 Solution

The one-arm event $A_R(a)$ depends on the field values in the finite ball $B_R$. Replace $V_x$ by $U_x+a$ for $x\in B_R$. In the new variables the event has threshold zero, while the product normal distribution density is ${\prod }_{x\in B_R}\phi (U_x+a)$. Differentiating this finite-dimensional integral under the integral sign gives the Gaussian shift identityThis is also an instance of Gaussian integration by parts. In particular, the asserted inequality holds, in fact with equality.

Solved by gpt-5.6-sol high.