1999 USAMO Problems/Problem 4: Difference between revisions
problem and solution |
I like pie (talk | contribs) mNo edit summary |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
Let <math>a_{1}, a_{2}, \dots, a_{n}</math> (<math>n > 3</math>) be real numbers such that | Let <math>a_{1}, a_{2}, \dots, a_{n}</math> (<math>n > 3</math>) be real numbers such that | ||
<cmath> | <cmath> | ||
| Line 11: | Line 10: | ||
== Solution == | == Solution == | ||
First, suppose all the <math>a_i</math> are positive. Then | First, suppose all the <math>a_i</math> are positive. Then | ||
<cmath> \max(a_1, \dotsc, a_n) \ge \sqrt{\frac{a_1^2 + | <cmath> \max(a_1, \dotsc, a_n) \ge \sqrt{\frac{a_1^2 + | ||
| Line 29: | Line 27: | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
Since <math>k<n</math>, <math>4(n-k) > 4</math>. It follows that <math>\max(a_1, \dotsc, a_n) \ge \sqrt{4} = 2</math>, as desired. <math>\blacksquare</math> | Since <math>k<n</math>, <math>4(n-k) > 4</math>. It follows that <math>\max(a_1, \dotsc, a_n) \ge \sqrt{4} = 2</math>, as desired. <math>\blacksquare</math> | ||
{{alternate solutions}} | {{alternate solutions}} | ||
== Resources == | == Resources == | ||
* [[1999 USAMO Problems]] | * [[1999 USAMO Problems]] | ||
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=63591#63591 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=63591#63591 Discussion on AoPS/MathLinks] | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
Revision as of 10:30, 24 April 2008
Problem
Let 
(
) be real numbers such that
![\[a_{1} + a_{2} + \cdots + a_{n} \geq n \qquad \mbox{and} \qquad a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \geq n^{2}.\]](http://latex.artofproblemsolving.com/b/4/e/b4e04b54c2d1f0dad88a502d7cd3313552e78b32.png)
Prove that 
.
Solution
First, suppose all the 
are positive. Then
![\[\max(a_1, \dotsc, a_n) \ge \sqrt{\frac{a_1^2 + \dotsb + a_n^2}{n}} \ge \sqrt{n} \ge 2 .\]](http://latex.artofproblemsolving.com/3/5/7/3577ed2f8c34f0c8fad982e812743fd964a1ca09.png)
Suppose, on the other hand, that without loss of generality,
![\[a_1 \ge a_2 \ge \dotsb \ge a_k \ge 0 > a_{k+1} \ge \dotsb \ge a_n,\]](http://latex.artofproblemsolving.com/e/e/0/ee078bb1a2ad809ed0ae9b32af1a26c713de7de6.png)
with 
. If 
we are done, so suppose that 
. Then
, so
![\[\sum_{i=k+1}^n -a_i \le 2k-n .\]](http://latex.artofproblemsolving.com/f/3/e/f3e5202c11b585e83d51b3a687e17520aa1f6255.png)
Since 
is a positive real for all 
, it follows that
\[\sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n} -a_i \right)^2 \le (2k-n)^2 .\] (Error compiling LaTeX. Unknown error_msg)
Then
Since 
, 
. It follows that 
, as desired. 
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
