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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.

$$y_i=sum_j a_{ij} x_jgeqslant prod_j x_j^{a_{ij} }$$
by Jensen inequality for logarithm. Now take the product over $i=1,2,dots,n$.

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 $bxinbR^n$ 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 $bxsuccby$ if

$$sum_{i=1}^k bar{x}_igeq sum_{i=1}^k bar{y}_i,;;forall k=1,dotsc, n-1, $$

$$sum_{i=1}^n bar{x}_i= sum_{i=1}^n bar{y}_i.$$

The symmetric group $S_n$ acts on $bR^n$ by permuting the coordinates of a vector. For $bxinbR^n$ we denote by $S_ncdotbx$ 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 $Ssubset bR^n$ 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 $bx,byinbR^n$. The following statements are equivalent.

1. 
   $bxsucc by$.


2. 
   $byin conv( S_ncdotbx)$.


3. There exists a doubly stochastic $ntimes n$ matrix $A$ such that $by=Abx$.

Fix an interval $(a,b)subset bR$. A symmetric $C^1$-function $f:(a,b)^ntobR$ 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 $bxin (a,b)^n$ and any $j,k=1,dotsc, n$.

The Schur majorization inequality states that

$$ bxsucc by implies f(bx)geq f(by), $$

for any Schur convex function $f:(a,b)^ntobR$ and any $bx,byin(a,b)^n$.

The function

$$ p:[0,infty)^ntobR,;;p(bx)=-x_1cdots x_n $$

is Schur convex. The inequality $p(bx)geq p(by)$ is the inequality that interests you.

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 $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.