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.



Bound on sum of $n$th super-diagonal entries in a $2n$ by $2n$ PSD matrix
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.


