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.