$a,b\in A$ are selfadjoint elements of $C^*$-algebras, such that $a\le b$, why is $\|a\|\le \|b\|$ Let $A$ be a unital $C^\ast$-algebra with unit $1_A$.

a) Why is $a\le \|a\|1_A$ for all selfadjoint $a\in A$ and

b) If $a,b\in A$ are selfadjoint such that $a\le b$, why is $\|a\|\le \|b\|$?

I would try it with the continuous functional calculus $$\phi:C(\sigma(a))\to A,\; f\mapsto f(a),$$ for a) with $f(x)=|x|-x$, which is a positive function on $\sigma(a)\subseteq \mathbb{R}$, is it correct?

for b) It is $a\le \|a\|$ and $b\le \|b\|$, therefore $a\le \|b\|$. How can I continue?

Regards
a) If $A=C(X)$, where $X$ is a compact Hausdorff space, for any $a,b\in A_{sa}$,
$$
a\leq b \Leftrightarrow a(x) \leq b(x) \quad\forall x\in X
$$
So if $f\in A_{sa}$ then , $f\leq \|f\|_{\infty}1_A$.

Now, if $a$ is self-adjoint, then $C^{\ast}(a,1_A)$ is abelian and unital, so by Gelfand's theorem, it is isomorphic to $C(X)$ for some compact Hausdorff $X$.

b) Now if $a \leq b$, then by (a), $a\leq \|b\|1_A$. Again look at the abelian $C^{\ast}$-algebra $C^{\ast}(a,1_A)$, and conclude that $\|a\| \leq \|b\|$

Edit: As @TrialandError points out below, part (b) only holds if $0 \leq a$ [it fails trivially even in a commutative $C^{\ast}$-algebra otherwise]
let A be a C*-algebra and a,x in A with 0≤xx*≤a.
questions:
1-  Put bn=(((1/n)+a)^(-1/2))(a^(1/4))x. show that (bn) is a cauchy sequence in A.
2-  Show that the limit b in A satisfies x=(a^(1/4))b and ∥b∥≤∥a∥^(1/4)

for question 1- it is easy but for question 2- how i show that ∥b∥≤∥a∥^(1/4)?