Difference between revisions of "1999 USAMO Problems/Problem 4"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
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 + | ||
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<cmath> \sum_{i=k+1}^n -a_i \le 2k-n . </cmath> | <cmath> \sum_{i=k+1}^n -a_i \le 2k-n . </cmath> | ||
Since <math>-a_i</math> is a positive real for all <math>k+1 \le i \le n</math>, it follows that | Since <math>-a_i</math> is a positive real for all <math>k+1 \le i \le n</math>, it follows that | ||
− | <cmath> \sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n | + | <cmath> \sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n-a_i \right)^2 \le (2k-n)^2 . </cmath> |
Then | Then | ||
<cmath> \begin{align*} | <cmath> \begin{align*} | ||
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\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> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Assume the contrary and suppose each <math>a_i</math> is less than 2. | ||
+ | |||
+ | Without loss of generality let <math>a_1 \le a_2 \le a_3 \le ... \le a_n < 2</math>, and let <math>k</math> be the largest integer such that <math>a_k \le 0</math> and <math>0 \le a_{k+1}</math> if it exists, or 0 if all the <math>a_i</math> are non-negative. If <math>k = 0</math>, then (as <math>n \ge 4</math>) <math>\sum_{i=1}^n a_i^2 < \sum_{i=1}^n 4 = 4n \le n^2</math>, a contradiction. Hence, assume <math>n \ge k \ge 1</math>. Then | ||
+ | <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 | ||
+ | <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 | ||
+ | <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. | ||
{{alternate solutions}} | {{alternate solutions}} | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
[[Category:Olympiad Inequality Problems]] | [[Category:Olympiad Inequality Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 09:47, 20 July 2016
Problem
Let (
) be real numbers such that
Prove that
.
Solution
Solution 1
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
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.