Let $A$ and $B$ be positive (but not strictly positive) matrices. $B^{+}$ is the Moore-Penrose inverse of $B$ and $X^{*}$ denotes the transpose of $X$.
I'm trying to show that $A \geq XB^{+}X^*$ is equivalent to $$ \begin{bmatrix} A & X\\ X^* & B \end{bmatrix} \geq 0, $$
What I tried doing was using matrix similarity $$ \begin{bmatrix} A & X\\ X^* & B \end{bmatrix} \sim \begin{bmatrix} I & -XB^{+}\\ O & I \end{bmatrix} \begin{bmatrix} A & X\\ X^* & B \end{bmatrix} \begin{bmatrix} I & O\\ -B^{+}X^* & I \end{bmatrix} $$
but the RHS is
\begin{bmatrix} A-XB^{+}X^{*} & X(I-B^{+}B)\\ (I-BB^{+})X^* & B \end{bmatrix}
If $(I-BB^{+})X^*$ and $X(I-B^{+}B)$ vanished, then I'd be done since then I could look at the upper-left corner with $A-XB^{+}X^*$ and see it must be nonnegative.
But since I'm dealing with pseudoinverses, how do I show this will still be the case even if $(I-BB^{+})X$ and $X(I-B^{+}B)$ don't vanish?