Show that $\|B^{-1/2}AB^{-1/2} - I\| >1/2$, when $\|A\|>2\|B\|$ Let $A,B$ be two positive-definite matrices. Suppose that $\|A\|>2\|B\|$. Is it possible to show that $\|B^{-1/2}AB^{-1/2} - I\| >1/2$, where $I$ is an identity matrix and the norm is the supremum norm. 
 A: Recall for matrices $C,D$ that $\sigma(CD)=\sigma(DC)$, where $\sigma(F)$ denotes the spectrum of a matrix $F$. Assuming that both $CD$ and $DC$ are normal, we conclude in particular $\|CD\|=\| DC\|$.
Since $A-B$ is self-adjoint, we may assume that it is diagonal, $A-B=\mbox{diag }(\lambda_1,\ldots,\lambda_n)$, with $\lambda_1\leq\ldots\leq \lambda_n$. Let $U=\mbox{diag }(\mbox{sign }\lambda_1,\ldots,\mbox{sign }\lambda_n)$ and $P=\mbox{diag }(|\lambda_1|,\ldots,|\lambda_n|)$. Then $A-B=UP$. If we let
$$
v_j=\begin{cases}1 &\mbox{if }\mbox{sign } \lambda_j=1,\\
0 &\mbox{if }\mbox{sign } \lambda_j=0,\\
i &\mbox{if }\mbox{sign } \lambda_j=-1,
\end{cases}
$$
and put $V=\mbox{diag }(v_1,\ldots,v_n)$ and $Q=P^{1/2}=\mbox{diag }(|\lambda_1|^{1/2},\ldots,|\lambda_n|^{1/2})$, then we have $(VQ)^2=UP=A-B$.
We see that
\begin{align*}
\|B^{-1/2}AB^{-1/2} -I\|=\|B^{-1/2}(A-B)B^{-1/2}\|=\|B^{-1/2}VQQVB^{-1/2}\|
=\|VQB^{-1}QV\|.
\end{align*}
We now use the relation $B^{-1}\geq 2/\|A\|$ to obtain
$$
VQB^{-1} QV\geq (2/\|A\|)VQQV=2(A-B)/\|A\|.$$
In particular, by an application of the triangle inequality,
\begin{align*}
\|B^{-1/2}AB^{-1/2} -I\|&\geq \frac{2\|A-B\|}{\|A\|}\geq \frac{2(\|A\|-\|B\|)}{\|A\|}>\frac{2\|A\|/2}{\|A\|}=1.
\end{align*}
EDIT: Here is a much simpler derivation. Pick $x$ such that $Ax=\|A\|x$ and $\|B^{1/2}x\|=1$. Then
\begin{align*}
\|(B^{-1/2}AB^{-1/2}-I)\|&\geq\|(B^{-1/2}AB^{-1/2}-I)B^{1/2}x\|\geq \langle B^{1/2}x,(B^{-1/2}AB^{-1/2}-I)B^{1/2}x\rangle\\
&=\langle x,(\|A\|-B)x\rangle\geq\langle x,(\|A\|-\|B\|)x\rangle\\
&>\langle x,\|B\|x\rangle \geq \langle x,Bx\rangle = \|B^{1/2}x\|^2=1.
\end{align*}
