sets of orthonormal Hermitian operators on the Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively. As we have seen earlier, $\rho_{AB}$ in Eq. (\[eq:sd\_svd\]), can be written in the Schmidt decomposed-like form (in operator space) via the singular value decomposition as $$\rho_{AB} = \sum_{k=1}^{d_B^2} \lambda_k G^A_k \otimes G^B_k,$$ where singular values $\lambda_k$ are real and non-negative, and we have assumed that $d_A \geq d_B$.

**Corollary 3**\
*Let $\rho_{AB}$ be a separable quantum state shared between two subsystems $A$ and $B$. Then $$2\sum_i |\lambda_i - \lambda_i^2 g_i^A g_i^B| \leq 2 - \sum_i \lambda_i^2[(g_i^A)^2 + (g_i^B)^2],
\label{eq:col3}$$ where $g_i^{A(B)} = \mbox{tr} [ G_i^{A(B)}]$.*\
Proof:\
Let $
\gamma^S = 
\begin{pmatrix}
\mathcal{A} && \mathcal{X} \\
\mathcal{X}^{T} && \mathcal{B}
\end{pmatrix}
$ be the symmetric block covariance matrix of the bipartite state $\rho_{AB}$. Using the orthogonality of the matrices $\{G_i^A\}$ and $\{G_i^B\}$, one can deduce $\mathcal{X}_{i,j} = \left(\lambda_i - \lambda_i^2 g_i^A g_i^B\right)$. Further, one can get $\mbox{tr}(\rho_A^2) = \sum_i \lambda_i^2 (g_i^B)^2$, and $\mbox{tr}(\rho_B^2) = \sum_i \lambda_i^2 (g_i^A)^2$. Together with Corollary 2, we can prove the claim. $\square$

Now using the relations $|a-b| \geq |a| - |b|$ and $a^2 + b^2 \geq 2|ab|$ we can have, $$\begin{aligned}
2\sum_i |\lambda_i - \lambda_i^2 g_i^A g_i^B| &\geq & 2\sum_i (\lambda_i - \lambda_i^2 