Lemma 8.2.88673FC834527

Proof. First prove the unnormalized Newton inequality for an arbitrary real list $\mu $ of length $N$. For $1\leq q<N$, put $R(x)=\prod _{a=1}^ N(x+\mu _ a)$ and $G=R^{(N-q-1)}$. Repeated Rolle's theorem makes $G$ a real-rooted polynomial of degree $d=q+1$. Away from its roots $r_ a$, logarithmic differentiation and Cauchy–Schwarz give

Continuity extends this polynomial inequality to every real $x$. At zero,

Consequently $S_{q-1}(\mu )S_{q+1}(\mu )\leq c_{N,q}S_ q(\mu )^2$. No sign restriction on $\mu $ or on these coefficients was used. For $q=N$, its left side is zero and the same assertion holds with $c_{N,N}=0$.

Apply this to the list with $i,j$ deleted. Put $A=S_{q-1;ij}$, $B=S_{q;ij}$, $C=S_{q+1;ij}$ and $D_ i=S_{q;i}$, $D_ j=S_{q;j}$. Deleted-coordinate positivity gives $A,D_ i,D_ j\geq 0$: the required degrees are $q+1\leq m$. Splitting in the two omitted coordinates gives the exact identity

Since $S_{q+1}(\lambda )\geq 0$ and $AC\leq c_{N,q}B^2$, this is at least $(1-c_{N,q})B^2$. There is no division by $A$ or $D_ i$; all degenerate endpoints are therefore included. Equivalently one may first add $\epsilon \mathbf1$ and pass to the closed cone. $\square$