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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.)

Helpful would be

$$e_{j+1}(\alpha,a_1/(1{-}a_1),\dots,a_n/(1{-}a_n))\ge 0$$

for any $\alpha\in\left]0,1\right[$. For $k\in {n{-}1,n}$ this turns out to be true for $\alpha=1-k/n$.

The supposed case of equality is $a_1=\dots=0$ and the case $a_1=\dots=a_k \nearrow 1, a_{k+1}=\dots=a_n$ seems to be critical to find the minimal $\alpha$.

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 $b_1b_2=0$).

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Proof for the special case $k=1$

The function \begin{align} f: \left]-\infty,1\right[&\to \mathbb R\\ a&\mapsto a/(1-a) \end{align} has the derivatives $f'(a)=1/(1{-}a)^2$ and $f''(a)=2/(1{-}a)^3>0$, especially $f(0)=0,f'(0)=1$ and $f$ is convex and therefore $f(a)\ge a$.

This shows already that $(n-1)/n+f(a_1)+\dots+f(a_n)> a_1+\dots+a_n\ge 0$, i.e. the statement for $j=1$.

We use the above substitution $b_i=1{-}a_i$, and find that \begin{align} e_2((n{-}1)/n,f(a_1),\dots,f(a_n))&=e_2(1/b_1,\dots,1/b_n)-(n{-}1)^2/n\sum_i 1/b_i+(n{-}1)(n{-}2)/2\\ &=:A(b_1,\dots,b_n) \end{align} with the above-mentioned constraints.

Assuming that $b_1 \le \dots\le b_n$, we show now that

$$A(b_1,\dots,b_n)\ge A\left(b_1,\frac{b_2{+}\dots{+}b_n}{n-1},\dots,\frac{b_2{+}\dots{+}b_n}{n-1}\right).$$

We calculate $$\frac{\partial}{\partial b_j} A(b_1,\dots,b_n)=-\left(\sum_i \frac1{b_i}-\frac{(n{-}1)^2}{n}-\frac1{b_j}\right){b_j}^{-2}\\ =:(1/b_j-B)b_j^{-2}$$

and claim that this is monotonically increasing in $j$ for $j\in{2,\dots,n}$. Fix B and derivate once again, to obtain $$ 2Bb_{j}^{-3}-3b_{j}^{-4}, $$ hence our claim turns into $2B-3/b_j\ge0$. This is estimated, using $b_{1}\leq b_{2}\leq b_{j}\leq\dots\leq b_{n}$, by $$ 2B-3/b_j\geq-2(n{-}1)^{2}+1/2/b_{1}+1/2/b_{2}+\sum_{i=3}^{n}2/b_{i}=:C(b), $$ which is a strictly convex function in $b$ with a unique minimizer in the half space $\{S_{1}(b)\ge1\}$. The minimizer with this constraint reads

$$ b_{1}=b_{2}=\frac{1}{2n-2},b_{3}=\dots=b_{n}=\frac{1}{n{-}1} $$

with the minimum 0. Now, for $t\in[0,1]$ consider the functions

\begin{align} b(t)&:=(1{-}t)b+t\left(b_{1},\frac{b_{2}+\dots+b_{n}}{n{-}1}\right),\\ g(t)&:=A\left(b(t)\right). \end{align}

As $b$, $b(t)$ satisfies the constraint $S_{1}(b(t))\leq 1$ for all $t$ and $b_{1}(t)\le\dots\le b_{n}(t)$.

To show that $g$ decreases, we derivate $$ g'(t)=\sum_{i=2}^{n}\left(\frac{b_{2}+\dots+b_{n}}{n{-}1}-b_{i}\right)\partial_{i}A(b(t))=:\sum_{i=2}^{n}c_{i}\partial_{i}A(b(t)). $$ Since $i\leq j$ implies $c_{i}\ge c_{j}$ and $\partial_{i}A(b(t))\ge\partial_{j}A(b(t))$, Chebyshev's sum inequality implies $$ g'(t)=\sum_{i=2}^{n}c_{i}\partial_{i}A(b(t))\leq\left(\frac{c_{2}+\dots+c_{n}}{n{-}1}\right)\sum_{i=2}^{n}\partial_{i}A(b(t))=0. $$ This shows that $g$ decreases mononotically and that $$ A(b)=g(0)\leq g(1)=A\left(b_1,\frac{b_2{+}\dots{+}b_n}{n-1},\dots,\frac{b_2{+}\dots{+}b_n}{n-1}\right). $$ It remains to show the inequality for $b_{1}\leq b_{2}=\dots=b_{n}$. The constraint turns into $b_{2}=(1{-}b_{1})/(n{-}1)$ and we have to minimize

\begin{align} -(n{-}1)^{2}\left(\frac{1}{b_{1}}+\frac{(n{-}1)^{2}}{1{-}b_{1}}\right)+\frac{1}{b_{1}}\frac{(n{-}1)^{2}}{1{-}b_{1}}+\frac{\left(n{-}1\right)\left(n{-}2\right)}{2}\frac{\left(n{-}1\right)^{2}}{1{-}b_{1}} =\frac{(n{-}1)^{2}(n{-}2)}{1{-}b_{1}}\left(-n+\frac{n{-}1}{2}\frac{1}{1{-}b_{1}}\right). \end{align}

The minimizer is attained where $1/(1{-}b_{1})=n/(n{-}1)$, which leads to $b=(1,\dots,1)/n$, $a=0$ and $S_{2}((n{-}1)/n,f(a_1),\dots,f(a_n))=0$.

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The identity $$ k!(1{-}a_1)\cdots(1{-}a_n)e_k\left(\frac{a_1}{1{-}a_1},\dots,\frac{a_n}{1{-}a_n}\right)=\partial_t^k \prod_{i=1}^n (s+ta_i)|_{s=1,t=-1} $$ transforms the conjecture into the question Inequality of elementary symmetric polynomials written in terms of derivatives of $(s+ta_1)\dotsm(s+ta_n)$ . For $2k\geq n$, my answer to that question yields a number $\alpha<1$.

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Solution applying Rolle's theorem on $(1+ta_1)\cdots(1+ta_n)$

The function \begin{align} f: \left]-\infty,1\right[&\to \mathbb R\\ a&\mapsto a/(1-a) \end{align} has the derivatives $f'(a)=1/(1{-}a)^2$ and $f''(a)=2/(1{-}a)^3>0$, especially $f(0)=0,f'(0)=1$ and $f$ is convex and therefore $f(a)\ge a$. Let us abbreviate the tuple $(a_1,\dots,a_n)$ by $a$, $(f(a_1),\dots,f(a_n))$ by $f(a)$, etc.

The set $$ \Gamma^n_k=\{a\in \mathbb R^n | S_1(a),\dots,S_k(a)>0\}, $$ the so-called Gårding cone of $S_k$ has the property that $S_j(a)\le S_j(b)$ for all $b\ge a\in \bar\Gamma^n_k$ (i.e. $b_i\ge a_i$ forall $i$). This shows that $e_j(\alpha,f(a))\geq e_j(a)\ge 0$ for $j\le k$ already. It remains to show $S_{k+1}(\alpha,f(a))\ge 0$ for a suitable $\alpha$.

For the time being, assume $a\in\Gamma^n_k$ (i.e. not in the boundary of the cone). Let $$ p_a(t):=(1+ta_1)\cdots (1+ta_n). $$ Note that $e_k(a/(1+ta))$ can be expressed by the derivative $$ p_a^{(k)}(t)=k!p_a(t)e_k(\frac a{1+ta}), $$ Since $e_{k+1}(\alpha,f(a))=\alpha e_k(f(a))+e_{k+1}(f(a))$, our question can be expressed as \begin{align} \frac{n{-}1}{n}\ge \alpha_k^n:&=-\inf_{a\in \Gamma_k^n,a<1} \frac{e_{k+1}(f(a))}{e_k(f(a))}\\ &=-\inf_{a\in \Gamma_k^n,a<1} \frac 1{k+1} \frac{p_a^{(k+1)}(-1)}{p_a^{(k)}(-1)}. \end{align} Now, let us study the first derivative: $$ p_a'(t)=1 e_1(a)(t)+\dots+ne_n(a)t^{n-1}. $$ We are going to write it in the form $p_a'(t)=e_1(a)(1+b_1t)\dots(1+b_{n-1}t)$. If $a_i=0$, it turns into 1, i.e. it can be omitted and we can assume $a_i\neq 0$ for all $i$ in the following.

Assuming the order $a_1\le\dots a_p < 0< a_{p+1} \le \dots\le a_n$, the roots of $p_a$ are $$-\frac{1}{a_{p+1}}\le\dots\le -\frac{1}{a_n}<0<-\frac{1}{a_1}\le -\frac{1}{a_p}.$$ According to Rolle's theorem, $p_a'(t)$ has roots $c_i$ satisfying $$ -\frac{1}{a_{p+1}}\le c_{n-1} \le -\frac{1}{a_{p+2}} \le \dots \le -\frac{1}{a_n} \le c_p \le \underbrace{-\frac{1}{a_1}\le \dots \le c_1 \le -\frac{1}{a_p}}_{\text{negative }a_i} $$ and we have the equality of the coefficients \begin{align} p'_a(t)&=1e_1(a)+\dots+ne_n(a)t^{n-1}\\ &=ne_n(a)\left(e_{n-1}(-c)+\dots+e_0(-c)t^{n-1}\right). \end{align} Since we assumed $a_i\neq 0$, this implies $e_{n-1}(c)=c_1\cdots c_{n-1}\neq 0$. Therefore, let us substitute $b_i=-1/c_i$ and we get \begin{align} p'_a(t)&=1e_1(a)+\dots+ne_n(a)t^{n-1}\\ &=e_1(a)(1+tb_1)\cdots(1+tb_n)\\ &=e_1(a)\left(e_0(b)+\dots+e_0(b)t^{n-1}\right) \end{align} and by comparison of coefficients $$e_1(a)e_{i-1}(b)=ie_{i}(a).$$ In particular, the negative factors in $$ b_p=\frac{na_1\cdots a_n}{e_1(a)b_1\cdots b_{p-1}b_{p+1}\cdots b_{n-1}} $$ are $a_1\cdots a_p/b_1/\cdots/b_{p-1}$, their number is odd and $b_p<0$.

Altogether, $b$ satisfies similar assumptions as $a$, namely (the above conditions for $c_i$ by Rolle's theorem convert into) $b_i<1$ and $e_1(b),\dots,e_k(b)>0$. Therefore, we can decrement $k$ and $n$ as follows: \begin{align} \alpha&:=-\inf_{a\in \Gamma^n_k,a<1}\frac{1}{k{+}1}\frac{p_a^{(k+1)}(-1)}{p_a^{(k)}(-1)}\\ &\le-\inf_{b\in \Gamma^{n-1}_{k-1},b<1}\frac{1}{k{+}1}\frac{p_b^{(k)}(-1)}{p_b^{(k-1)}(-1)}\\ &\le\dots\\ &\le -\inf_{c\in\Gamma^{n-k+1}_1,c<1}\frac{1}{k{+}1}\frac{p_c''(-1)}{p_c'(-1)}\\ &= -\inf_{c\in\Gamma^{n-k+1}_1,c<1}\frac{2}{k{+}1}\frac{e_2(f(c))}{e_1(f(c))}. \end{align} Application of the answer https://mathoverflow.net/a/485640/536759 concludes that $$\alpha\le\frac{2}{k{+}1}\frac{n{-}k}{n{-}k{+}1}.$$

If, this case was left, $a\in \partial\Gamma^n_k$, then the assertion is shown by a sequence in $\Gamma^n_k$ converging to $a$.

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