Difference between revisions of "2003 IMO Problems/Problem 5"

Latest revision as of 04:25, 7 October 2025

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

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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 \\ &= \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*}

@@ Line 9: / Line 9: @@
 ==Solution==
 {{solution}}
-\[
-\text{We first make use of symmetry to rewrite the inequality as}
+\begin{align*}\left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2 &=\left(2\sum_{1\le i\le j\le n}(x_j-x_i)\right)^2 \\
-\]
+&= \left((2n-2)x_n+(2n-6)x_{n-1}+\dots +(2-2n)x_1\right)^2 \\
-\[
+&\le ((2n-2)^2+(2n-6)^2+(2n-10)^2+\dots + (2-2n)^2)(x_1^2+x_2^2+\dots + x_n^2) \\
-\left(\sum_{1\le i<j\le n}|x_i-x_j|\right)^2\le\frac{n^2-1}3<br="">\left(\sum_{1\le i<j\le n}|x_i-x_j|^2\right)<br="">\]
+&= \frac{4(n-1)(n)(n+1)}{3}(x_1^2+x_2^2+\dots + x_n^2) \\
-\[
+&= \frac{2(n^2-1)}{3}\cdot 2(nx_1^2 + nx_2^2 + \dots + nx_n^2) \\
-\text{WLOG that } x_1\le x_2\le\dots\le x_n \text{ and let } x_{i-1}-x_i=a_i.
+&= \frac{2(n^2-1)}{3}\cdot 2\left((n-1)\left(\sum_{i=1}^{n}{x_i^2}\right) + \left(\sum_{i=1}^{n}{x_i}\right)^2 - 2\sum_{1\le i<j\le n}x_ix_j\right) \\
-\]
+&= \frac{2(n^2-1)}{3}\cdot 2\left(\sum_{1\le i<j\le n}(x_i-x_j)^2\right) \\
-\[
+&= \frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2
-\text{The inequality is equivalent to}
+\end{align*}
-\]
-\[
-\left(\sum_{1\le i<j\le n}\left(a_i+\dots+a_{j-1}\right)="" \right)^2\le\frac{n^2-1}3<br="">\left(\sum_{1\le i<j\le n}\left(a_i+\dots+a_{j-1}\right)^2\right)="" <br="">\]
-\[
-\text{for all } a_1,\dots,a_{n-1}. \text{But this can be rewritten as}
-\]
-\[
-\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right) \right)^2\le\frac{n^2-1}3
-\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right)
-\]
-\text{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)&amp;\ge\left(\sum_{l=1}^{n-1}\sum_{j-i=l}l\left(a_i+\dots+a_j\right)\right)^2\\ &amp;=\left(\sum_{l=1}^{n-1}l\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2
-\end{align*}
-\text{We claim that }
-\[
-\sum_{j-i=l}(a_i+\dots+a_j)=\sum_{j-i=(n-l)}(a_i+\dots+a_j).
-\]
-\text{Indeed, we may consider the } <math>l\times(n-l)</math> \text{ matrix:}
-\[
-\left( \begin{array}{cccc}
-a_1 &amp; a_2 &amp; \dots &amp; a_l \\
-a_2 &amp; a_3 &amp; \dots &amp; a_{l+1} \\
-\vdots &amp; \vdots &amp; \ddots &amp; \vdots\\
-a_{n-l} &amp; a_{n-l+1} &amp; \dots &amp; a_n
-\end{array} \right)
-\]
-\text{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.}
-\text{Hence:}
-\begin{align*}
-\left(\sum_{l=1}^{n-1}l\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2&amp;=\left(\sum_{l=1}^{n-1}\frac n2\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2\\ &amp;=\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*}
-\text{We may also check that }
-\[
-\sum_{l=1}^{n-1}\sum_{j-i=l}l^2=\sum_{l=1}^{n-1}(n-l)l^2=\frac{n^4-n^2}{12}.
-\]
-\text{Thus we have proven that }
-\[
-\frac{n^4-n^2}{12}\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_{j}\right)^2\right)\ge\frac{n^2}4\left(\sum_{l=1}^{n-1}\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2
-\]</j\le></j\le></j\le></j\le>
 ==See Also==
 {{IMO box|year=2003|num-b=4|num-a=6}}

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Difference between revisions of "2003 IMO Problems/Problem 5"

Latest revision as of 04:25, 7 October 2025

Problem

Solution

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