Let $A$ and $B$ be two matrices with $\det(A)=\det(B)=1$. Does it follow that

$\sqrt{\mathrm{tr}(A^TB^TBA-I)}\le\sqrt{A^TA-I}+\sqrt{B^TB-I}$

I suspect that this can be shown using the singular value decomposition, but I've not been able to write a proof yet. Scaling arguments suggest that the determinant condition is really needed.

If needed, it may be assumed that all singular values of both $A$ and $B$ are positive.

Let $A$ and $B$ be two matrices with $\det(A)=\det(B)=1$. Does it follow that

$\sqrt{\mathrm{tr}(A^TB^TBA-I)}\le\sqrt{\mathrm{tr}(A^TA-I)}+\sqrt{\mathrm{tr}(B^TB-I)}$

I suspect that this can be shown using the singular value decomposition, but I've not been able to write a proof yet. Scaling arguments suggest that the determinant condition is really needed.

If needed, it may be assumed that all singular values of both $A$ and $B$ are positive.