Given real symmetric positive definite (SPD) matrices $A_i$, let
$$ f(x) := \det \left( \sum_i x_i A_i \right) $$
Is the function $f$ convex?
Minkowski's inequality doesn't seem to do the trick. I suppose there are other relevant ones.
EDIT: It may be necessary to restrict the domain of $x$ to $0 \leq x_i \leq 1, \forall \; i \in \{1,\ldots,n\}$
It is not convex. Consider the case
$$ A_1 = \pmatrix{10 & 0\cr 0 & 1\cr},\ A_2 = \pmatrix{1 & 0\cr 0 & 10\cr}$$ $$ f(x) = (10 x_1 + x_2)(x_1 + 10 x_2)$$ The Hessian matrix is $$\pmatrix{20 & 101\cr 101 & 20\cr}$$ which is indefinite, so this is nowhere convex.
EDIT: Somewhat more generally, for symmetric $2 \times 2$ matrices $$A = \pmatrix{a_{11} & a_{12}\cr a_{12} & a_{22}\cr}$$ $\det(A) = a_{11} a_{22} - a_{12}^2$ (as a function of $a_{11}, a_{12}, a_{22}$) has Hessian matrix $$ \pmatrix{0 & 0 & 1\cr 0 & -2 & 0\cr 1 & 0 & 0\cr}$$ which has one positive and two negative eigenvalues. Any two-dimensional linear subspace of the symmetric matrices will have nontrivial intersection with the span of the eigenvectors for negative eigenvalues, and thus includes directions in which the function is strictly concave.