Does the following trace inequality hold?
$\mathrm{tr}\{[AB^{\dagger}+(A^{\dagger})^{\top}B^{\top}-2AB^{\top}]C\}\geq \mathrm{tr}(2I-A^{\top}A-B^{\top}B)\min c_{i}$,
where $A,B\in \mathbb{R}^{n\times m}$ ($n\geq m$) are of full column rank (i.e., $\mathrm{rank}(A)=\mathrm{rank}(B)=m$) with singular values $\sigma_{i}(A),\sigma_{j}(B)\in[0,1]$, $C=\mathrm{diag}(c_{1},c_{2},\ldots,c_{n})$ with $c_{i}\geq 0$, and $I$ represents the identity matrix of appropriate size (i.e., $m\times m$). Specifically $A,B$ could be of the form $A=X^{\top}Y$ (where $X,Y$ are orthonormal).
Currently I can only see that this inequality holds for the scalar case, i.e., $n=m=1$, under the assumption that $AB>0$. We may assume similar conditions for the matrix version, e.g., $\mathrm{tr}(A^{\top}B)>0$.