$\frac{1}{{1 + {\left\| A \right\|} }} \le {\left\| {{{(I - A)}^{ - 1}}} \right\|}$ Let a matrix norm $ {\left\| . \right\|}$ have the property that $ {\left\| I \right\|}  = 1$ and $ {\left\| A \right\|}  < 1$. Why does the following inequality hold?
$$\frac{1}{{1 +  \left\| A \right\| }} \le \left\| {{{(I - A)}^{ - 1}}} \right\|.$$ ($A \in {M_n}$)
 A: The answer by @MotylaNogaTomkaMazura shows why this holds if $\|\cdot\|$ is submultiplicative and satisfies $\|I\|=1$. However, if it is not submultiplicative, this does not need to be true.
Consider, e.g., the $\max$-norm
$$
\|X\|:=\max_{i,j}|x_{ij}|.
$$
Clearly, $\|\cdot\|$ is a matrix norm (matrix analogue of the vector $\infty$-norm).
Also, $\|X\|=1$. But with
$$
X=\pmatrix{0&1\\1&1},\quad Y=\pmatrix{1&1\\0&1},
$$
$$
\|XY\|=2\not\leq 1=\|X\|\|Y\|,
$$
so $\|\cdot\|$ is not submultiplicative.
With
$$
A=\frac{1}{5}\left(\begin{array}{rr}-3&4\\-4&-3\end{array}\right),
$$
we have $\|A\|=3/5<1$. But
$$
\frac{1}{1+\|A\|}=\frac{5}{9}\not\leq\frac{1}{2}=\|(I-A)^{-1}\|.
$$
A: $$1=||(I-A)^{-1} (I-A)||\leq  ||(I-A)^{-1} ||\cdot|| (I-A)||\leq ||(I-A)^{-1} || (||I|| +||A||)= ||(I-A)^{-1} || (1 +||A||)$$
A: you can use this methods to solve it:   since $I-A$ is nonsingular, by definition $\|(I-A)^{-1}\| = \sup_{\|y\|=1} \|(I-A)^{-1}y\| = \sup_{\|y\|=1} \|x_y\|$ where $x_y := (I-A)^{-1}y$, i.e., $y = (I-A) x_y$. Now
$$
1 = \|y\| = \|(I-A) x_y\| = \|x_y - A x_y\| \le \|x_y\| +\|A x_y\|
\le  \|x_y\| + \|A\| \|x_y\| = (1 +\|A\|) \|x_y\|.
$$
Finally, $\|x_y\| \geq 1/(1 +\|A\|)$ and so is the $\sup$.
ref:[1] 
$\left \| \cdot \right \|$ is an induced norm. If $\left \| A \right \|<1$, how to show that $I-A$ is nonsingular and ...?
