Difference between revisions of "1999 USAMO Problems/Problem 4"
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<cmath>\sum_{i=1}^k a_i \ge n - \sum_{i=k+1}^n a_i > n - 2(n-k) = 2k - n.</cmath> | <cmath>\sum_{i=1}^k a_i \ge n - \sum_{i=k+1}^n a_i > n - 2(n-k) = 2k - n.</cmath> | ||
Because <math>a_i \le 0</math> for <math>i \le k</math>, both sides of the inequality are non-positive, so squaring flips the sign. But we also know that <math>a_i a_j \ge 0</math> for <math>i, j \le k</math>, so | Because <math>a_i \le 0</math> for <math>i \le k</math>, both sides of the inequality are non-positive, so squaring flips the sign. But we also know that <math>a_i a_j \ge 0</math> for <math>i, j \le k</math>, so | ||
− | <cmath>\sum_{i= | + | <cmath>\sum_{i=1}^k a_i^2 \le \left(\sum_{i=1}^k a_i \right)^2 < (2k - n)^2 = 4k^2 - 4kn + n^2,</cmath> |
which results in | which results in | ||
− | <cmath>\sum_{i= | + | <cmath>\sum_{i=1}^n a_i^2 = \sum_{i=1}^k a_i^2 + \sum_{i=k+1}^n a_i^2 < 4k^2 - 4kn + n^2 + 4(n-k) = n^2 - 4(k-1)(n-k) \le n^2,</cmath> |
a contradiction to our given condition. The proof is complete. | a contradiction to our given condition. The proof is complete. | ||
Revision as of 16:29, 28 September 2014
Contents
Problem
Let (
) be real numbers such that
Prove that
.
Hint
Remember that the problem said real numbers, which can be negative!
Solution
First, suppose all the are positive. Then
Suppose, on the other hand, that without loss of generality,
with
. If
we are done, so suppose that
. Then
, so
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.
Solution 2
Assume the contrary and suppose each is less than 2.
Without loss of generality let , and let
be the largest integer such that
and
if it exists, or 0 if all the
are non-negative. If
, then (as
)
, a contradiction. Hence, assume
. Then
Because
for
, both sides of the inequality are non-positive, so squaring flips the sign. But we also know that
for
, so
which results in
a contradiction to our given condition. The proof is complete.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1999 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.