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Difference between revisions of "2003 IMO Problems/Problem 5"
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==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
−
We first make use of symmetry to rewrite the inequality as
+
have
−
<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>.
+
\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 \\
−
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
+
&= \left((2n-2)x_n+(2n-6)x_{n-1}+\dots +(2-2n)x_1\right)^2 \\
−
<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>.
+
&\le ((2n-2)^2+(2n-6)^2+(2n-10)^2+\dots + (2-2n)^2)(x_1^2+x_2^2+\dots + x_n^2) \\
−
But this can be rewritten as
+
&= \frac{4(n-1)(n)(n+1)}{3}(x_1^2+x_2^2+\dots + x_n^2) \\
−
<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>
+
&= \frac{2(n^2-1)}{3}\cdot 2(nx_1^2 + nx_2^2 + \dots + nx_n^2) \\
−
By Cauchy-Schwarz:
+
&= \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) \\
−
\begin{align*}
+
&= \frac{2(n^2-1)}{3}\cdot 2\left(\sum_{1\le i<j\le n}(x_i-x_j)^2\right) \\
−
\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\\
+
&= \frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2
−
&=\left(\sum_{l=1}^{n-1}l\sum_{j-i=l}\left(a_i+\dots+a_j\right)\right)^2
 
 
\end{align*}
 
\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}}

Latest revision as of 04:08, 26 March 2024

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. 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
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