## A proposed problem of IMO

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A proposed problem of IMO
If $c_{1},c_{2},....,c_{n}$ are real numbers ($n>=2$) such that
$(n-1)(c_{1}^{2}+c_{2}^{2}+.....+c_{n}^{2})=(c_{1}+c_{2}+.....+c_{n})^{2}$
Show that either all of them are non-negative,or all of them are non positive.[Proposed by czechoslovakia in 1977, was unused]
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### Re: A proposed problem of IMO

A nice place to use induction.
My Proof :
Last edited by Phlembac Adib Hasan on Fri Jan 27, 2012 9:55 am, edited 1 time in total.
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sourav das
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### Re: A proposed problem of IMO

Adib, try to find bug in your solution (if it has any) and reply why you are right or wrong...

Solution:
For the sake of contradiction let $c_{i}c_j<0$. So it follows that $\sqrt{c_{i}^2+c_{j}^2}>|c_{i}+c_{j}|$

Now using Cauchy-Schwartz's inequality for $n-1$ terms:

$(n-1)(\sum^{n-2}c_{k}^2 + c_{i}^2+c_{j}^2)\geq (\pm(\sum^{n-2}c_{k}) + \sqrt{c_{i}^2+c_{j}^2})^2$
notice that either, $(\sum^{n-2}c_{k})+|c_{i}+c_{j}|\geq |\sum c_i|$ or $-(\sum^{n-2}c_{k})+|c_{i}+c_{j}|\geq |\sum c_i|$
As, $\sqrt{c_{i}^2+c_{j}^2}>|c_{i}+c_{j}|$
$(n-1)(\sum^{n-2}c_{k}^2 + c_{i}^2+c_{j}^2)>( \sum c_i)^2$

So our assumption was wrong.
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### Re: A proposed problem of IMO

Yes.There is.But it is not a bug of my proof.I'll say it "typing mistake"-I wrote $(n-1)$ instead of $k$ and $(k-1)$ unmindfully.Now it is edited.
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### Re: A proposed problem of IMO

Phlembac Adib Hasan wrote:A nice place to use induction.
My Proof :
Ok, then explain me, in your induction step you assume that IF$(k-1) \sum_{p=1}^k c_p^2=(\sum_{p=1}^k c_p)^2$.....(i) holds ONLY THEN All $C_i$ satisfy the condition. But in second step
$k \sum_{p=1}^{k+1} c_p^2=(\sum_{p=1}^{k+1} c_p)^2$....... (ii) doesn't need to satisfy (i). So you can't say "As all $c_p$s(without $c_{k+1}$) are either non-negative or non-positive".
You spin my head right round right round,
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### Re: A proposed problem of IMO

No.No.You may misunderstand (ii).I used same $c_p$s in (ii), as in (i).I only added such a number $c_{k+1}$ in $c_p$s so that they fulfills (ii).I defined (i) before.So I can say all $c_p$s (without $c_{p+1}$) are of same sign.I only showed if the statement is true for $k$ variables, it will also be true for $k+1$ variables.
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sourav das
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### Re: A proposed problem of IMO

Phlembac Adib Hasan wrote:No.No.You may misunderstand (ii).I used same $c_p$s in (ii), as in (i).I only added such a number $c_{k+1}$ in $c_p$s so that they fulfills (ii).I defined (i) before.So I can say all $c_p$s (without $c_{p+1}$) are of same sign.I only showed if the statement is true for $k$ variables, it will also be true for $k+1$ variables.
Actually, I still don't understand. (I'm very dull in understanding) . Just answer the following questions:
(i)In your induction you assume that for any $k$ variables that satisfies the given condition must be all same in sign. Next you take a set of {$c_1,c_2.....c_k$} that satisfy the given condition(***). Next you work with adding a $c_{k+1}$ (So that {$c_1,c_2.....c_k,c_{k+1}$ also satisfy the condition})and showing $c_{k+1}$ also same sign as others,that's why you demand that for any $k+1$ element it'll be true. Am i right?

(ii)If you agree with (i) then tell me one thing. It may possible that we can find a set A={$c_1,c_2....c_{k+1}$}
such that it satisfy the given condition but any $k$ element subset of A doesn't satisfy the given condition. So your induction clearly doesn't work for those type $k+1$ element sets.

Actually i think you are choosing the value of variables.
You spin my head right round right round,
When you go down, when you go down down......
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### Re: A proposed problem of IMO

No, you may missed something.Just look at my definition of $c_{k+1}$.
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Corei13
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### Re: A proposed problem of IMO

Your Induction hypothesis is not enough to prove the problem, it just proves a case of the given problem.

Note that, the problem asked you to prove for all $c_i$ with only one condition $(n-1)\sum_{i\le n}{c_i^2} = \left ( \sum_{i\le n}{c_i} \right )^2$. That is, you can't add a extra constraint $(n-2)\sum_{i<n}{c_i^2} = \left ( \sum_{i < n}{c_i} \right )^2$ unless you prove that, $(n-1)\sum_{i\le n}{c_i^2} = \left ( \sum_{i\le n}{c_i} \right )^2 \Longrightarrow (n-2)\sum_{i<n}{c_i^2} = \left ( \sum_{i < n}{c_i} \right )^2$(which is actually not true). But you added this extra constraint with your hypothesis. So, you just proved a case.
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