Difference between revisions of "2003 USAMO Problems/Problem 5"
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So <math>\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} \le \sum \left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8</math>, as desired. | So <math>\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} \le \sum \left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8</math>, as desired. | ||
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| − | + | == See also == | |
| − | + | {{USAMO newbox|year=2003|num-b=4|num-a=6}} | |
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
Revision as of 20:52, 6 April 2013
Problem
Let 
, 
, 
be positive real numbers. Prove that
Solution
Since all terms are homogeneous, we may assume WLOG that 
.
Then the LHS becomes 
.
Notice 
, so 
.
So 
, as desired.
See also
| 2003 USAMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
