Determinant inequality and positive definite matrix

Hints for 2: because $B$ is positive definite, you can write $$ B+C=B^{1/2}(I+B^{-1/2}CB^{-1/2})B^{1/2}\implies\det(B+C)=\det(B)\det(I+B^{-1/2}CB^{-1/2}). $$ Now argue that the eigenvalues of $I+B^{-1/2}CB^{-1/2}$ are no less than $1$. What then can you say about the product of those eigenvalues?

---

Hints for 3: Let's prove a more general problem.

(a) First show that if $A$ is psd, then $I-A$ is psd implies $A^{-1}-I$ is psd.

Proof: $$ v'(A^{-1}-I)v=(A^{-1/2}v)'(I-A)(A^{-1/2}v)\geq0.\quad\square $$

(b) Suppose that $X$ and $Y$ are positive definite and $X-Y$ is psd. We claim that $Y^{-1}-X^{-1}$ is psd.

Proof: Use (a) along with the following observation: $$ 0\preceq X-Y=X^{1/2}(I-X^{-1/2}YX^{-1/2})X^{1/2}\implies 0\preceq I-X^{-1/2}YX^{-1/2}.\quad\square $$

---

Can you find the appropriate $X$ and $Y$ for your problem?