A sharp Gaussian harmonic-mean inequality for Neumann eigenvalues of the Ornstein-Uhlenbeck operator

Let $N\ge 2$ and let $\mathrm{\Omega }\subset {\mathbb{R}}^N$ be a connected Lipschitz domain, possibly unbounded, symmetric with respect to the origin, and such that $0<\text{gammaN}(\mathrm{\Omega })<1$. We assume that the Gaussian Sobolev embedding $H^1(\mathrm{\Omega },\text{gammaN})\hookrightarrow L^2(\mathrm{\Omega },\text{gammaN})$ is compact; a sufficient condition is the existence of a bounded Gaussian Sobolev extension operator from $\mathrm{\Omega }$ to ${\mathbb{R}}^N$. Denote by $$0={\mu }_0(\mathrm{\Omega })<{\mu }_1(\mathrm{\Omega })\le {\mu }_2(\mathrm{\Omega })\le \cdots$$ the Neumann eigenvalues of the positive Ornstein–Uhlenbeck operator $-\mathrm{\Delta }+x\cdot \mathrm{\nabla }$ in $\mathrm{\Omega }$. We prove the sharp reciprocal-sum inequality $$\sum _{k=1}^N\frac{1}{{\mu }_k(\mathrm{\Omega })}\ge \frac{N}{{\mu }_1(B_R)},$$ where $B_R$ is the Euclidean ball centred at the origin and satisfying $\text{gammaN}(B_R)=\text{gammaN}(\mathrm{\Omega })$. Equality holds if and only if $\mathrm{\Omega }=B_R$. The proof combines a coupled $N$-dimensional Ritz argument with a Gaussian raywise rearrangement. The angular imbalance is encoded by a symmetric trace-free matrix, whose contribution is controlled by a finite-dimensional convexity inequality.