2003 IMO Problem 5

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2003 IMO Problem 5
Let all $x_i$ are real for all $i=1,2,...,n$ and $x_i \geq x_j$ if and only if $i\geq j$
(a)
Prove that $\left ( \sum_{i,j=1}^{n}\left  x_ix_j \right  \right )^2\leq \frac{2(n^21)}{3}\sum_{i,j=1}^{n}\left ( x_ix_j \right )^2$
(b)
Prove that equality holds if and only if $\{x_i\}$ is an arithmetic progression .
Comment: 100th post
(a)
Prove that $\left ( \sum_{i,j=1}^{n}\left  x_ix_j \right  \right )^2\leq \frac{2(n^21)}{3}\sum_{i,j=1}^{n}\left ( x_ix_j \right )^2$
(b)
Prove that equality holds if and only if $\{x_i\}$ is an arithmetic progression .
Comment: 100th post
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )

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Re: 2003 IMO Problem 5
I'm also posting my solution. If there is
any bug; please inform me.
any bug; please inform me.
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
Re: 2003 IMO Problem 5
Well, the problem statement says , equality appears when $\{x_i\}$ is an arithmetic progression so I think that the first move should consist of Cauchy Schwarz . Though , the solution here appears alright, and so congratulations for completing such a tedious one!
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Nur Muhammad Shafiullah  Mahi
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Re: 2003 IMO Problem 5
Tell me Mahi, how can you assume using any specific kind of tactics at the very beginning? Please give us an example. And this question is for all who have a lot of experience in solving problems. Let's discuss and restart to solve this problem with our discussion.It'll help us all to learn choosing tactics. (I'm still a learner of solving problems )*Mahi* wrote:Well, the problem statement says , equality appears when $\{x_i\}$ is an arithmetic progression so I think that the first move should consist of Cauchy Schwarz .
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
 FahimFerdous
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Re: 2003 IMO Problem 5
I completely agree with Sourav. So Mahi, please tell how you started thinking about the first step.
Your hot head might dominate your good heart!

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Re: 2003 IMO Problem 5
Is it because of CauchySchwarz equality condition has similarity with the difference between two elements of an arithmetic progression?
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
Re: 2003 IMO Problem 5
You got it, all I tried was multiplying different arithmetic progression with $\sum_{i,j=1}^{n}\left (  x_ix_j  \right )^2$, first with easy ones like $\sum 1$ and then others, which you can find out. And trying well known techniques sometimes has it's benefits too.
I'm a learner too , so we should all try together!
I'm a learner too , so we should all try together!
Last edited by *Mahi* on Fri Oct 28, 2011 9:49 pm, edited 1 time in total.
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Re: 2003 IMO Problem 5
Mahi, Fahim, any progress? For using CauchySchwarz inequality we need product of two sum of square sequences. But since in our left side, there are terms which are not square. Confusing....
You spin my head right round right round,
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......($from$ "$THE$ $UGLY$ $TRUTH$" )
Re: 2003 IMO Problem 5
In CauchySchwarz LHS has two sum of square sequences, right? And remember that the one greater should be the LHS here.sourav das wrote:Mahi, Fahim, any progress? For using CauchySchwarz inequality we need product of two sum of square sequences. But since in our left side, there are terms which are not square. Confusing....
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 zadid xcalibured
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Re: 2003 IMO Problem 5
By cauchy schawrtz inequality(£xixj)^2<n(n1)/2{£(xixj)^2}all we need to prove is that n(n1)/2<2/3(n^21).this is true because n^2+3n4>0.here £ means sigma.how can this solution be so simple?