Difference between revisions of "2003 IMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
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\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 \\ | \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 \\ | &= \left((2n-2)x_n+(2n-6)x_{n-1}+\dots +(2-2n)x_1\right)^2 \\ | ||
Latest revision as of 04:25, 7 October 2025
Problem
Let
be a positive integer and let
be real numbers. Prove that
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.
\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 | ||