For Hermitian matrices $X, Y$, I write $X\ge Y\ge 0$ to mean $X-Y$ and $Y$ are positive semidefinite. 


In Lemma 2.5 of [Linear Algebra Appl. 452 (2014) 1-6] I proved that if $X + Y\ge W + Z$,
$X\ge W\ge Y\ge 0$ and $X\ge Z \ge Y\ge 0$, then $$\det X+\det Y\ge \det W+\det Z.$$ 


I was thinking of relaxing the condition a little bit. 


Is the same inequality true under only the condition $X + Y\ge W + Z$, $X\ge Y\ge 0$, $X\ge W\ge 0$, $X\ge Z\ge 0$?



On an inequality among determinants
The answer is yes. I thank fedja for discussion. 


WLOG, assume $X=I$ (we may do so by pre-post multiplying both sides by $\det X^{-1/2}$, with a standard continuity argument). 


After this, we may further assume $Y=D$ to be diagonal (by unitary similarity). 


So the question is equivalent to showing if $I+D\ge W+Z$ and $I\ge D, W, Z\ge 0$, then $1+\det D\ge \det W+\det Z$.


Note that $I+D\ge W+Z$ implies $I+D\ge diag(W)+diag(W)$, where $diag(\cdot)$ means the diagonal part. 


By a simple induction on the size of the matrix, one gets $1+\det D\ge \det (diag(W))+\det (diag(Z))$. The conclusion follows by applying the Hadamard inequality, $\det (diag(W))\ge \det W$.


