using the arithmetic-geometric mean inequality, $$\begin{split}
    \frac{d-1}{d+1}&\geq\sum_{i>1}|\lambda_i(\alpha^{-1}\Phi^{-1})|^2\\
    &\geq(d^2-1)\left(\prod_{i>1}|\lambda_i(\alpha^{-1}\Phi^{-1})|^2\right)^{\frac{1}{d^2-1}}\\
    &=(d^2-1)|\det\alpha^{-1}\Phi^{-1}|^{\frac{2}{d^2-1}}\;,
\end{split}$$ which implies $$\begin{split}
    |\det\alpha^{-1}\Phi^{-1}|&\leq\left(\frac{d-1}{(d+1)(d^2-1)}\right)^{\frac{d^2-1}{2}}\\
    &=\left(\frac{1}{d+1}\right)^{d^2-1}=\det\Phi^{-1}_\text{SIC}\;.
\end{split}$$ From Eq. , we can write $$\begin{split}
\tr E_iE_j&=\sum_{kl}\alpha_{ik}\alpha_{jl}\tr H_kH_l\iff G=\alpha G_{\rm SIC}\alpha^T\\
&\iff \det G=(\det \alpha)^2\det G_{\rm SIC}\;,
\end{split}$$ where $G$ is the MIC Gram matrix and $G_{\rm SIC}$ is the SIC Gram matrix. Recall the definition of the $A$ matrix from the proof of Lemma \[det\]. The arithmetic-geometric mean inequality shows $\det A\leq(1/d)^{d^2}$ with equality iff $h_i=1/d$. Then, since $G=\Phi_{\rm p}^{-1}A$, Lemma \[det\] shows $$\begin{aligned}
\label{maxdet}
    \det G =(\det\Phi_{\rm p}^{-1})(\det A)&\leq& (\det\Phi^{-1}_{\rm SIC})(1/d)^{d^2}\nonumber\\
    &=&\det G_{\rm SIC}\;,\end{aligned}$$ with equality iff the MIC is a SIC. This implies $(\det\alpha)^2\leq 1$, and so $|\det\alpha|\leq 1$. Since $|\det\alpha^{-1}\Phi^{-1}|=|\det\alpha^{-1}||\det\Phi^{-1}|$, we conclude that $$|\det\Phi^{-1}|\leq\det\Phi^{-1}_\text{SIC}\;.$$ Equivalently, $\det\Phi_\text{SIC}\leq |\det\Phi|$. Theorem 3.3.2 in [@Horn:1994] shows $s(A)\succ_{\log} |\lambda(A)|$ for an arbitrary matrix $A$. To show the desired weak log majorization result, we wish to prove $|\lambda(\Phi)|\succ_{w\log}\lambda(\Phi_\text{SIC})$. For this we show weak majorization of the log of the entries. $$\begin{split}
            \log|\lambda(\Phi)|&\succ\left(\frac{\sum_{i=1}^{d^2}\log|\lambda_i(\Phi)|}{d^2-1},\dots,\frac{\sum_{i=1}^{d^2}\log|\lambda_i(\Phi)|}{d^2-1},0\right)\\
            &=\left(\frac{\log|\det\Phi|}{d^2-1},\ldots,\frac{\log|\det\Phi|}{d^2-1},0\right)\\
            &\succ_w\left(\frac{\log\det\Phi_\text{SIC}}{d^2-1},\dots,\frac{\log\det\Phi_\text{SIC}}{d^2-1},0\right)\\
            &=(\log(d+1),\ldots,\log(d+1),0)=\lambda(\log\Phi_\text{SIC})\;.
    \end{split}$$ Thus, $$s(\Phi)\succ_{\log}|\lambda(\Phi)|\succ_{w\log}\lambda(\Phi_\text{SIC})=s(\Phi_{\rm SIC})\;.$$ If $\{H_i\}$ and $\{\sigma_j\}$ are SICs, $\Phi^{-1}_{ij}=\frac{1}{d}\tr\Pi_i\Pi_j'$, where $\{\Pi_i\}$ and $\{\Pi_j'\}$ are SIC projectors in dimension $d$. Then $$\begin{split}
        &\left[\Phi^{-1}\Phi^{{-1}\dag}\right]_{ij}=\frac{1}{d^2}\sum_k(\tr\Pi_i\Pi_k')(\tr\Pi_j\Pi_k')\\
        &=\frac{1}{d^2}\tr\left[(\Pi_i\otimes\Pi_j)\left(\sum_k\Pi_k'\otimes\Pi_k'\right)\right]\\
        &=\frac{1}{d^2}\tr\left[(\Pi_i\otimes\Pi_j)\left(\frac{2d}{d+1}P_\text{sym}\right)\right]\\
        &=\frac{1}{d(d+1)}\tr\left[(\Pi_i\otimes\Pi_j)\left(I\otimes I+\sum_{kl}^d\ketbra{k}{l}\otimes\ketbra{l}{k}\right)\right]\\
        &=\frac{1+\tr\Pi_i\Pi_j}{d(d+1)}=\frac{d\delta_{ij}+d+2}{d(d+1)^2}=\left[\Phi_{\rm SIC}^{-2}\right]_{ij}\;,
\end{split}$$ where $P_{\rm sym}$ is the projector onto the 