In my previous post I demonstrated that
$$
\det (t A + (1-t) B) \geq (\det A)^t (\det B)^{1-t}
$$
reduces to
$$
nt + \ln \det ( \mathbb{1} + t^{-1} (1-t) A^{-1} B) \geq (1-t) \ln \det A^{-1} B \qquad (*)
$$
A positive definite matrix has a positive definite square-root matrix. Proof:
The matrix ##B## is positive definite and so there is a similarity transformation such that ##S^{-1} B S = D## where ##D## is the diagonal matrix. Where the matrix ##S## is chosen to be the matrix with the ##n## eigenvectors as columns. The elements of ##D## are eigenvalues of ##B## and so are positive. When a matrix is symmetric, the diagonalizing matrix ##S## can be made an orthogonal matrix by suitably choosing the eigenvectors (then ##S^{-1} = S^T##). As such we can define a positive definite square-root matrix ##B^{1/2} = S D^{1/2} S^T## (##B^{1/2}## has the same eigenvectors as ##B## but with eigenvalues that are the square-root of the eigenvalues of ##B##).
Q.E.D.
Now, the matrix ##A^{-1} B## can be written as ##B^{-1/2} (B^{1/2} A^{-1} B^{1/2}) B^{1/2}##. As such the matrix ##A^{-1} B## has the same eigenvalues as the positive definite matrix ##B^{1/2} A^{-1} B^{1/2}## because:
$$
\det (\mathbb{1} \lambda - A^{-1} B) = \det (\mathbb{1} \lambda - B^{-1/2} (B^{1/2} A^{-1} B^{1/2}) B^{1/2}) = \det (\mathbb{1} \lambda - B^{1/2} A^{-1} B^{1/2})
$$
As such the matrix ##A^{-1} B## has positive eigenvalues, let us denote them as ##\nu_i##.
The matrix ##A^{-1} B## is not necessarily symmetric and so can't in general be diagonalised, but there is a similarity transformation that puts it into Jordan normal form. So that ##(*)## becomes
$$
nt + \ln \det ( \mathbb{1} + t^{-1} (1-t) J) \geq (1-t) \ln \det J
$$
where the diagonal elements of ##J## are the ##\nu_i > 0## (##i = 1 , \dots , n##). This reduces to
$$
nt + \ln \prod_{i=1}^n ( 1 + t^{-1} (1-t) \nu_i) \geq (1-t) \ln (\prod_{i=1}^n \nu_i)
$$
These eigenvalues are positive and so the logarithm makes sense. The above inequality can be rewritten as
$$
nt + \sum_{i=1}^n ( 1 + t^{-1} (1-t) \nu_i) \geq (1-t) \sum_{i=1}^n \nu_i
$$
or
\begin{align*}
nt + \sum_{i=1}^n 1 & \geq [(1-t) - t^{-1} (1-t)] \sum_{i=1}^n \nu_i
\nonumber \\
& = - (1-t)^2 t^{-1} \sum_{i=1}^n \nu_i
\end{align*}
which obviously holds as the RHS is non-positive for ##t \in (0,1]## as ##\nu_i## (##i = 1 , \dots , n##) is positive.