Difference between revisions of "2009 USAMO Problems/Problem 4"
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− | Assume without loss of generality that <math>a_i \ge a_{i+1}</math> for all <math>1\ | + | 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> | 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> |
Revision as of 11:02, 18 June 2009
Problem
For let
,
, ...,
be positive real numbers such that
Prove that max
4 min
.
Solution
Assume without loss of generality that for all
. Now we seek to prove that
Now by Cauchy-Schwartz,
Since
, clearly
,
so dividing yields:
as desired.