Difference between revisions of "2003 IMO Problems/Problem 5"
(Created page with "==Problem== Let <math>n</math> be a positive integer and let <math>x_1 \le x_2 \le \cdots \le x_n</math> be real numbers. Prove that <cmath>\left( \sum_{i=1}^{n}\sum_{j=i}^{...") |
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==Solution== | ==Solution== | ||
{{solution}} | {{solution}} | ||
| + | have | ||
| + | \begin{align*}\left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2 &=\left(2\sum_{1\le i\le j\le n}(x_j-x_i)\right)^2 \\ | ||
| + | &= \left((2n-2)x_n+(2n-6)x_{n-1}+\dots +(2-2n)x_1\right)^2 \\ | ||
| + | &\le ((2n-2)^2+(2n-6)^2+(2n-10)^2+\dots + (2-2n)^2)(x_1^2+x_2^2+\dots + x_n^2) \\ | ||
| + | &= \frac{4(n-1)(n)(n+1)}{3}(x_1^2+x_2^2+\dots + x_n^2) \\ | ||
| + | &= \frac{2(n^2-1)}{3}\cdot 2(nx_1^2 + nx_2^2 + \dots + nx_n^2) \\ | ||
| + | &= \frac{2(n^2-1)}{3}\cdot 2\left((n-1)\left(\sum_{i=1}^{n}{x_i^2}\right) + \left(\sum_{i=1}^{n}{x_i}\right)^2 - 2\sum_{1\le i<j\le n}x_ix_j\right) \\ | ||
| + | &= \frac{2(n^2-1)}{3}\cdot 2\left(\sum_{1\le i<j\le n}(x_i-x_j)^2\right) \\ | ||
| + | &= \frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2 | ||
| + | \end{align*} | ||
==See Also== | ==See Also== | ||
{{IMO box|year=2003|num-b=4|num-a=6}} | {{IMO box|year=2003|num-b=4|num-a=6}} | ||
Latest revision as of 03:08, 26 March 2024
Problem
Let 
be a positive integer and let 
be real numbers. Prove that
![\[\left( \sum_{i=1}^{n}\sum_{j=i}^{n} |x_i-x_j|\right)^2 \le \frac{2(n^2-1)}{3}\sum_{i=1}^{n}\sum_{j=i}^{n}(x_i-x_j)^2\]](http://latex.artofproblemsolving.com/a/d/0/ad091920cb01959f69cc0f1b7ce863992acab98e.png)
with equality if and only if 
form an arithmetic sequence.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it. have \begin{align*}\left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2 &=\left(2\sum_{1\le i\le j\le n}(x_j-x_i)\right)^2 \\ &= \left((2n-2)x_n+(2n-6)x_{n-1}+\dots +(2-2n)x_1\right)^2 \\ &\le ((2n-2)^2+(2n-6)^2+(2n-10)^2+\dots + (2-2n)^2)(x_1^2+x_2^2+\dots + x_n^2) \\ &= \frac{4(n-1)(n)(n+1)}{3}(x_1^2+x_2^2+\dots + x_n^2) \\ &= \frac{2(n^2-1)}{3}\cdot 2(nx_1^2 + nx_2^2 + \dots + nx_n^2) \\ &= \frac{2(n^2-1)}{3}\cdot 2\left((n-1)\left(\sum_{i=1}^{n}{x_i^2}\right) + \left(\sum_{i=1}^{n}{x_i}\right)^2 - 2\sum_{1\le i<j\le n}x_ix_j\right) \\ &= \frac{2(n^2-1)}{3}\cdot 2\left(\sum_{1\le i<j\le n}(x_i-x_j)^2\right) \\ &= \frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2 \end{align*}
See Also
| 2003 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
