TITLE: If $A$ and $B$ are positive definite, then is $B^{-1} - A^{-1}$ positive semidefinite?
QUESTION [20 upvotes]: I've found this while googling some properties of positive semidefinite matrices. (Unfortunately, I cannot remember where I've discovered it.) If this is true, it'll greatly save my time in my work. Is it true? How can you prove it?

Let's say I have two real symmetric and positive definite matrices $\mathbf A$ and $\mathbf B$ of the same size. Also, assume that $\mathbf A - \mathbf B$ is positive semidefinite too. Then, $\mathbf B^{-1} - \mathbf A^{-1}$ is also positive semidefinite.

REPLY [2 votes]: Let me offer a constructive proof.  
Lemma 1. For invertible and symmetric matrix $A$, and vectors $x$ and $y$, we have 
$$
x^{'}A^{-1}x = \underset{y}{max\, \, } 2{x}'y - {y}'Ay
$$
Lemma 2. For invertable matrices $A$ and $B$, we have
$$
(A + C) ^{-1} = A^{-1} - A^{-1} (A ^{-1} + C ^{-1}) ^{-1} A ^{-1}
$$
Theorem. If $A > 0$, $B > 0$, $A$ and $B$ are symmetric matricex, and $A > B$, then $ B ^{-1} > A ^{-1} $.
Proof. By Lemma 1, we have  
$$x ^{'} ( B ^{-1} - A ^{-1} ) x  = \underset{y}{max\, \, } 2{x}'y - {y}' ( B ^{-1} - A ^{-1} ) ^{-1} y$$
Let $ y = A ^{-1} x $, we have
$$ x ^{'} ( B ^{-1} - A ^{-1} ) x \ge 2x ^{'} A ^{-1} x - x ^{'} A ^{-1} ( B ^{-1} - A ^{-1} ) ^ {-1} A ^{-1} x $$
Then by Lemma 2, we have
$$ x ^{'} ( B ^{-1} - A ^{-1} ) x \ge x ^{'} A ^{-1} x + x ^{'} ( A - B ) ^{-1} x $$
By assumption, $ A > 0 $, so $ A ^{-1} > 0 $. Also $ A > B $, i.e. $ A - B > 0 $, so $ (A - B ) ^ {-1} > 0$.
Thus the LHS of above equation is greater than 0, so $ B ^{-1} - A ^{-1} > 0 $, i.e. $ B ^{-1} > A ^{-1} $.
EDIT: typos.