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.
 A: The inequality is false. Here is a random counterexample:
$$
\begin{aligned}
A&=\pmatrix{
 6& 3&-2&-8\\
 3& 2& 0&-4\\
-2& 0& 4& 4\\
-8&-4& 4&16},
\ B=\pmatrix{
 5&-2&-1& 4\\
-2&10& 6&-5\\
-1& 6& 6&-6\\
 4&-5&-6& 9},\\
X=AB&=\pmatrix{
-6& 46& 48&-51\\
-5& 34& 33&-34\\
 2&  8&  2&  4\\
28&-80&-88&108}.
\end{aligned}
$$
We have $\|X-X^{-1}\|_F = 191.7695 < 192.0703 = \|X-I\|_F$ here.
However, the similar inequality $\|AB-\color{red}{A^{-1}B^{-1}}\|_F\ge\|AB-I\|_F$ is true. Since $A$ and $B$ are positive definite, the eigenvalues $x_1,x_2,\ldots,x_n$ of $AB$ are positive. Therefore
$$
\begin{aligned}
&\|AB-A^{-1}B^{-1}\|^2-\|AB-I\|^2\\
&=\|A^{-1}B^{-1}\|^2+2\operatorname{tr}(AB)-3\operatorname{tr}(I)\\
&\ge\sum_i|\lambda_i(A^{-1}B^{-1})|^2+2\operatorname{tr}(AB)-3\operatorname{tr}(I)\\
&=\sum_i\lambda_i(B^{-1}A^{-1})^2+2\operatorname{tr}(AB)-3\operatorname{tr}(I)\\
&=\sum_i(x_i^{-2}+2x_i-3)\\
&=\sum_i(x_i^{-1}-1)^2(2x_i+1)\ge0.
\end{aligned}
$$
Since a real matrix is the product of two symmetric positive definite matrices if and only if it has a positive spectrum, the inequality in the OP can be rephrased as
$$
\|X-X^{-1}\|\ge\|X-I\|
\quad\text{when $X$ is a real matrix with a positive spectrum.}
$$
Observe that
$$
\|X-X^{-1}\|^2-\|X-I\|^2
=\operatorname{tr}\left((I-X^{-1})^T(I+X+X^T)(I-X^{-1})\right).
$$
Since $I+X+X^T$ is in not necessarily positive definite (except in the scalar case), $\operatorname{tr}\left((I-X^{-1})^T(I+X+X^T)(I-X^{-1})\right)$ may not be always positive. Indeed, it is not, as the counterexample above shows.
A: 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.
