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If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\|K\| < 1$) 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, we have $\newcommand{tr}{\mathrm{tr}}$ $$ \tr(B) = \tr(A^{1/2} K C^{1/2}) \leq \|K\|\tr(A^{1/2} C^{1/2})\leq \sum_{i=1}^n \sqrt{\lambda_i(A)\lambda_i(C)} $$