BdMO Online Forum

Problem $\boxed {12}$(may be posted before):
Let $n\geq 2$ be an integer and $a_1.a_2,...,a_n$ be real numbers. Prove that for any non empty subset $S\in \{1,2,\dots,n\}$ the following inequality holds:
$(\displaystyle\sum_{i\in S}a_i)^2\leq \displaystyle\sum_{1\leq i\leq j\leq n}(a_i+...+a_j)^2$

Source:Romania 2004

SANZEED wrote:Problem $\boxed {12}$(may be posted before):
Let $n\geq 2$ be an integer and $a_1.a_2,...,a_n$ be real numbers. Prove that for any non empty subset $S\in \left \{ a_1,a_2,...,a_n \right \}$ the following inequality holds:
$(\displaystyle\sum_{i\in S}a_i)^2\leq \displaystyle\sum_{1\leq i\leq j\leq n}(a_i+...+a_j)^2$

Source:Romania 2004

Shouldn't $S$ be a subset of $\{ 1,2,...,n \}$? I mean, what is $a_{a_i}$ where $a_i$ is real?

Sorry please edit the typo
$S$ is a subset of $\left \{ 1,2,...,n\right \}$.

well, let's proceed on. Solve this one.
Solve this functional equation for $f:\mathbb N_0\to \mathbb N_0$ for distinct $x,y$:
\[f(x^2-y^2)=f(x)f(y)\]

Firstly notice $f(x)=f(-x)$. So we may work with only the right half of the number line. Substitute $x=y$ to derive $f(x)^2=f(0)$. So $f(x)=c$ or $-c$ at every point. So $\pm c=f(x^2-y^2)=\pm c^2\Rightarrow c^2\pm c=0\Rightarrow c=-1,0,1$.
So the possible functions are:
$f(x)=0$
$f(x)=1$
$f(x)=-1$
$f(x)=\pm 1$ and $f(0)=1$
It is easy to see the third does not satisfy the equation. I think the fourth is correct. But today I don't have much time to verify. (বার্ষিক পরীক্ষা শুরু হইসে ) So it should be left for tomorrow.

there was a little description change.

(Just trying .. It may have bug)

Let $P(x,y): f(x^2-y^2)=f(x)f(y)$
Now $P(x,-x): f(0)=f(x)f(-x)$ .

For distinct $x,y$; $P(x,y),P(x,-y)$ imply $f(x)f(y)=f(x)f(-y)$ … $(St1)$

Consider some cases .
Case 1 : $f(x)=0 \forall x$

Case 2 : There is only one integer $i$ so that $f(i) \neq 0$
If $i$ is of form $x^2-y^2$; then
$f(i)=f(x^2-y^2)=f(x)f(y)=0$.
So $i$ is not of form $x^2-y^2$ . Spacifically $i$ is of form $2n$ with $n$ odd.
That’s why solution for this case is
$f(i)=c \neq 0$ [ $i=2n$ with $n$ any odd number ]
$f(x)=0 \forall x \in { \mathbb{Z}/i}$

Case 3 : There is at least two integer $i,j$ with $ f(i),f(j)\neq 0$
Plugging $i$ instead $x$ in $(St1)$, for all $y\neq i$ , $f(y)=f(-y)$.
Similarly plugging $j$ , for all $y\neq j$ , $f(y)=f(-y)$.
As $i,j$ are distinct ,We can say for all $y\in \mathbb{Z}$ , $f(y)=f(-y)$.

So $f(x)^2=f(0)$.
Squaring both side of the main equation , $f(0)=f(0)^2$
Or $f(0)=0,1$ . ( $f(0)\neq 0$ as that will repeat Case 1 )
So $f(x)^2=1$

$f(x)=1$ is indeed a solution .
$f(x)=-1$ is not a solution .
________________________________________________________________

Another solution may be
$f(a_i)=1, f(b_i)=-1$ for $i=1,2..$ and $\mathbb{Z}={a_i} \cup {b_i}$ .
But how to charectarize ${a_i},{b_i}$?? (Need a lil help )

পুরাটা পড়ি নাই, কিন্তু এখনো বেশী সোজা। তাই আরেকটু পরিবর্তন।

মাসুম ভাই দোন্ত উই ডিসারভ আ লিটল হিন্ট ?

You've almost killed the marathon

হায় হায় আমারে এই অভিযোগে দোষী করলা? উলটা এইটা তো মনে হয় বেশী সোজা হইয়া গেছে। যাহাই হোক, এই হইলো তোমার কাংক্ষিত হিন্ত।
Set $y=x-1$ and $x=y+2$. আগে প্রমাণ কর যে $f(1)=0$, actually $f(x^2)=0$