When does $\| T^{-1} \| = \inf_{x \neq 0} \frac{\| x \|}{\| T x \|}$ hold? Let $T: X \to Y$ be a linear operator between normed Banach spaces $X$ and $Y$.
The definition of the operator norm
$$
\| T \|
:= \sup_{x \neq 0} \frac{\| Tx \|}{\| x \|}
$$
is well known.
Now, let $T$ be bijective.
When can we say that
$$
\| T^{-1} \| 
= \inf_{x \neq 0} \frac{\| x \|}{\| T x \|}
$$
holds?
 A: This fails in finite dimensions. 
Take
$$
T=\begin{pmatrix} 2&0\\0&3  \end{pmatrix}
$$
So $||T||=3$ and 
$$
T^{-1}=\begin{pmatrix} 1/2&0\\0&1/3  \end{pmatrix}
$$
and $||T^{-1}||=1/2$. 
But, 
$$
\inf_{x\ne0}\frac{||x||}{||Tx||}\geq\frac{1}{||T||}=\frac13
$$
while $x=(0,1)$ achieves this lower bound, showing that 
$$
\inf_{x\ne0}\frac{||x||}{||Tx||}=\min_{x\ne 0}\frac{||x||}{||Tx||}=\frac13\ne \frac12
$$ 
A: First of all, you need to assume that $T^{-1}$ is bounded. If $X$ and $Y$ are Banach spaces, this holds automatically, but not for normed spaces. For example, consider $X=Y=c_{00}(\mathbb N)$, the space of sequences with finite support, and let $T$ be defined by $(Tx)(n)=\frac{1}{n}x(n)$.  Then $T$ a bounded linear bijection (under any $p$-norm), but it's inverse is not bounded.
Under this additional hypothesis, your question is equivalent to asking when $\|T^{-1}\|=\|T\|^{-1}$.  I am not aware of any well-known conditions which guarantees this, but it doesn't happen a lot. In fact, if wikipedia is to be believed, this only happens in finite-dimensional spaces when $T$ is a scalar multiple of an isometry.   
