I recently came across a result in a Swiss IMO training script for advanced techniques in inequalities for which I couldn't find a proof anywhere.
Let $d_k$ be the $k$-th normalized elementary-symmetric polynomial in variables $X_i$, where $ i $ ranges from $1$ to $n$. That is:
$$d_k(X_1 , \ldots , X_n )=\binom{n}{k}^{-1}\sum_{1\le j_1 < j_2 < \cdots < j_k \le n} X_{j_1} \dotsm X_{j_k}$$
The statement in the script was the following:
If the inequality $f(d_1,d_2,\dots d_n)\geq 0$ holds for the normalized elementary-symmetric polynomials with $n$ positive variables $X_1,\, X_2,\ldots , X_n$, then the same inequality also holds for elementary symmetric polynomials in $n+1$ variables $d^{'}_k$, that is $f(d^{'}_1,d^{'}_2,\dots d^{'}_n)\geq 0$ holds true also.
This result seems so powerful that it seemed odd that I could not find a proof for it. Does anyone have a proof or could provide a reference?
Claim: Suppose $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0\in \mathbb{R}[x]$, $a_n\not=0$ is a polynomial whose roots are all positive. Then all roots of $g(x)=\sum_{i=0}^{n} \binom{n}{i}a_ix^i$ are also positive.
Proof: This post tells us that all roots of $g(x)$ are real.
Since all roots of $p(x)$ are positive, $a_i$ has the same sign as $(-1)^{n-i}a_n$. Hence $g(x)$ does not have a negative root. $g(0)=a_0\not=0$. Hence all roots of $g(x)$ are positive.
Let us prove the statement in the question.
Suppose $f(d_1(X_1, \ldots , X_n),d_2(X_1,\ldots , X_n),\cdots, d_n(X_1, \ldots , X_n))\geq 0$ holds for arbitrary positive numbers $X_1,\ldots , X_n$.
Let $p(x)=\sum_{i=0}^n(-1)^{i}{n\choose i}d_{i}(X_1, \ldots , X_n)x^{n-i}$, with $d_0=1$. Then $p(x)$ has $n$ positive roots $X_1, X_2, \cdots, X_n$.
Let $q(x)=\sum_{i=0}^n{n\choose i}(-1)^{i}{n\choose i}d_{i}(X_1, \ldots , X_n)x^{n-i}$. Thanks to the Claim, $q(x)$ also have $n$ positive roots. Denote these positive roots by $Y_1, Y_2, \cdots, Y_n$. Then $${n\choose i}(-1)^i{n\choose i}d_i(X_1, \ldots , X_n)=(-1)^id_i'(Y_1,Y_2,\cdots, Y_n)$$ which means $$d_i'(X_1, \ldots , X_n)=d_i(Y_1,Y_2,\cdots, Y_n)$$
Hence $$\begin{aligned}&\quad f(d_1'(X_1, \ldots , X_n),d_2'(X_1,\ldots , X_n),\cdots, d_n'(X_1, \ldots , X_n))\\ &=f(d_1(Y_1, \ldots , Y_n),d_2(Y_1,\ldots , Y_n),\cdots, d_n(Y_1, \ldots , Y_n))\\ &\ge0 \end{aligned}$$
The statement is still true if we drop the condition that $X_i$'s be positive.