Reversing the usual inequality involving the determinant of the sum of positive definite matrices Given positive definite matrices $A$ and $B$, of dimension $n$, is it possible to derive an inequality of the form $$\det(A+B)\le f(\det(A),\det(B)),$$ where $f$ is some linear function (perhaps involving n)?.

The Minkowski inequality goes in the other direction, with $f(X,Y)=X+Y$. How about this one, though?

EDIT: I'm also open to allowing $f$ to contain information about the spectral norms of $A$ or $B$, or information of this kind.
This is false even for $2\times2$ diagonal matrices. For these, what
you want reduces to
\begin{eqnarray*}
\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) &amp; = &amp; \det\left(\left(\begin{array}{cc}
a_{1}\\
 &amp; a_{2}
\end{array}\right)+\left(\begin{array}{cc}
b_{1}\\
 &amp; b_{2}
\end{array}\right)\right)\\
 &amp; \overset{!}{\leq} &amp; \alpha\cdot\det\left(\begin{array}{cc}
a_{1}\\
 &amp; a_{2}
\end{array}\right)+\beta\cdot\det\left(\begin{array}{cc}
b_{1}\\
 &amp; b_{2}
\end{array}\right)\\
 &amp; = &amp; \alpha a_{1}a_{2}+\beta b_{1}b_{2}
\end{eqnarray*}
with suitable $\alpha,\beta \in \Bbb{R}$ and all $a_1,a_2,b_1,b_2 &gt;0$.

Now consider $a_{1}=b_{2}=n$ and $a_{2}=b_{1}=\frac{1}{n}$. Then
the desired inequality becomes
$$
n^{2}\leq\left(n+\frac{1}{n}\right)^{2}\leq\alpha+\beta
$$
for all $n\in\mathbb{N}$, which is absurd.
You cannot. Basically, you want some constants $a,b,c$, which are possibly dependent on $n$, such that $$\det(A+B)\leq a\det (A) + b\det (B) + c .$$

However, take $A=\begin{bmatrix}1 &amp; 0 &amp;\dots &amp;0\\
0&amp;0&amp;\dots &amp;0\\
\vdots &amp;\vdots &amp;\ddots &amp;\vdots\\
0&amp;0&amp;\dots&amp;0\end{bmatrix}$
and $B=I_n - A$.

Then, $\det(\alpha (A+B)) = \alpha$, and $\det(A)=\det(B)=0$, which means that for every real value $\alpha$, you have $\alpha \leq c$. Obviously, no such $c$ exists.