2003 IMO Problems/Problem 5

Problem

Let $n$ be a positive integer and let $x_1 \le x_2 \le \cdots \le x_n$ be real numbers. Prove that

$\left( \sum_{i=1}^{n}\sum_{j=i}^{n} |x_i-x_j|\right)^2 \le \frac{2(n^2-1)}{3}\sum_{i=1}^{n}\sum_{j=i}^{n}(x_i-x_j)^2$

with equality if and only if $x_1, x_2, ..., x_n$ form an arithmetic sequence.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. We first make use of symmetry to rewrite the inequality as $\left(\sum_{1\le i<j\le n}|x_i-x_j|\right)^2\le\frac{n^2-1}3\left(\sum_{1\le i<j\le n}|x_i-x_j|^2\right)$ . WLOG that $x_1\le x_2\le\dots\le x_n$ and let $x_{i-1}-x_i=a_i$ . The inequality is equivalent to $\left(\sum_{1\le i<j\le n}\left(a_i+\dots+a_{j-1}\right) \right)^2\le\frac{n^2-1}3\left(\sum_{1\le i<j\le n}\left(a_i+\dots+a_{j-1}\right)^2\right)$ for all $a_1,\dots,a_{n-1}$ . But this can be rewritten as $\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right) \right)^2\le\frac{n^2-1}3\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right)$ By Cauchy-Schwarz: \begin{align*} \left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right)\left(\sum_{l=1}^{n-1}\sum_{j-i=l}l^2\right)&\ge\left(\sum_{l=1}^{n-1}\sum_{j-i=l}l\left(a_i+\dots+a_j\right)\right)^2\\ &=\left(\sum_{l=1}^{n-1}l\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2 \end{align*}

We claim that $\sum_{j-i=l}(a_i+\dots+a_j)=\sum_{j-i=(n-l)}(a_i+\dots+a_j)$ . Indeed, we may consider the $l\times(n-l)$ matrix: \[ \left( \begin{array}{cccc} a_1 & a_2 & \dots & a_l \\ a_2 & a_3 & \dots & a_{l+1} \\ \vdots & \vdots & \ddots & \vdots\\ a_{n-l} & a_{n-l+1} & \dots & a_n \end{array} \right)\] The first sum corresponds to summing the matrix row by row, and the second corresponds to summing it column by column. Thus the two sums are equal, as claimed.

Hence: \begin{align*} \left(\sum_{l=1}^{n-1}l\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2&=\left(\sum_{l=1}^{n-1}\frac n2\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2\\ &=\frac{n^2}4\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2 \end{align*}

We may also check that $\sum_{l=1}^{n-1}\sum_{j-i=l}l^2=\sum_{l=1}^{n-1}(n-l)l^2=\frac{n^4-n^2}{12}$ . Thus we have proven that $\frac{n^4-n^2}{12}\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right)\ge\frac{n^2}4\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2$ Dividing $\frac{n^2}4$ yields $\frac{n^2-1}{3}\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right)\ge\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2$ as desired.

Furthermore, from Cauchy's equality condition, equality holds if and only if $a_1=a_2=\dots=a_{n-1}$ - that is, when the $x_i$ form an arithmetic sequence.

2003 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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