Prove/disprove: $\text{det}(A+B)\geq\text{det}(A)+\text{det}(B)$ Let $A,B$ be non-neg $n$ by $n$ matrices over $\mathbb{C}$. Show $\text{det}(A+B)\geq\text{det}(A)+\text{det}(B)$.
I attempted this problem in preparation for qualifying exams, but I'm a little unsure of the solution. My thought is to use the minmax theorem. 
Let $\eta_i,\lambda_i$ and $\zeta_i$ are the ith eigenvalues of $A$, $B$, and $A+B$. The determinant of each matrix is the product of the respective eigenvalues, and since the matrices are all positive, each eigenvalue is $>0$ and so each determinant is as well. Using the Rayleigh quotient, we have $R_{A+B}(v)=\frac{<v,Av>+<v,Bv>}{<v,v>}=R_A(v)+R_B(v)$ for each $v\neq 0$. Since the Rayleigh coefficients determinant the eigenvalues in increasing order, I would think this means that $\eta_i+\lambda_i=\zeta_i$. Hence, $\text{det}(A+B)=\prod_{i=1}^n\zeta_i=\prod_{i=1}^n(\eta_i+\lambda_i)\geq\prod_{i=1}^n\eta_i+\prod_{i=1}^n\lambda_i=\text{det}(A)+\text{det}(B)$.
Does anyone see a flaw in this argument? If it is correct, does anyone know perhaps an easier argument?

Write
$$
\det (A+B) = \det A^{1/2} (1+ A^{-1/2}BA^{-1/2}) A^{1/2} = \det A \det (1+A^{-1/2}BA^{-1/2}) ,
$$
assuming for the moment that $A$ is invertible. We now want to show that the second determinant is
$$
\ge 1+\frac{\det B}{\det A} = 1 + \det A^{-1/2}BA^{-1/2} .
$$
In other words, we've reduced matters to the special case $A=1$, which can be handled with your method since $A,B$ commute now:
$$
\det (1+B) = \prod (1+\lambda_j(B)) \ge 1 + \prod \lambda_j(B) = 1 + \det B
$$
The general case ($A$ not necessarily invertible) follows from this by approximation.
