You know what you want to prove : it is that for all $x = (x_1,...,x_m)\in \mathbb C^m$, we have:
$$
\max_{1 \leq i \leq m} |x_i| \leq \left(\sum_{i=1}^m |x_i|^2\right)^{\frac 12} \leq \sqrt m \max_{1 \leq i \leq m} |x_i|
$$
You have to prove this statement, from the statements that you already know are true, and some definitions.
Let us try to do the first one.
For the first one, we note that $|x_i| \geq 0$ for all $i$, by definition of the absolute value. Consequently, let us fix $1 \leq I \leq m$, then $|x_I|^2 \leq \sum_{i=1}^m |x_i|^2$, since the right hand side is $|x_I|^2$ plus something non-negative. By taking square roots (and noting that (positive) square roots preserve inequality) we get $|x_I| \leq \left(\sum_{i=1}^M |x_i|^2\right)^{\frac 12}$.
Now , the point is that we can choose any $I$ above that we want , since the choice of $I$ did not affect the calculation. Take $I$ such that $|x_I| = \max_{1 \leq i \leq m} |x_i|$,and then the conclusion follows.
For the second one, note that for each $I$ we have $|x_I| \leq \max_{1 \leq i \leq m} |x_i|$, so we get $\sum_{i=1}^m |x_i|^2 \leq m \times (\max_{1 \leq i \leq m} |x_i|)^2$ by applying the inequality for each $|x_I|$ and then summing separately. Now look above at the second inequality, and see how it is similar to the statement I have just written. Can you work out the second inequality from here?
In general $\|B^{-1}\|\ge\frac{1}{\|B\|}$ since $1=\|BB^{-1}\|\le\|B\|\|B^{-1}\|$.
Also, $\|I-A\|\le1+\|A\|$. Hence combining the two, $$\|(I-A)^{-1}\|\ge\frac{1}{\|I-A\|}\ge\frac{1}{1+\|A\|}$$
Best Answer
Observe that
$\lVert x \rVert = \lVert (x -y) +y \rVert \leq \lVert (x -y) \rVert + \lVert y \rVert$
which gives
$\lVert x \rVert - \lVert y \rVert \leq \lVert x -y \rVert$ ... $(1)$
Further,
$-(\lVert x \rVert - \lVert y \rVert ) \leq \lVert (y -x) \rVert = \lVert (x -y) \rVert $... $(2)$
From $(1)$ and $(2)$ result follows.