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Difference between revisions of "2009 USAMO Problems/Problem 4"

(New page: == 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 <cmath> (a_1+a_2+ ... +a_n)\big( {1 \over a_1} + {1...)
 
(→‎Problem)
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<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>
 
<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>
 
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)</math> <math> \le </math> 4 min <math>(a_1, a_2, ... , a_n)</math>.
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 +
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== Solution ==
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Assume without loss of generality that <math>a_i \ge a_{i+1}</math> for all <math>1\lei\le n-1</math>. Now we seek to prove that <math>a_1 \le 4a_n</math>
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 +
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>
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<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>
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<cmath> n+ {1 \over 2} \ge n-2 + \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n}</cmath>
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<cmath>{5 \over 2} \ge \sqrt{a_n \over a_1} + \sqrt{a_1 \over a_n}</cmath>
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<cmath>{17 \over 4} \ge {a_n \over a_1} + {a_1 \over a_n}</cmath>
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<cmath> 0 \ge (a_1 - 4a_n)(a_1 - {a_n \over 4})</cmath>
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Since <math>a_1 \ge a_n</math>, clearly <math>(a_1 - {a_n \over 4}) > 0</math>,
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so dividing yields:
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<cmath>0 \ge (a_1 - 4a_n)</cmath>
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<math>4a_n \ge a_1</math> as desired.

Revision as of 09:55, 12 June 2009

Problem

For $n \ge 2$ let $a_1$, $a_2$, ..., $a_n$ be positive real numbers such that \[(a_1+a_2+ ... +a_n)\big( {1 \over a_1} + {1 \over a_2} + ... +{1 \over a_n} \big) \le \big(n+ {1 \over 2} \big) ^2\] Prove that max $(a_1, a_2, ... ,a_n)$ $\le$ 4 min $(a_1, a_2, ... , a_n)$.


Solution

Assume without loss of generality that $a_i \ge a_{i+1}$ for all $1\lei\le n-1$ (Error compiling LaTeX. Unknown error_msg). Now we seek to prove that $a_1 \le 4a_n$

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

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