A proposed problem of IMO

[SANZEED]

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]

[Phlembac Adib Hasan]

A nice place to use induction.
My Proof :

Surely it's true for $n=2,3$.Now let it is true for $n=k$.Means \[(k-1) \sum_{p=1}^k c_p^2=(\sum_{p=1}^k c_p)^2 \]
We have to prove it for $n=k+1$.
Let a new number be $c_{k+1} $ so that it follows $k \sum_{p=1}^{k+1} c_p^2=(\sum_{p=1}^{k+1} c_p)^2$.As all $c_p$s(without $c_{k+1}$) are either non-negative or non-positive, .So if $c_{k+1}$ is of opposite sign, the inequality follows: \[ k\sum_{p=1}^{k+1}c_p^2\geq \left ( k-1 \right )\sum_{p=1}^{k}c_p^2=\left ( \sum_{p=1}^{k}c_p \right )^2\geq \left ( \sum_{p=1}^{k+1}c_p \right )^2\] with equality iff $c_{k+1}=0$.So our proof is complete.

[sourav das]

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.

[Phlembac Adib Hasan]

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.

[sourav das]

A nice place to use induction.
My Proof :

Surely it's true for $n=2,3$.Now let it is true for $n=k$.Means \[(k-1) \sum_{p=1}^k c_p^2=(\sum_{p=1}^k c_p)^2 \]
We have to prove it for $n=k+1$.
Let a new number be $c_{k+1} $ so that it follows $k \sum_{p=1}^{k+1} c_p^2=(\sum_{p=1}^{k+1} c_p)^2$.As all $c_p$s(without $c_{k+1}$) are either non-negative or non-positive, .So if $c_{k+1}$ is of opposite sign, the inequality follows: \[ k\sum_{p=1}^{k+1}c_p^2\geq \left ( k-1 \right )\sum_{p=1}^{k}c_p^2=\left ( \sum_{p=1}^{k}c_p \right )^2\geq \left ( \sum_{p=1}^{k+1}c_p \right )^2\] with equality iff $c_{k+1}=0$.So our proof is complete.

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".

[Phlembac Adib Hasan]

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.

[sourav das]

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.

[Phlembac Adib Hasan]

No, you may missed something.Just look at my definition of $c_{k+1}$.

[Corei13]

@Adib:
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.

[Phlembac Adib Hasan]

Ok,I want to end all argument.That proof has been withdrawn.Here I'm giving another proof.

Let $a_1,a_2,...,a_m,-a_{m+1},-a_{m+2},...,-a_n\; \; \; \; \; $ be a permutation of $c_1,c_2,...,c_n$ with all non-negative $a_i$s (and all of them are not equal to $0$).Now I'll show they can cot follow the given condition.(A proof using contradiction).Let they follow given condition: \[(n-1) \left ( \sum_{p=1}^{n} a_p^2\right )=\left ( \sum_{p=1}^n a_p \right )^2\]
\[\Rightarrow (n-1) \left ( \sum_{p=1}^{m} a_p^2\right )+(n-1) \left ( \sum_{p=m+1}^{n} a_p^2\right )=\left ( \sum_{p=1}^m a_p- \sum_{p=m+1}^n a_p\right )^2\]
\[\Rightarrow (n-1) \left ( \sum_{p=1}^{m} a_p^2\right )+(n-1) \left ( \sum_{p=m+1}^{n} a_p^2\right )=\left ( \sum_{p=1}^m a_p\right )^2 \] \[ -2\left ( \sum_{p=1}^m a_p\right ) \left (\sum_{p=m+1}^n a_p\right )+\left (\sum_{p=m+1}^n a_p\right )^2\]
\[\Rightarrow (n-1) \left ( \sum_{p=1}^{m} a_p^2\right )+(n-1) \left ( \sum_{p=m+1}^{n} a_p^2\right )+ 2\left ( \sum_{p=1}^m a_p\right ) \left (\sum_{p=m+1}^n a_p\right ) \] \[-\left (\sum_{p=m+1}^n a_p\right )^2=\left ( \sum_{p=1}^m a_p\right )^2 \]
As\[(n-1) \left ( \sum_{p=m+1}^{n} a_p^2\right )-\left (\sum_{p=m+1}^{n} a_p\right )^2\] \[\geq (n-m) \left ( \sum_{p=m+1}^{n} a_p^2\right )-\left (\sum_{p=m+1}^{n} a_p\right )^2\geq 0\; \; \; [Cauchy-Schwartz].....(1)\]
Again from cauchy,\[(n-1) \left ( \sum_{p=1}^{m} a_p^2\right )\geq m \left ( \sum_{p=1}^{m} a_p^2\right )\geq \left (\sum_{p=1}^{m} a_p\right )^2....(2)\]
From $(1)$ and $(2)$ we can say \[\Rightarrow (n-1) \left ( \sum_{p=1}^{m} a_p^2\right )+(n-1) \left ( \sum_{p=m+1}^{n} a_p^2\right ) \ge \left ( \sum_{p=1}^m a_p- \sum_{p=m+1}^n a_p\right )^2\]
With equality iff \[\frac {a_1}{1}=\frac{a_2}{1}=...=\frac{a_n}{1}=0\]A contradiction comes which completes the proof.