# An inequality related to Power sum and elementary symmetric polynomial and majorizes

• Power sum and elementary symmetric polynomial

Let $x_1,. . . , x_n$ be variables, denote for $k \ge 1$ by $p_k(x_1,\dots,x_n)$ the $k-th$ power sum:

$$p_k(x_1,\dots,x_n)=\sum\nolimits_{i=1}^nx_i^k = x_1^k+\cdots+x_n^k,$$

and for $k \ge 1$ denote by $e_k(x_1,\dots, x_n)$ the elementary symmetric polynomial (that is, the sum of all distinct products of k distinct variables), so

: \begin{align} e_0(x_1, \ldots, x_n) &= 1,\\ e_1(x_1, \ldots, x_n) &= x_1 + x_2 + \cdots + x_n,\\ e_2(x_1, \ldots, x_n) &= \textstyle\sum_{1\leq i<j\leq n}x_ix_j,\\ e_n(x_1, \ldots, x_n) &= x_1 x_2 \cdots x_n,\\ e_k(x_1, \ldots, x_n) &= 0, \quad\text{for}\ k>n.\\ \end{align}

• Majorizes

If $x_1,. . . , x_n$ and $y_1, . . . , y_n$ are numbers, such that $(x_1,. . . , x_n)$ majorizes $(x_1,. . . , x_n)$ if only if

$x_1+x_2+\dots+x_n = y_1+y_2+\dots+y_n$ and $x_{1}\geq x_{2}\geq \cdots \geq x_{n}$ and $y_{1}\geq y_{2}\geq \cdots \geq y_{n}$

$x_{1}+\cdots +x_{i}\geq y_{1}+\cdots +y_{i}$ for all $i \in \{1,..., n − 1\}$.

I am looking for a proof of the inequality related to Power sum and elementary symmetric polynomial and majorizes as follows:

Let $n$ be an integer number $n \ge 2$ and $x_1,. . . , x_n$ and $y_1, . . . , y_n$ are nonegative real numbers such that $x_1+x_2+\dots+x_n = y_1+y_2+\dots+y_n$ then $(x_1,. . . , x_n)$ majorizes $(x_1,. . . , x_n)$ if only if then

$$e_k(x_1, \ldots, x_n) \leq e_k(y_1, \ldots, y_n)$$

and $$p_k(x_1,\dots,x_n) \ge p_k(y_1,\dots,y_n)$$

for all $k \in \{2, \cdots, n \}$.

• See comment
• The $p_k$ inequality follows from Karamata's inequality, since the function $x \mapsto x^k$ is convex on $\mathbb{R}_+$. – darij grinberg Jun 16 '18 at 17:32
• As for the $e_k$ inequality, I think the best way is to generalize. If $A$ is a partially ordered abelian group (written multiplicatively), then we say that a function $f : \mathbb{R}_+ \to A$ is log-concave if and only if it satisfies $f\left(a\right) f\left(b\right) \leq f\left(c\right) f\left(d\right)$ whenever $a$ and $b$ are two nonnegative reals and $c$ and $d$ are two elements of the interval $\left[a,b\right]$ satisfying $c + d = a + b$. Now, ... – darij grinberg Jun 16 '18 at 17:36
• ... an analogue (and generalization, if you think about it properly) of Karamata's inequality says that if an $n$-tuple $\left(x_1, x_2, \ldots, x_n\right)$ of nonnegative reals majorizes another such $n$-tuple $\left(y_1, y_2, \ldots, y_n\right)$, then $f\left(x_1\right) f\left(x_2\right) \cdots f\left(x_n\right) \leq f\left(y_1\right) f\left(y_2\right) \cdots f\left(y_n\right)$ for any partially ordered abelian group $A$ and any log-concave function $f : \mathbb{R}_+ \to A$. (This can be proven easily by recalling ... – darij grinberg Jun 16 '18 at 17:38
• ... the well-known fact that $\left(y_1, y_2, \ldots, y_n\right)$ can be obtained from $\left(x_1, x_2, \ldots, x_n\right)$ by a finite sequence of steps, where each step "moves" two entries of the $n$-tuple closer to each other while preserving their sum.) Now, apply this analogue to $A = \left(\mathbb{R}_+\left[t\right], \cdot\right)$ (the multiplicative monoid of all polynomials in one variable $t$ with nonnegative real coefficients, with coefficientwise ordering) and $f\left(x\right) = 1+tx$. [Correction: When I said "group" in the above comments, I meant "monoid".] ... – darij grinberg Jun 16 '18 at 17:42
• The "if" part fails for $n = 3$ and $\left(x_1,x_2,x_3\right) = \left(9,2,2\right)$ and $\left(y_1,y_2,y_3\right) = \left(6,6,1\right)$. – darij grinberg Jun 16 '18 at 18:28