Difference between revisions of "2006 IMO Shortlist Problems/A4"
(New page: == Problem == Prove the inequality <cmath> \sum_{i<j} \frac{a_ia_j}{a_i+a_j} \le \frac{n}{2(a_1 + a_2 + \dotsb a_n)} \sum_{i<j} a_i a_j </cmath> for positive real numbers <math>a_1, \dots...) |
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Note that | Note that | ||
− | <cmath> \sum_{i<j} \frac{a_ia_j}{a_i+a_j} = 1/2 \sum_{i\neq j} \frac{a_ia_j}{a_i + a_j} = 1/2 \sum_{ | + | <cmath> \sum_{i<j} \frac{a_ia_j}{a_i+a_j} = 1/2 \sum_{i\neq j} \frac{a_ia_j}{a_i + a_j} = 1/2 \sum_{j=1}^n \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} . </cmath> |
Suppose that <math>1 \le k \neq \ell \le n</math>. Note that <math>1/(1/a_i + 1/a_j)</math> is an increasing function of both <math>a_i</math> and <math>a_j</math>. It follows that if <math>a_k \le a_\ell</math>, then | Suppose that <math>1 \le k \neq \ell \le n</math>. Note that <math>1/(1/a_i + 1/a_j)</math> is an increasing function of both <math>a_i</math> and <math>a_j</math>. It follows that if <math>a_k \le a_\ell</math>, then | ||
<cmath> \sum_{j \neq k} \frac{1}{1/a_k + 1/a_j} \le \sum_{j\neq \ell} \frac{1}{a_\ell + a_j}, </cmath> | <cmath> \sum_{j \neq k} \frac{1}{1/a_k + 1/a_j} \le \sum_{j\neq \ell} \frac{1}{a_\ell + a_j}, </cmath> | ||
i.e., <math>\sum_{j\neq i} \frac{1}{1/a_i + 1/a_j}</math> is an increasing function of <math>j</math>. | i.e., <math>\sum_{j\neq i} \frac{1}{1/a_i + 1/a_j}</math> is an increasing function of <math>j</math>. | ||
+ | |||
+ | Since <math>\sum_{i\neq j}(a_j + a_i) = (n-2)a_j + \sum_{i=1}^n a_i</math> is also an increasing function of <math>j</math>, it follows from [[Chebyshev's Inequality]] that | ||
+ | <cmath> \frac{2n-2}{n} \sum_{j=1}^n a_j \cdot \sum_{j=1}^n \sum_{j\neq i} \frac{1}{1/a_i + 1/a_j} \le \sum_{j=1}^n \left[ \sum_{i\neq j} (a_i + a_j) \cdot \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} \right], </cmath> | ||
+ | or | ||
+ | <cmath> 1/2 \sum_{j=1}^n a_j \cdot \sum_{j=1}^n \sum_{j\neq i} \frac{1}{1/a_i + 1/a_j} \le \frac{n}{4(n-1)} \sum_{j=1}^n \left[ \sum_{i\neq j} (a_i + a_j) \cdot \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} \right] . </cmath> | ||
+ | Now, for fixed <math>j</math>, both <math>a_i + a_j</math> and <math>1/(1/a_i + 1/a_j)</math> are increasing functions of <math>a_i</math>. It follows again from Chebyshev's Inequality that | ||
+ | <cmath> \frac{1}{n-1} \sum_{i\neq j} (a_i + a_j) \sum_{i \neq j} \frac{1}{1/a_i + 1/a_j} \le \sum_{i\neq j} \frac{a_i+a_j}{1/a_i + 1/a_j} = \sum_{i\neq j} a_i a_j, </cmath> | ||
+ | or | ||
+ | <cmath> \sum_{i\neq j} (a_i + a_j) \sum_{i \neq j} \frac{1}{1/a_i + 1/a_j} \le (n-1) \sum_{i\neq j} a_ia_j, </cmath> | ||
+ | which in sum becomes | ||
+ | <cmath> \frac{n}{4(n-1)} \sum_{j=1}^n \left[ \sum_{i\neq j}(a_i + a_j) \sum_{i \neq j} \frac{1}{1/a_i + 1/a_j} \right] \le \frac{n}{4} \sum_{j=1}^n \sum_{i \neq j} a_i a_j = \frac{n}{2} \sum_{i<j} a_i a_j .</cmath> | ||
+ | If we denote <math>\sum_{i=1}^n a_i =S</math>, then in summary, we thus have | ||
+ | <cmath> \sum_{i < j} \frac{a_ia_j}{a_i + a_j} = (1/S) \cdot S/2 \cdot \sum_{j=1}^n \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} \le n/(2S) \cdot \sum_{i<j} a_i a_j, </cmath> | ||
+ | as desired. <math>\blacksquare</math> | ||
+ | |||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | == Resources == | ||
+ | |||
+ | * [[2006 IMO Shortlist Problems]] | ||
+ | * [http://www.mathlinks.ro/Forum/viewtopic.php?p=874975#p874975 Discussion on AoPS/MathLinks] | ||
+ | |||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] |
Latest revision as of 12:12, 29 December 2007
Problem
Prove the inequality for positive real numbers .
Solution
Note that Suppose that . Note that is an increasing function of both and . It follows that if , then i.e., is an increasing function of .
Since is also an increasing function of , it follows from Chebyshev's Inequality that or Now, for fixed , both and are increasing functions of . It follows again from Chebyshev's Inequality that or which in sum becomes If we denote , then in summary, we thus have as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.