Trace inequality for a product of p.s.d. matrices and their pseudo inverse. Let $A, B_i$ be positive semidefinite real matrices. Let $\dagger$ stand for the Moore-Penrose generalized inverse. I managed to prove that if
$\operatorname{Ran}B_1\subseteq\operatorname{Ker}B_2$ then
$$\operatorname{trace}\left((A + B_1 + B_2)^\dagger B_1 \right) \leq\operatorname{trace}\left(( A + B_1)^\dagger B_1\right) $$
Does it still hold without this assumption?
 A: Yes. In general, if $X,Y$ are positive semidefinite and $P=XX^\dagger$ denotes the orthogonal projection onto $\operatorname{ran}(X)$, then $P(X+Y)^\dagger P\preceq X^\dagger$. This can be easily proved by using Schur complements.
Now, if you put $X=A+B_1$ and $Y=B_2$, you get
$$
B_1^{1/2}P(A+B_1+B_2)^\dagger PB_1^{1/2}\preceq B_1^{1/2}(A+B_1)^\dagger B_1^{1/2}.
$$
Since $B_1^{1/2}P=PB_1^{1/2}=B_1^{1/2}$, the result follows.
Edit. If $X=0$, the inequality $P(X+Y)^\dagger P\preceq X^\dagger$ simply means $0\preceq0$. Suppose $X$ is PSD but nonzero. Since $\operatorname{ran}(X)\subseteq\operatorname{ran}(X+Y)$, by a change of orthonormal basis, we may assume that
$$
X=\pmatrix{X_1&0&0\\ 0&0&0\\ 0&0&0},\ P=\pmatrix{I&0&0\\ 0&0&0\\ 0&0&0},\ Y=\pmatrix{H&R&0\\ R^T&S&0\\ 0&0&0},\ X+Y=\pmatrix{X_1+H&R&0\\ R^T&S&0\\ 0&0&0}
$$
where $X_1$ and $Z:=\pmatrix{X_1+H&R\\ R^T&S}$ are the matrix representations of $X|_{\operatorname{ran}(X)}$ and $(X+Y)|_{\operatorname{ran}(X+Y)}$ respectively and they are positive definite.
As $Z\succ0$, we must have $S\succ0$. Yet $Y\succeq0$ by assumption. Therefore the Schur complement $H-RS^{-1}R^T$ must be $\succeq0$. It follows that $X_1+H-RS^{-1}R^T\succeq X_1\succ0$ and in turn $0\prec X_1^{-1}\preceq(X_1+H-RS^{-1}R^T)^{-1}$. But this means $P(X+Y)^\dagger P\preceq X^\dagger$, because
$$
X^\dagger=\pmatrix{X_1^{-1}&0&0\\ 0&0&0\\ 0&0&0},\ (X+Y)^\dagger=\pmatrix{(X_1+H-RS^{-1}R^T)^{-1}&\ast&0\\ \ast&\ast&0\\ 0&0&0},
$$
