I asked a question earlier but it wasn't quite correctly stated, so I'll reset.
Let $A$ be an $n\times n$ matrix of rank $k<n$ and let $X,Y$ be two symmetric positive definite matrices. Let $Z^+$ denote the Moore-Penrose pseudoinverse of a matrix $Z$.
Then I want to prove the following: $$ (AXA^T)^+ \geq (AXA^T + Y)^{-1} \textrm{ on the range of } A\tag{1} $$ or equivalently: $$ A^T (AXA^T)^+ A \geq A^T (AXA^T + Y)^{-1} A\tag{2} $$
Numerical simulations I've done suggest that this is true, but I'm not sure how to prove it.
Attempts/Ideas: I've attempted to look at the SVD/Eigendecomposition of $A$, but substituting this in seems to go nowhere: If $AXA^T = Q D Q^T$ for $D$ a diagonal matrix, $Q$ orthogonal, then we can rewrite (2) as: $$ A^TQD^+Q^T A \geq A^T Q (D + \hat{Y})^{-1}Q^T A $$ for $\hat{Y} = Q^T Y Q$.
I suspect some sort of partitioned matrices/Schur complement approach might work, in combination with my above idea but it seems very complicated.
Presumably the matrices have real entries. Note that $\operatorname{range}(AXA^T)\subseteq\operatorname{range}(A)$ and $\operatorname{nullity}(AXA^T)=\operatorname{nullity}(X^{1/2}A^T)=\operatorname{nullity}(A^T)=\operatorname{nullity}(A)$. Therefore $AXA^T$ and $A$ have the same range. It follows that \begin{align*} &A^T(AXA^T)^+A\ge A^T(AXA^T+Y)^{-1}A\\ &\Leftrightarrow AXA^T(AXA^T)^+AXA^T\ge AXA^T(AXA^T+Y)^{-1}AXA^T\\ &\Leftrightarrow AXA^T\ge (P-Y)P^{-1}(P-Y)\quad\text{where $P=AXA^T+Y>0$}\\ &\Leftrightarrow P-Y\ge P-2Y+YP^{-1}Y\\ &\Leftrightarrow Y\ge YP^{-1}Y\\ &\Leftrightarrow Y^{-1}\ge P^{-1}\\ &\Leftrightarrow P\ge Y\\ &\Leftrightarrow AXA^T\ge 0\\ \end{align*} and the last inequality is obviously true.