Is $|A| < |B|$ if $A-B$ is positive definite? I want to prove this. Say if $A-B$ is a positive definite matrix then can we find a relation between $\det(A)$ and $\det(B)$?
e.g. is $|A| < |B|$.

Consider $A=2$, $B=1$. Then $A-B=1$ is positive definite but det$(A)=2 \not<$ det$(B)=1$. If you have mistyped the direction of the $<$ sign for the determinants then with:
$$A=\begin{pmatrix}1& 0 \\ 0& 1\end{pmatrix}\qquad B=\begin{pmatrix}-2& 0 \\ 0& -2\end{pmatrix}$$
$$A-B=\begin{pmatrix}3& 0 \\ 0& 3\end{pmatrix}$$
Is positive definite and det$(A)=1$, det$(B)=4$.
It appears to me if $A, B$ are positive definite diagonalisable finite dimensional matrices then $A-B$ is positive definite implies det$(A)>$det$(B)$.
To see this note that if $A$ is positive definite it is invertible with positive definite inverse. The product of positive operators is positive so
$$1-A^{-1}B=A^{-1}(A-B)$$
is positive. For it to be positive all eigenvalues of $A^{-1}B$ must be smaller than 1. So det$(A^{-1}B)=\frac{det(B)}{det(A)}<1$, which implies det$(A)>$det$(B)$.
