Prove that $\|UVU^{-1}V^{-1}-I\|\leq 2\|U-I\|\|V-I\|$ $U,V$ are unitary $n\times n$ matrices, and the norm is the operator norm (so we can use $\|UV\|\leq\|U\|\|V\|$).
I've noticed that
\begin{align}
\|UVU^{-1}V^{-1}-I\|&= \|(UV-VU)U^{-1}V^{-1}\|\\
&\leq \|UV-VU\|\|U^{-1}V^{-1}\|
\end{align}
I can bound the first term by $\|UV\|+\|VU\|$, but I don't think this is useful. 
Hints (rather than complete answers) would be appreciated.
The question comes from here (exercise 1)

(I cannot prove the inequality as stated in the exercise, but here is some information that maybe will help you).
Note that $\|UVU^{-1}V^{-1}\|=1$ for all unitaries $U,V$. We would like to show that 
$$\tag{1}
\|UVU^{-1}V^{-1}-I\|\leq 2\|U-I\|\,\|V-I\|.
$$
The left-hand-side, as hinted in the exercise, is $\|UV-VU\|$. Now
$$\tag{2}
\|UV-VU\|\leq\|UV-V\|+\|V-VU\|=2\|U-I\|.
$$
As the roles of $U$ and $V$ are symmetric, we can get, by multiplying $(2)$ and the corresponding inequality for $V$,
$$
\|UV-VU\|\leq 2\|U-I\|^{1/2}\|V-I\|^{1/2},
$$
which is sharper than $(1)$ when $\|U-I\|>1$ and $\|V-I\|>1$. 
