I am looking for a proof of the inequality as follow:

Let $n$ be an integer number $n \ge 2$ and $x_1, \cdots, x_n$ and $y_1,\cdots, y_n$ are nonegative real numbers such that $(x_1,\cdots, x_n)$ majorizes $(y_1,\cdots, y_n)$; Let $0 \leq a_1, a_2,\cdots,a_n \leq 1$ then

$$\sum_{\text{sym}} {x_1}^{a_1}{x_2}^{a_2}\cdots {x_n}^{a_n}\leq \sum_{\text{sym}} {y_1}^{a_1}{y_2}^{a_2}\cdots {y_n}^{a_n}$$

Note: The inequality above is not Muirhead's Inequality.

Example: Let $0 \leq a_i \leq 1$ then

  • $4^{a_1}1^{a_2}+ 4^{a_2}1^{a_1} \le 3^{a_1}2^{a_2}+ 3^{a_2}2^{a_1}$

  • $5^{a_1}5^{a_2}2^{a_3}+5^{a_1}5^{a_3}2^{a_2}+5^{a_2}5^{a_1}2^{a_3}+5^{a_2}5^{a_3}2^{a_1}+5^{a_3}5^{a_1}2^{a_2}+5^{a_3}5^{a_2}2^{a_1} \leq 4.5^{a_1}4^{a_2}3.5^{a_3}+4.5^{a_1}4^{a_3}3.5^{a_2}+4.5^{a_2}4^{a_1}3.5^{a_3}+4.5^{a_2}4^{a_3}3.5^{a_1}+4.5^{a_3}4^{a_1}3.5^{a_2}+4.5^{a_3}4^{a_2}3.5^{a_1}$

See also:


1 Answer 1


$\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}} \newcommand{\tsi}{\tilde\si}$

For $x=(x_1,\dots,x_n)\in(0,\infty)$, let \begin{equation*} f(x):=2\sum_{\si\in S_n}\prod_{k=1}^n x_k^{a_{\si(k)}}, \end{equation*} where $S_n$ is the set of all permutations of the set $\{1,\dots,n\}$. By the symmetry and the Schur--Ostrowski criterion, it suffices to show that \begin{equation*} x_1\le x_2\overset{\text{(?)}}\implies f_1(x)-f_2(x)\ge0, \tag{1} \end{equation*} where $f_j(x)$ is the partial derivative of $f(x)$ in $x_j$. Consider the bijection $S_n\ni\si\leftrightarrow\tsi\in S_n$ defined by the formula \begin{equation*} \tsi(k):= \left\{ \begin{aligned} \si(2)&\text{ if }k=1,\\ \si(1)&\text{ if }k=2,\\ \si(k)&\text { otherwise}. \end{aligned} \right. \end{equation*} Then \begin{align*} f(x)&:=\sum_{\si\in S_n}x_1^{a_{\si(1)}}x_2^{a_{\si(2)}}\prod_{k=3}^n x_k^{a_{\si(k)}} + \sum_{\tsi\in S_n}x_1^{a_{\tsi(1)}}x_2^{a_{\tsi(2)}}\prod_{k=3}^n x_k^{a_{\tsi(k)}} =\sum_{\si\in S_n}Q P, \tag{2} \end{align*} where \begin{equation*} Q:=Q_\si(x):=x_1^{b_1}x_2^{b_2}+x_1^{b_2}x_2^{b_1},\quad P:=P_\si(x):=\prod_{k=3}^n x_k^{b_k}, \end{equation*} \begin{equation*} b_k:=b_{\si;k}:=a_{\si(k)}. \end{equation*} Denoting by $Q_j$ the partial derivative of $Q$ in $x_j$, we have \begin{equation*} Q_1-Q_2=b_1 (x_1 x_2)^{b_1-1}(x_2^{b_2-b_1+1}-x_1^{b_2-b_1+1}) +b_2 (x_1 x_2)^{b_2-1}(x_2^{b_1-b_2+1}-x_1^{b_1-b_2+1})\ge0. \end{equation*} Now noting that $P$ does not depend on $(x_1,x_2)$, we see that (1) follows from (2), as desired.

  • 1
    $\begingroup$ @ĐàoThanhOai: the first name of Iosif start with "I" not "L". see the last sentence of this comment of Iosif. $\endgroup$
    – Mahdi
    Jun 18, 2018 at 7:11
  • $\begingroup$ Oh, I am sorry; Thank You very much dear @Mahdi and IosifPinelis $\endgroup$ Jun 18, 2018 at 7:30

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