Let $A,B,C\in\mathbb{R}^{n\times n}$ be such that $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$. I would like to prove that $$\mathrm{trace}\,B \le \sum_{i=1}^n \sqrt{\lambda_i(A)\lambda_i(C)},$$ where for any symmetric $M\in\mathbb{R}^{n \times n}$, $\lambda_1(M) \le \lambda_2(M) \le \cdots \le \lambda_n(M)$ denote the sorted eigenvalues of $M$.

Using SVD, Schur complement and Von-Neumann's trace inequality I am able to show that the above by is true if $$\mathrm{trace}\,\left(\left[\Sigma G^2 \Sigma\right]^{1/2} G^{-1}\right) \ge \mathrm{trace}\,\Sigma$$ for every $G\succ 0$ and diagonal $\Sigma \succeq 0$; according to simulations random matrices seem to satisfy this. This inequality follows from a simple symmetry argument if $f(G) = \mathrm{trace}\,\left(\left[\Sigma G^2 \Sigma\right]^{1/2} G^{-1}\right)$ happens to be operator-convex, but I have so far not been able to prove this.