Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2003 USAMO Problems/Problem 5"

m (→‎Problem)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
solution by paladin8:
 +
 
 +
WLOG, assume <math>a + b + c = 3</math>.
 +
 
 +
Then the LHS becomes <math>\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} = \sum \frac {a^2 + 6a + 9}{3a^2 - 6a + 9} = \sum \left(\frac {1}{3} + \frac {8a + 6}{3a^2 - 6a + 9}\right)</math>.
 +
 
 +
Notice <math>3a^2 - 6a + 9 = 3(a - 1)^2 + 6 \ge 6</math>, so <math>\frac {8a + 6}{3a^2 - 6a + 9} \le \frac {8a + 6}{6}</math>.
 +
 
 +
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.
  
 
== Resources ==
 
== Resources ==
Line 10: Line 18:
 
* [[2003 USAMO Problems]]
 
* [[2003 USAMO Problems]]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53672 Discussion on AOPS/MathLinks]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53672 Discussion on AOPS/MathLinks]
 +
* [http://www.mathlinks.ro/Forum/viewtopic.php?t=46 Discussion 2]
 +
* [http://www.mathlinks.ro/Forum/viewtopic.php?t=28678 Discussion 3]
 +
* [http://www.mathlinks.ro/Forum/viewtopic.php?t=43278 Discussion 4]
 +
* [http://www.mathlinks.ro/Forum/viewtopic.php?t=48989 Discussion 5]
  
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]

Revision as of 18:58, 18 December 2008

Problem

Let $a$, $b$, $c$ be positive real numbers. Prove that

$\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.$

Solution

solution by paladin8:

WLOG, assume $a + b + c = 3$.

Then the LHS becomes $\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} = \sum \frac {a^2 + 6a + 9}{3a^2 - 6a + 9} = \sum \left(\frac {1}{3} + \frac {8a + 6}{3a^2 - 6a + 9}\right)$.

Notice $3a^2 - 6a + 9 = 3(a - 1)^2 + 6 \ge 6$, so $\frac {8a + 6}{3a^2 - 6a + 9} \le \frac {8a + 6}{6}$.

So $\sum \frac {(a + 3)^2}{2a^2 + (3 - a)^2} \le \sum \left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8$ as desired.

Resources

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_USAMO_Problems/Problem_5&oldid=28633"