The general question
For $1\leq k\leq n$, let $$e_k(a_1,\dots,a_n):=\sum_{j_1<\dots<j_k}a_{j_1}\cdots a_{j_k}$$ be the $k$-th elementary symmetric polynomial.
Let $a_1,\dots,a_n<1$ and $e_1(a_1,\dots,a_n),\dots,e_k(a_1,\dots,a_n)\ge 0$. I wonder if then also $$e_{j+1}((n{-}1)/n,a_1/(1{-}a_1),\dots,a_n/(1{-}a_n))\ge 0$$ for $j=0,\dots,k$. (Supposingly, the critical case is $j=1$ and for larger $j$ one can replace $(n{-}1)/n$ by smaller numbers.)
The supposed cases of equality are $a_1=\dots=0$ and $a_1 \nearrow 1, a_2=\dots=a_n \searrow -1/(n{-}1)$.
I have checked this for $n\leq 3$ and for $j=n{-}1$.
A special case
In the case of $k=1$, the substitution $b_i:=1{-}a_i$ transforms my conjecture to the following: Let $b_1,\dots,b_n$ be nonnegative reals. Then the $k$-th elementary symmetric means $S_j:=e_j(b_1,\dots,b_n)/\binom{n}{k}$ satisfy $$nS_{n-2}S_1^2+(n{-}2)S_n\ge 2(n{-}1)S_{n-1}S_1.$$ Can anybody prove this?
Now, as far as I know, equality holds if $b_1=\dots=b_n$ or ($b_2=\dots=b_n$ and $b_1b_2=0$).
