Suppose that $$ X = \begin{bmatrix} A & C^T \\ C & B \end{bmatrix} $$ where $X, A, B \succeq 0$ are (real) positive semidefinite matrices.
Is it true that the Cauchy-Schwarz like inequality:
$$\|C\|_\ast \leq \sqrt{\|A\|_\ast \|B\|_\ast}$$
holds, where $\|\cdot\|_\ast$ denotes the sum of the singular values?
By padding the matrices with zeroes if necessary, we may assume that $A,B$ and $C$ are square matrices of the same sizes. If $C=USV^T$ is a singular value decomposition, by considering $(V\oplus U)^TX(V\oplus U)$, we may further assume that $C$ is a nonnegative diagonal matrix $\operatorname{diag}(c_1,c_2,\ldots,c_n)$. Hence $A,B,C$ are positive semidefinite and their nuclear norms are precisely their traces.
Since $X$ is positive semidefinite, its $2\times2$ principal minors are nonnegative. Hence $c_i^2\le a_{ii}b_{ii}$ for each $i$. It follows that \begin{aligned} \|C\|_\ast^2=\left(\sum_ic_i\right)^2 &\le\left(\sum_i\sqrt{a_{ii}b_{ii}}\right)^2\\ &\le\left[\sum_i(\sqrt{a_{ii}})^2\right]\left[\sum_i(\sqrt{b_{ii}})^2\right] \ \text{(Cauchy-Schwarz inequality)}\\ &=\left(\sum_ia_{ii}\right)\left(\sum_ib_{ii}\right)=\|A\|_\ast\|B\|_\ast. \end{aligned}