Question about an inequality described by matrices Let $A=(a_{ij})_{1 \le i, j \le n}$ be a matrix such that $\sum_\limits{i=1}^{n} a_{ij}=1$ for every $j$, and $\sum_\limits{j=1}^n a_{ij} = 1$ for every $i$, and $a_{ij} \ge 0$. Let
$$\begin{equation}
  \begin{pmatrix}
  y_1 \\
  \vdots \\
  y_n \\
  \end{pmatrix}
  =\mathbf{A}
  \begin{pmatrix}
  x_1 \\
  \vdots \\
  x_n
  \end{pmatrix}
  \end{equation}$$
with non-negative $y_i$ and $x_i$. Prove that  $y_1 \cdots y_n \ge x_1 \cdots x_n$.
It may somehow matter to convex function.
 A: $$y_i=\sum_j a_{ij} x_j\geqslant \prod_j x_j^{a_{ij} }$$
by Jensen inequality for logarithm. Now take the product over $i=1,2,\dots,n$.
A: This is a special case  of the so called  Schur's  majorization inequality. Here are the details.
$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bx}{\boldsymbol{x}}$ $\newcommand{\by}{\boldsymbol{y}}$ 
Given $\bx\in\bR^n$ we denote   by $\bar{\bx}$ the vector obtained from $\bx$ by rearranging its coordinates in decreasing order.  We say that $\bx$ dominates $\by$ and we write  this $\bx\succ\by$ if 
$$\sum_{i=1}^k \bar{x}_i\geq  \sum_{i=1}^k \bar{y}_i,\;\;\forall k=1,\dotsc, n-1, $$
$$\sum_{i=1}^n \bar{x}_i=  \sum_{i=1}^n \bar{y}_i.$$
The symmetric group $S_n$ acts on $\bR^n$ by permuting the coordinates of a vector. For $\bx\in\bR^n$  we denote by $S_n\cdot\bx$ the orbit of $\bx$ with respect to this action, i.e., the set of vectors that can be obtained from $\bx$ by permuting its coordinates. $\DeclareMathOperator{\conv}{conv}$ 
For a set $S\subset \bR^n$ we denote by $\conv(S)$ its convex hull.  
The next nontrivial theorem   characterizes the dominance relation. For a nice presentation of this theorem and   Schur's majorization inequality  I refer to Chapter 13 of 

J.Michael Steele: The Cauchy-Schwarz Master Class, Cambridge University Press, 2004.

Theorem. Let $\bx,\by\in\bR^n$. The following statements are equivalent. 


*

*$\bx\succ \by$.  

*$\by\in \conv( S_n\cdot\bx)$.

*There exists a doubly stochastic $n\times n$ matrix $A$ such that $\by=A\bx$.
Fix  an interval $(a,b)\subset  \bR$.  A symmetric $C^1$-function $f:(a,b)^n\to\bR$ is called Schur convex if $\newcommand{\pa}{\partial}$
$$ (x_j-x_k)\left( \frac{\pa f}{\pa x_j}(\bx)-\frac{\pa f}{\pa x_k}(\bx)\right)\geq  0,$$
for any $\bx\in (a,b)^n$ and any $j,k=1,\dotsc, n$.
The Schur majorization inequality   states that
$$ \bx\succ \by \implies f(\bx)\geq f(\by), $$
for any Schur convex function $f:(a,b)^n\to\bR$ and any $\bx,\by\in(a,b)^n$. 
The function
$$ p:[0,\infty)^n\to\bR,\;\;p(\bx)=-x_1\cdots x_n $$
is Schur convex. The inequality $p(\bx)\geq p(\by)$ is the inequality that interests you.
A: By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,\dots,y_n]^T$ and $x=[x,\dots,x_n]^T$ is equivalent to the condition that $\sum_1^n g(y_i)\ge\sum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=\ln$ to get the desired inequality $y_1 \cdots y_n \ge x_1 \cdots x_n$. 
