If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\lambda_n(K)=\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see Theorem IX.5.9 of Bhatia's book: Matrix Analysis). By [Von-Neumann's trace inequality](https://en.wikipedia.org/wiki/Trace_inequalities#Von_Neumann's_trace_inequality), we have
$\newcommand{tr}{\mathrm{tr}}$
$$
\tr(B) = \tr(A^{1/2} K C^{1/2}) \leq\sum_{i=1}^n \lambda_i(K)\lambda_i(C^{1/2}A^{1/2} ) 
\leq \sum_{i=1}^n \lambda_i(C^{1/2}A^{1/2} ) .$$

By the weak majorization property of the singular values of the product of matrices (e.g. see Theorem 3 of [this paper](https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1063207/pdf/pnas01556-0014.pdf) of Horn), we have

$$
\sum_{i=1}^n \lambda_i(C^{1/2}A^{1/2} ) 
 \leq \sum_{i=1}^n \lambda_i(A^{1/2})\lambda_i(C^{1/2}).
$$

Now, the result at hand.

Note that here $\lambda_i(\cdot)$ stands for the singular values, which is equal to the eigenvalues for positive semidefinite matrices.