Bound on sum of $n$th super-diagonal entries in a $2n$ by $2n$ PSD matrix 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.
 A: 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, 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 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.
A: A slightly more precise result is noted below. You may like it given that you tried Schur complements, and convexity arguments.

Observation.
  $\DeclareMathOperator{tr}{tr}$
  \begin{equation*}
\max\{|\tr B| : A \succeq BC^{-1}B^T\} \le \min_{X \succ 0}\sqrt{\tr(AX)\tr(CX^{-1})}.
\end{equation*}

From this inequality one sees that the maximum is attained when $B=(AC)^{1/2}$, and equals the value of the min on the right, yielding the value $\tr[(A^{1/2}CA^{1/2})^{1/2}]$. 
As a corollary, we also obtain (which is the first set of inequalities in Mahdi's answer):
\begin{equation*}
|\tr B| \le \tr[(A^{1/2}CA^{1/2})^{1/2}] = \sum_i\lambda_i^{1/2}(A^{1/2}CA^{1/2})=
\sum_i\lambda_i^{1/2}(AC),
\end{equation*}
where $\lambda_i(\cdot)$ denotes eigenvalues.
