= Solution
Put $u=F_N/\gamma_N$, where the Gaussian density is strictly positive. Extend $u\log u$ continuously at zero by $0\log0=0$. Both densities have <integral> one, so the <relative entropy> can be written as
$$
\begin{aligned}
H_N(F_N)&=\int_{\mathbb R^N}\gamma_N u\log u\,d\mathbf v\\
&=\int_{\mathbb R^N}\gamma_N(u\log u-u+1)\,d\mathbf v.
\end{aligned}
$$
The bracket is nonnegative and vanishes only at $u=1$: its derivative for $u>0$ is $\log u$, with a unique minimum at one. Hence
$$
\boxed{H_N(F_N)\geq0,\qquad H_N(F_N)=0\ \Longleftrightarrow\ F_N=\gamma_N\ \text{a.e.}}
$$
The inequality holds also for infinite entropy. Its negative integrand part is integrable, since $u\log u\geq-1/e$ and $\gamma_N$ has <integral> one, so the extended-value <integral> is well defined. This is <relative entropy in Kac's model>; no differentiation is needed for nonnegativity.
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