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

Revision as of 10:02, 18 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 $1\le i \le n-1$ . 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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 == Solution ==
-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>
+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>
 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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Difference between revisions of "2009 USAMO Problems/Problem 4"

Revision as of 10:02, 18 June 2009

Problem

Solution