that <math>a_1 \le a_2 \le \cdots \le a_n</math>. Now we seek to prove that <math>a_1 \le 4a_n</math>.


== Solution ==
By the [[Cauchy-Schwarz Inequality]], <cmath>\begin{align*}
Assume without loss of generality that <math>a_i \ge a_{i+1}</math> for all <math>1\le i \le n-1</math>. Now we seek to prove that <math>a_1 \le 4a_n</math>
(a_n+a_2+ a_3 + ... +a_{n-1}+a_1)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) &\ge \left( \sqrt{a_n \over a_1} + n-2 + \sqrt{a_1 \over a_n} \right)^2 \\
(n+ {1 \over 2})^2 &\ge \left( \sqrt{a_n \over a_1} + n-2 + \sqrt{a_1 \over a_n} \right)^2 \\
n+ {1 \over 2} &\ge n-2 + \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n} \\
{5 \over 2} &\ge \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n} \\
{17 \over 4} &\ge {a_n \over a_1} + {a_1 \over a_n} \\
0 &\ge (a_1 - 4a_n)\left(a_1 - {a_n \over 4}\right) \end{align*}</cmath>
Since <math>a_1 \ge a_n</math>, clearly <math>(a_1 - {a_n \over 4}) > 0</math>, dividing yields:
 
<cmath>0 \ge (a_1 - 4a_n) \Longrightarrow 4a_n \ge a_1</cmath>  
 
as desired.
 
== See Also ==
{{USAMO newbox|year=2009|num-b=3|num-a=5}}


Now by Cauchy-Schwartz, <cmath>(a_n+a_2+ a_3 + ... +a_{n-1}+a_1)({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}) \ge \big( \sqrt{a_n \over a_1} + n-2 + \sqrt{a_1 \over a_n} \big)^2</cmath>
[[Category:Olympiad Algebra Problems]]
<cmath>(n+ {1 \over 2})^2 \ge \big( \sqrt{a_n \over a_1} + n-2 + \sqrt{a_1 \over a_n} \big)^2 </cmath>
<cmath> n+ {1 \over 2} \ge n-2 + \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n}</cmath>
<cmath>{5 \over 2} \ge \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n}</cmath>
<cmath>{17 \over 4} \ge {a_n \over a_1} + {a_1 \over a_n}</cmath>
<cmath> 0 \ge (a_1 - 4a_n)(a_1 - {a_n \over 4})</cmath>
Since <math>a_1 \ge a_n</math>, clearly <math>(a_1 - {a_n \over 4}) > 0</math>,
so dividing yields:
<cmath>0 \ge (a_1 - 4a_n)</cmath>
<math>4a_n \ge a_1</math> as desired.

Revision as of 11:23, 18 July 2009

Problem

For $n \ge 2$ let $a_1$, $a_2$, ..., $a_n$ be positive real numbers such that

$(a_1+a_2+ ... +a_n)\left( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \right) \le \left(n+ {1 \over 2} \right) ^2$

Prove that max $(a_1, a_2, ... ,a_n) \le 4 \text{min}\, (a_1, a_2, ... , a_n)$.

Solution

Assume without loss of generality that $a_1 \le a_2 \le \cdots \le a_n$. Now we seek to prove that $a_1 \le 4a_n$.

By the Cauchy-Schwarz Inequality, \begin{align*} (a_n+a_2+ a_3 + ... +a_{n-1}+a_1)\left({1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n}\right) &\ge \left( \sqrt{a_n \over a_1} + n-2 + \sqrt{a_1 \over a_n} \right)^2 \\ (n+ {1 \over 2})^2 &\ge \left( \sqrt{a_n \over a_1} + n-2 + \sqrt{a_1 \over a_n} \right)^2 \\ n+ {1 \over 2} &\ge n-2 + \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n} \\ {5 \over 2} &\ge \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n} \\ {17 \over 4} &\ge {a_n \over a_1} + {a_1 \over a_n} \\ 0 &\ge (a_1 - 4a_n)\left(a_1 - {a_n \over 4}\right) \end{align*} Since $a_1 \ge a_n$, clearly $(a_1 - {a_n \over 4}) > 0$, dividing yields:

\[0 \ge (a_1 - 4a_n) \Longrightarrow 4a_n \ge a_1\]

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
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