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


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


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



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


$$\prod\_{i=1}^n b\_i \leq \prod\_{1}^{n} a\_i .$$



Is a generalization of Padoa inequality correct?
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.


