TITLE: Show that if $X \succeq Y$, then $\det{(X)}\ge\det{(Y)}$
QUESTION [8 upvotes]: Assume that two symmetric positive definite matrices $X$ and $Y$ are such
  that $X-Y$ is a positive semidefinite matrix. Show that $$\det{(X)}\ge\det{(Y)}$$

I felt this result is clear, but I can't use mathematics methods explain this why?

REPLY [8 votes]: Lemma 1 If $A$ is positive semi-definite, then $\det(A + I) \ge 1$. 

Proof: Since $A$ is positive semi-definite, there exists an orthogonal matrix $S$ and
  a diagonal matrix $D$ whose entries are non-negative such that $A = SDS^{-1}$. 
  Since $A + I = S(D + I)S^{-1}$ and $D +I$ is a diagonal matrix whose eigenvalues 
  are all $\ge 1$, $\det(A+I) = \det(D+I) \ge 1$.

Lemma 2 If $B$ is positive definite, $B$ has a positive definite square root.

Proof: Since $B$ is positive definite, there exists an orthogonal matrix $T$ and a
  digonal matrix $E$ whose diagonal entries are positive such that $B = TET^{-1}$.
  Let $E^{1/2}$ be the diagonal matrix whose entries are positive square roots of those of $E$. It is easy to check $T E^{1/2} T^{-1}$ is positive define and $(T E^{1/2} T^{-1})^2 = B$.

By Lemma 2, $Y$ has a positive definite square root. Let us call it $R$.
We have
$$\det(X) = \det(R (R^{-1}XR^{-1}) R )
= \det(R)^2 \det(R^{-1}XR^{-1})\\
= \det(Y) \det( (R^{-1}(X-Y)R^{-1} + I )$$
It is easy to see $X - Y$ positive semi-definite implies $R^{-1} (X-Y) R^{-1}$  positive semi-definite. By Lemma 1, we have $\det( R^{-1}(X-Y)R^{-1} + I ) \ge 1$ and we are done.