Is a generalization of Padoa inequality correct?

Question

Padoa's inequality is named after Alessandro Padoa (1868-1937):

Let $a$, $b$, $c$ be sidelengths of a given triangle $\triangle ABC$ then

$$(b+c-a)(c+a-b)(a+b-c) \le abc .$$

My question: Is the following generalization of Padoa's inequality corect?

Let $a_i>0$ for $1\le i\le n$ and let $$S:=a_1+a_2+....+a_n.$$ Suppose that $$b_i:=S-(n-1)a_i \ge 0\quad\text{ for} \quad 1\le i\le n.$$ Then

$$\prod_{i=1}^n b_i \leq \prod_{1}^{n} a_i .$$

Answer 1

Since $a_i=\frac{S-b_i}{n-1}=\frac{\sum_j b_j -b_i}{n-1}$ we can write $$\prod_{i=1}^n a_i=\prod_{i=1}^n \left(\frac{\sum_{j\neq i}b_i}{n-1}\right)\geq \prod_{i=1}^n \left(\prod_{j\neq i}b_i\right)^{1/n-1}=\prod_{i=1}^n b_i$$ by the Arithmetic-Geometric mean inequality.