For $n\times n$ symmetric matrices $A_1, \ldots, A_n$, the mixed discriminant $D(A_1, \ldots, A_n)$ can be defined as $1/n!$ times the coefficient of $t_1\ldots t_n$ in the homogeneous polynomial $\det(t_1A_1 + \ldots t_n A_n)$, i.e. $$ D(A_1, \ldots, A_n) = \frac{\partial^n}{n!\partial t_1\ldots \partial t_n}\det(t_1A_1 + \ldots t_n A_n). $$

This is closely related to mixed volumes. If $K_1, \ldots, K_n$ are $n$ compact convex sets in $\mathbb{R}^n$, then the mixed volume $V(K_1, \ldots, K_n)$ is $1/n!$ times the coefficient of $t_1\ldots t_n$ in $\mathrm{vol}(t_1 K_1 + \ldots + t_n K_n)$, the latter being a homogeneous polynomial of degree $n$ by Minkowski's theorem. (Different authors sometimes omit the normalizing $1/n!$ term).

Both mixed discriminants and mixed volumes satisfy a *generalized Brunn-Minkowski* inequality. I.e. both functions below are concave on $\alpha \in [0,1]$:
$$
f(\alpha) = D(\alpha A + (1-\alpha)A', i; B_{i+1}, \ldots, B_n)^{1/i},\\
g(\alpha) = V(\alpha K + (1-\alpha)K', i; L_{i+1}, \ldots, L_n)^{1/i}.
$$
Here I have used the notation $D(A, i; B_{i+1}, \ldots, B_n)$ for $D(A, \ldots, A, B_{i+1}, \ldots, B_n)$ with $i$ copies of $A$, and similarly for $V$. The concavity of these two functions is equivalent to the *Aleksandrov-Fenchel inequalities*. **The point of this question is to understand how far this kind of concavity can be pushed, or is it the best possible.**

Let me use the analogous notation $D(A, i; B, j; C_{i+j+1}, \ldots, C_n) = D(A, \ldots, A, B, \ldots, B, C_{i+j+1}, \ldots, C)$ with $i$ copies of $A$ and $j$ copies of $B$. My question(s) follow.

Is the function $h(\alpha) = D(\alpha A + (1-\alpha)A', i; \alpha B + (1-\alpha)B', j; C_{i+j+1}, \ldots, C_n)^{1/(i+j)}$ concave?

Is it concave if $A, A', B, B', C_{i+j+1}, \ldots, C_n$ are all positive semidefinite?

What about the analogous function with mixed volume in the place of mixed discriminant?

I suspect negative answers to the above. In case I am correct, I'd be happy to see an argument/example.