 Sudakov-Fernique inequality

For centered multivariate normal distributions, if $\mathbb{E}(U_i-U_j{)}^2\le \mathbb{E}(V_i-V_j{)}^2$ for every pair, then the displayed comparison holds. For a quick proof, make $U,V$ independent and set $Z_r=\sqrt{1-r}U+\sqrt{r}V$. Apply Gaussian integration by parts to $F_{\tau }(z)={\tau }^{-1}\log {\sum }_i{e}^{\tau z_i}$. With softmax function weights $p_i$ and $D=\mathrm{Cov}(V)-\mathrm{Cov}(U)$, the derivative of $\mathbb{E}{F}_{\tau }(Z_r)$ is $\frac{\tau }{4}\mathbb{E}{\sum }_{i,j}{p}_i{p}_j(D_{ii}+D_{jj}-2D_{ij})\ge 0$. Since $\max z_i\le F_{\tau }(z)\le \max z_i+{\tau }^{-1}\log n$, let $\tau \to \infty$. Singular covariance matrices follow by adding identical small independent normal noise to both vectors and taking a limit.