For matrices $A$ and $B$, $B-A\succeq 0$ (i.e. psd) implies $\text{Tr}(B)\geq \text{Tr}(A)$ and $\det(B)\geq\det(A)$ Suppose that $B$ and $A$ are symmetric matrices of the same dimensions.
Claim: $B-A\succeq 0$ implies $\text{Tr}(B)\geq \text{Tr}(A)$ and $\det(B)\geq\det(A)$
Attempt: Let $C=B-A$, then $C$ has nonnegative eigenvalues so that 
$$
0\leq\text{Tr}(C)=\text{Tr}(B)-\text{Tr}(A)\implies \text{Tr}(B)\geq \text{Tr}(A).
$$
I don't know how to proceed with the second claim.
The context of this claim is that $A$ and $B$ are covariance matrices so maybe we need to assume that they both are positive semidefinite. But even with that, I'm still stuck. 
Thank you very much!
 A: As you said, you need $A\succeq0$, otherwise your claim isn't true (e.g. when $B=-I_2$ and $A=-2I_2$).
Suppose $B\succeq A\succeq0$, then for any $\epsilon>0$, we have $B+\epsilon I\succeq A+\epsilon I\color{red}{\succ}0$ and hence
$$
(A+\epsilon)^{-1/2}(B+\epsilon)(A+\epsilon)^{-1/2}\succeq I\succ0.
$$
Therefore all eigenvalues of $(A+\epsilon)^{-1/2}(B+\epsilon)(A+\epsilon)^{-1/2}$ are at least 1. You may continue from here.
A: My approach is similar to that of user1551, perhaps a bit more pedestrian.
As above let $C=B-A$. As noted by user1551, it is necessary to assume that $A\succeq 0$. If it is not the case that $A\succ 0$, then $\det(A)=0$ so that $\det(B)\geq 0=\det(A)$ holds trivially. So let us assume $A\succ 0$. This allows us to define: 
$$
D=A^{-1/2}CA^{-1/2}. 
$$
Because $C$ is psd, $D$ is also psd and thus has nonnegative eigenvalues. This means $I+D$ has eigenvalues that are no less than $1$ so we have $\det(I+D)\geq 1$. Now our claim follows:
$$
B=A+C=A^{1/2}(I+D)A^{1/2}\implies\det(B)=\det(A)\det(I+D)\geq\det(A).
$$
