matrix of the MIC elements and $A_{ij}:=h_i\delta_{ij}$. Note that $\Phi^{-1}_{\rm p}$ has real, positive eigenvalues because it has the same spectrum as the positive definite matrix $A^{-1/2}GA^{-1/2}$. Also note that $$\sum_i\frac{1}{\lambda_i(\Phi_{\rm p})}=\tr\Phi_{\rm p}^{-1}=\sum_ih_i\tr\rho_i\sigma_i\leq\sum_ih_i=d\;.$$ One of the eigenvalues of $\Phi_{\rm p}$, which we denote $\lambda_{d^2}(\Phi_{\rm p})$, must equal $1$ because an equal-entry row vector is always a left eigenvector with eigenvalue $1$ of a matrix with columns summing to unity. Therefore, we may write $$\label{reciprocalbound}
    \sum_{i<d^2}\frac{1}{\lambda_i(\Phi_{\rm p})}\leq d-1.$$ The reciprocal of this expression is proportional to the harmonic mean of the first $d^2-1$ eigenvalues of $\Phi_{\rm p}$. Thus, because the geometric mean is always greater than or equal to the harmonic mean, $$\left(\prod^{d^2-1}_{i=1}\lambda_i(\Phi_{\rm p})\right)^{\!\frac{1}{d^2-1}}\geq\left(\frac{1}{d^2-1}\sum_{i=1}^{d^2-1}\frac{1}{\lambda_i(\Phi_{\rm p})}\right)^{\!-1}\geq d+1\;,$$ which, noting that $\lambda_{d^2}(\Phi_{\rm p})=1$, implies $$\label{mindet}
    \det\Phi_{\rm p}\geq(d+1)^{d^2-1}=\det\Phi_\text{SIC}\;.$$ Equality is achieved in this iff all the $\lambda_i(\Phi_{\rm p})$ are equal, so Eq.  is saturated iff $\lambda(\Phi_{\rm p})=\lambda(\Phi_{\rm SIC})$. We next show this implies that in fact the MIC is a SIC.

For any $\Phi_{\rm p}^{-1}$, we may write $\Phi_{\rm p}^{-1}=P^{-1}DP$ where the rows of $P$ are the left-eigenvectors of $\Phi_{\rm p}^{-1}$ and $D$ is the diagonal matrix of eigenvalues of $\Phi_{\rm p}^{-1}$. Since $\Phi_{\rm p}^{-1}$ is column-stochastic, the row vector $(1/d,\dots,1/d)$ is the (scaled) 