A direct proof of the result Beenakker cites:$\newcommand\tr{\operatorname{tr}}\newcommand\E{\mathbb{E}}$
 
Take $x\sim\mathcal{N}(0,I)$ and recall that $\tr(C) = \E[x^*Cx]$. If $B = P \Lambda P^{-1}$ is diagonalizable let $\Lambda^{1/2}$ be a (possibly complex) square root of $\Lambda$. Then using the cyclic property of the trace
$$
\begin{align}
\tr(AB) &= \tr(\Lambda^{1/2} P^{-1}AP\Lambda^{1/2})\\
&= \E[(\Lambda^{1/2}x)^* P^{-1}A P (\Lambda^{1/2}x)]\\
&= \sum_{i\neq j} \lambda_i^{1/2}\lambda_j^{1/2}\E[x_ix_j](P^{-1}A P)_{ij} + \sum_{i} \lambda_i\E[x_i^2](P^{-1}A P)_{ii}\\
&= \sum_{i} \lambda_i(P^{-1}A P)_{ii}.
\end{align}
$$
From here it follows that
$$\tr(AB) \leq \lambda_{\max}\sum_{i} (P^{-1}A P)_{ii} = \lambda_{\max}\tr(P^{-1}AP) = \lambda_{\max}\tr(A)$$
and
$$\tr(AB) \geq \lambda_{\min}\sum_{i} (P^{-1}A P)_{ii} = \lambda_{\min}\tr(P^{-1}AP) = \lambda_{\min}\tr(A)$$
as desired. Density will tell us this also holds for non-diagonalizable $B$.