I'm interested by the following problem :
Let $x_i>0$ be $n$ real numbers and $y_i>0$ be $n$ real numbers such that :
$1)$ $\forall i$ and $\forall j$ indices and $i\neq j$ we have : $$|x_i-x_j|\leq |y_i-y_j|$$
$2)$$\sum_{i=1}^{n}x_i\geq \sum_{i=1}^{n}y_i$$
Then we have : $$\prod_{i=1}^{n}x_i\geq \prod_{i=1}^{n}y_i$$
I can prove this using majorization tools it's easy but without this how to do ?
I think we can use contradiction to prove the case $n=3$ but I'm not able to find this proof.
So if you have ideas...
Thanks a lot
I will show you main steps. I suggest you try to think why these work.
(1) Enough to show when $\sum x_i=\sum y_i$ and $x_1,...,x_n$ are distinct. (Try to use limit to treat the case that some of them are equal)
(2) The set of $(y_1,...y_n)\in\mathbb{R}_{\ge 0}^n$ satisfying prop (1) is compact, so you would have a pair $(z_1,...z_n)$ that gives the maximum you can get.
(3) Assume that $|x_i-x_j|<|z_i-z_j|$ for some $i,j$ that $z_i,z_j$ are adjacent among $z_1,...,z_n$ and $z_i<z_j$(remember that $z_1,...z_n$ are distinct by prop(1)). Check that a new pair $(z_1',...z_n')$ that $z_r'=z_r$ for $r\neq i,j$ and $z_i'=z_i+d$, $z_j'=z_j-d$ with $z_j'-z_i'=|x_i-x_j|$ also satisfies prop (1), so such $(z_1,...z_n)$ does not give the maximum.
(4) Now you observe that the maximum is obtained only when $\{x_1,...x_n\}$ and $\{y_1,...,y_n\}$ are the same set, and so we are done.