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Consider the following inequality.

Let $x_1>x_2>...>x_n>0$ be some positive numbers. Then

$\sum_{i=1}^n x_i+\sum_{i=1}^n\sum_{j\neq i} \frac{x_ix_j}{x_i+x_j}\leq M\left [\sum_{i=1}^n i\cdot x_i\right]$

It is not hard to show that the bound holds for M=2 (by adding and subtracting $(i-1)x_i$ to the LHS and using $\frac{x_i-x_j}{x_i+x_j}\leq 1$). However, after some experiments I am convinced it also holds for M=4/3, but I have no clue how to demonstrate such a bound. Any ideas or advice would be greatly appreciated!

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For $x_1=\ldots =x_n=1$ (or, as you required strict order, approximately for $x_i=1-i\epsilon$), you want $$ M\left[n+n\cdot(n-1)\cdot \frac 12\right]\le 1+2+\ldots +n$$ As $1+2+\ldots +n=\frac{n(n+1)}2$, this implies $M\le 1$.

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  • $\begingroup$ But if inequality holds with $x_i=1$ it does imply that it holds for all choices of x (but I claim it does hold for all ordered x if M=4/3) $\endgroup$ – Simba_loo May 16 at 21:32
  • $\begingroup$ Apologies, M should've been on the RHS $\endgroup$ – Simba_loo May 16 at 21:57
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This is true for $M=1$ and it is the smallest possible constant. To check that $M\geq 1$ plug $x_i\rightarrow1$ as Hagen suggested. For $M=1$ we have:

$\sum\limits_{i=1}^n x_i+\sum\sum \frac{x_ix_j}{x_i+x_j}\leq\sum\limits_{i=1}^n x_i+\sum\sum \frac{(x_i+x_j)^2}{4(x_i+x_j)}=\sum\limits_{i=1}^n x_i+\frac{2(n-1)}{4}\sum\limits_{i=1}^n x_i =\frac{n+1}{2}\sum\limits_{i=1}^n x_i\leq \sum\limits_{i=1}^n ix_i $

The last one is true because of rearrangement inequality. For any permutations of $\{b_i\}=1,2..n$ we have

$\sum\limits_{i=1}^n b_{k_i}x_i\leq \sum\limits_{i=1}^n ix_i $

Adding all possible such inequalities we get:

$(n-1)!\sum\limits_{i=1}^n \left(\sum\limits_{j=1}^nj\right)x_i\leq n!\sum\limits_{i=1}^n ix_i$

$\frac{n+1}{2}\sum\limits_{i=1}^n x_i\leq \sum\limits_{i=1}^n ix_i$ and we are done

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    $\begingroup$ Thanks for the response, but it doesn't hold for M=1! Try n=2, x_1 = 2, x_2 = 1, LHS = 1+2+2/3+2/3 = 13/3, RHS = 2+1*2 = 12/3 The bounds you use though might be useful, I'll try playing around with them. $\endgroup$ – Simba_loo May 17 at 12:32
  • $\begingroup$ I think the proof doesn't work because you don't account for the fact that in there are two $x_ix_j$ terms, e.g. for n=2 you get $LHS = x_1+x_2 + \frac{2x_i x_j}{x_i+x_j}$ $\endgroup$ – Simba_loo May 17 at 12:42

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