Let $f, K, \phi$ be as in the parent entry.

The proof uses the nonnegativity of relative entropy $D(f||\phi)$ and an interesting property of quadratic forms. If $A$ is a quadratic form and $p,q$ are probability distributions each with mean $\mv{0}$ and covariance matrix $\mv{K}$ we have

\begin{equation} \int p\ x_i x_j\ dx_i dx_j = K_{ij} = \int q\ x_i x_j\ dx_i dx_j \end{equation}and thus \begin{equation} \int A p = \int A q \end{equation} Now note that since \begin{equation} \phi(\mv{x}) = \left((2\pi)^n |\mv{K}| \right)^{-\frac{1}{2}} \exp{(- \frac{1}{2} \mvt{x} \mv{K}^{-1} \mv{x})}, \end{equation}we see that $\log \phi$ is a quadratic form plus a constant.

\begin{eqnarray*} \textbf{0} &\le D(f||\phi)\\ &= \int f \log \frac{f}{\phi}\\ &= \int f \log f - \int f \log \phi\\ &= -h(f) - \int f \log \phi\\ &= -h(f) - \int \phi \log \phi \qquad \text{by the quadratic form property above}\\ &= -h(f) + h(\phi) \end{eqnarray*}and thus $h(\phi) \ge h(f)$