Suppose $A$ and $B$ are two $n \times n$ symmetric positive definite real matrices and let $M$ be an $n \times m$ real matrix of full rank. Suppose that $A - B$ is positive definite. Then show that the difference, $$\biggl(A - AM(M^TAM)^{-1}M^TA\biggr) - \biggl(B - BM(M^TBM)^{-1}M^TB\biggr) $$ is positive semi-definite.
What I've tried so far is re-writing the expression as,
$$\biggl(A - B\biggr) + \biggl(BM(M^TBM)^{-1}M^TB - AM(M^TAM)^{-1}M^TA\biggr) $$
We know that the first term in parentheses is P.D. by assumption. If we can show that the second term is P.D., then the result would hold since the sum of two P.D. matrices is again P.D. However, I'm stuck on showing that the second term is P.D. Since $A > B$, we can show that $$M(M^TBM)^{-1}M^T > M(M^TAM)^{-1}M^T$$ But unfortunately this isn't quite the result we need. Am I on the right track here? Some hints/guidance would be greatly appreciated.
The statement is false. E.g. when $m=n$ and $M=I$, the difference in question is zero for all $A\succ B$.
The correct conclusion is that the difference is positive semidefinite. For any $t\ge0$ and any positive definite matrix $S$, define $$ P_S(t)=tI+\pmatrix{S&SM\\ M^TS&M^TSM}=tI+\pmatrix{S^{1/2}\\ M^TS^{1/2}}\pmatrix{S^{1/2}&S^{1/2}M}. $$ When $t>0$, clearly $P_S(t)$ is positive definite. It follows that $P_A(t)-P_B(t)=P_{A-B}(t)\succ0$. Thus $P_A(t)^{-1}\prec P_B(t)^{-1}$ and this inequality is also inherited by the leading principal $n\times n$ submatrices of $P_A(t)^{-1}$ and $P_B(t)^{-1}$. Now these principal submatrices can be calculated using Schur complements and you may continue from here.