$A=(a_{ij})_{1 \le i, j \le n}$ is a matrix 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 $a_{ij} \ge 0$.And $$\begin{equation} \begin{pmatrix} y_1 \\ \vdots \\ y_n \\ \end{pmatrix} =\mathbf{A} \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} \end{equation}$$ $y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 \cdots y_n \ge x_1 \cdots x_n$

It may somehow matter to convex function.

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.