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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}^{...")
 
Line 9: Line 9:
 
==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
 +
We first make use of symmetry to rewrite the inequality as
 +
<cmath>\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)</cmath>.
 +
WLOG that <math>x_1\le x_2\le\dots\le x_n</math> and let <math>x_{i-1}-x_i=a_i</math>. The inequality is equivalent to
 +
<cmath>\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)</cmath>for all <math>a_1,\dots,a_{n-1}</math>.
 +
But this can be rewritten as
 +
<cmath>\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)</cmath>
 +
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
 +
<cmath>\sum_{j-i=l}(a_i+\dots+a_j)=\sum_{j-i=(n-l)}(a_i+\dots+a_j)</cmath>. Indeed, we may consider the <math>l\times(n-l)</math> 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
 +
<cmath>\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}</cmath>. Thus we have proven that
 +
<cmath>\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</cmath>
 +
Dividing <math>\frac{n^2}4</math> yields
 +
<cmath>\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</cmath>as desired.
 +
 +
Furthermore, from Cauchy's equality condition, equality holds if and only if <math>a_1=a_2=\dots=a_{n-1}</math> - that is, when the <math>x_i</math> form an arithmetic sequence.
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=2003|num-b=4|num-a=6}}
 
{{IMO box|year=2003|num-b=4|num-a=6}}

Revision as of 14:26, 3 December 2023

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.

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
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