Suppose $x\_1\ge x\_2\ge \cdots \ge x\_n\ge 0$ and $y\_1\ge y\_2\ge\cdots\ge y\_n\ge0$ be reals such that for any positive **integer** $p$,
$$
\sum\_{i=1}^n x\_i^p \geq \sum\_{i=1}^n y\_i^p.
$$



> 
> **Question:** Is there a constant $c\ge 1$ such that for all $x,y$ satisfying the above,
>  we have $cx\succ\_w y$.
> 
> 
> 


Here $cx$ denotes the sequence $cx\_1, cx\_2,\ldots, cx\_n$ and $a \succ\_w b$ means that $a$ weakly majorizes $b$ from below.


Recall that a sequence $a\_1\ge a\_2\ge \cdots\ge a\_n$ is said to weakly majorize from below another sequence $b\_1\ge b\_2\ge\cdots\ge b\_n$ if for all $1\le k \le n$
$$
\sum\_{i=1}^k a\_i\ge \sum\_{i=1}^k b\_i.
$$


**Update:** [user35593](https://mathoverflow.net/users/35593/user35593)'s example $x=(4,1,1)$ and $y=(3,3,0)$ shows that if there is such a $c$, then $c\ge 1.2$.


**Remark 1:** If $\sum f(x\_i)\ge \sum f(y\_i)$ for all convex non-decreasing functions, then we have $x\succ\_w y$, a result due to Karamata; Hardy, Littlewood and Pólya. In fact testing against all functions of the form $z\mapsto (z-t)^+$ is enough.



Sufficient conditions for weak majorization
I think there is no such $c$. In fact, for $k$ and $\epsilon \gt 0$ given, there are $n \gg k$ and $x\_1\ge x\_2\ge \cdots \ge x\_n\ge 0$ and $y\_1\ge y\_2\ge\cdots\ge y\_n\ge0$ with
$$\sum\_{i=1}^n x\_i^p \geq \sum\_{i=1}^n y\_i^p.$$ for all $p\ge 1$ but
$$k\cdot \sum\_{i=1}^k (x\_i - \epsilon) \le\sum\_{i=1}^k y\_i$$
So $c$ had to be at least $k$.


The sequences are as follows:
$$x\_1 = 1 + \epsilon, x\_2 = \cdots = x\_n = \epsilon$$
$$ y\_1 = \cdots = y\_k = 1,
y\_{k+1} = \cdots = y\_n = 0.$$
Now
$$k\cdot \sum\_{i=1}^k (x\_i - \epsilon) = \sum\_{i=1}^k y\_i = k$$
So we only have to choose $n \gg k$ s. t.
$$\sum\_{i=1}^n x\_i^p \geq \sum\_{i=1}^n y\_i^p = k \quad \forall p \ge 1$$ This is clearly possible, since $x\_1^p \ge k$ for all but a finite number of $p$, and if $p$ is fixed, $n \ge 1 + \frac{k-1}{\epsilon^p}$ also does the job.


