Show that if $X \succeq Y$, then $\det{(X)}\ge\det{(Y)}$ 
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?
 A: 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.
A: Eigenvalues only complicate this problem.
The question asks for a proof of $\det\left(Y+Z\right) \geq \det Y$ where $Y$ is a positive definite $n\times n$-matrix and $Z$ is a positive semidefinite $n\times n$-matrix. (My $Z$ is your $X-Y$.)
Write $Y$ as $U^T U$ for $U$ an invertible square matrix (this is possible since $Y$ is positive definite), and write $Z$ as $U^T V U$ for $V$ a positive semidefinite matrix (this is possible since we can take $V = \left(U^{-1}\right)^T Z U^{-1}$). Then, what we want to prove rewrites as $\det\left(U^T U + U^T V U\right) \geq \det\left(U^T U\right)$. In other words, what we want to prove rewrites as $\left(\det U\right)^2 \det\left(I_n + V\right) \geq \left(\det U\right)^2$. After cancelling $\left(\det U\right)^2$, this becomes $\det\left(I_n + V\right) \geq 1$. But $\det\left(I_n + V\right)$ is the sum of all principal minors of $V$ (to see this, recall the formula for the characteristic polynomial of $V$ in terms of its principal minors, and substitute $-1$ into this characteristic polynomial). One of these principal minors is the $0\times 0$-minor $1$, while all others are nonnegative (since $V$ is positive semidefinite). Hence, their sum is $\geq 1$, which is precisely what we needed to show.
When $n > 0$, this argument actually shows the stronger claim $\det\left(I_n + V\right) \geq 1 + \det V$ (since $\det V$, too, is a principal minor -- namely, the $n\times n$ minor), which translates back into $\det\left(Y+Z\right) \geq \det Y + \det Z$. By continuity, we can see that this inequality $\det\left(Y+Z\right) \geq \det Y + \det Z$ also holds when $Y$ and $Z$ are both positive semidefinite (rather than $Y$ being positive definite and $Z$ positive semidefinite).
