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?
 A: 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.
A: If $A$ and $B$ are not invertible then $\det(A)=\det(B)=0$ but $A\geq0$ and $B\geq 0$ implies that $A+B\geq0$ so 
$$
\det(A+B)\geq 0= \det(A)+\det(B)
$$
if we assume now that $B$ is invertible (or $A$ this not change anything !! ) then $B$ will be positive definite in particular it can define a inner-product in $\mathbb{C^n}$ using Simultaneous diagonalization we can find an invertible matrix $P\in \mathcal{M}_n(\mathbb{C})$, and a diagonal matrix $D=Diag(\lambda_1,\dots,\lambda_n)$ such that :
$$
PDP^*=A\\
PP^*=B
$$
where $(\lambda_i)_{i\leq n}=\sigma(A)$ are the eigenvalues of $A$ $(\lambda_i\geq 0)$.
\begin{eqnarray}
\det(A+B)&=& \det(PDP^*+PP^*)\\
&=& \det(P(D+I_n)P^*)\\
&=& \det(PP^*)\det(D+I_n)\\
&=& \det(P)^2 \prod_{i=1}^n (1+\lambda_i)\\
&\geq&  \det(P)^2( 1+\prod_{i=1}^n \lambda_i)\\
&=& \det(P)^2( 1 +\det(D))\\
&=& \det(PP^*)+\det(PP^*)\det(D)\\
&=& \det(B)+\det(PDP^*)\\
&=& \det(B)+\det(A)
\end{eqnarray}
where the inequality 
$$
 \prod_{i=1}^n (1+\lambda_i)\geq  1+\prod_{i=1}^n \lambda_i
$$
hold because all $\lambda_i$ are non negative.
In the case whene $A$ or $B$ are not positive semidefinite the relation do not hold :
$$
A=\left(\begin{matrix} 
1& 0\\
0& 1
\end{matrix}
\right)
\text{ and } 
B=\left(\begin{matrix} 
-1& 0\\
0 &-1
\end{matrix}
\right)
$$
and even in the case when $A$  and $B$ are not invertible as shown this example :
$$
A=\left(\begin{matrix} 
0& 0\\
0& 1
\end{matrix}
\right)
\text{ and } 
B=\left(\begin{matrix} 
-1& 0\\
0 &0
\end{matrix}
\right)
$$
