How to prove that $\|AB-B^{-1}A^{-1}\|_F\geq\|AB-I\|_F$ when $A$ and $B$ are symmetric positive definite? Let $A$ and $B$ be two symmetric positive definite $n \times n$ matrices. Prove or disprove that
$$\|AB-B^{-1}A^{-1}\|_F\geq\|AB-I\|_F$$
where $\|\cdot\|_F$ denotes Frobenius norm. I believe it is true but I have no clue how to prove it.
Thanks for your help.

For the Froebenius Norm:
Since $A$ and $B$ are positive definite, we can write $C=AB=QDQ^\dagger$, with $D$ being the a diagonal matrix with the $n$ positive eigenvalues $\lambda_k$ and $Q$ a hermitian matrix ($QQ^\dagger=QQ^{-1}=I$). So we obtain
$$ 
||C - C^{-1}|| \geq ||C - I||
$$
Since the Froebenius Norm is invariant under coordinate rotations, i.e. $||QA||=||A||$, we can simplify this expression to
$$
||C - C^{-1}|| = || QDQ^\dagger - QD^{-1}Q^\dagger || = ||D - D^{-1}||
= \sqrt{\sum_{k=1}^n \left(\lambda_k-\lambda_k^{-1}\right)^2}
$$ and
$$
||C -I|| = ||QDQ^\dagger - I || = ||D -I|| = \sqrt{\sum_{k=1}^n \left(\lambda_k-1\right)^2}
$$
For all $\lambda_k>0$,
$$
\sqrt{\sum_{k=1}^n \left(\lambda_k-\lambda_k^{-1}\right)^2}
\geq \sqrt{\sum_{k=1}^n \left(\lambda_k-1\right)^2}
$$ holds.
