What can we say about the spectrum of the difference of positive elements?

Question

Let $\mathcal{A}$ be a unital $C^*$-algebra with $a,b\in\mathcal{A}_+$ such that $\|a\|<\|b\|$. Does it follow that $a-b\not\in\mathcal{A}_+$?

I can find very little about the spectrum of the sum or difference of two elements. But I believe there should at least be some useful observations to be made for positive elements. For example, I did prove the following.

Let $\mathcal{A}$ be a unital $C^*$-algebra with $a\in\mathcal{A}$ and $b\in\mathcal{A}_+$ such that $\|a\|<\|b\|$. If $C^*(a,b)$ is abelian, then $a-b\not\in\mathcal{A}_+$?

Proof: Since $b\in\mathcal{A}_+$ we have $\|b\|=r(b)$, so we find $x\in\sigma(b)$ with $x>\|a\|$. Then there exists a homomorphism $h$ on $C^*(a,b)$ such that $h(b)=x$. We find $h(a-b)=h(a)-h(b)\leq\|a\|-h(b)<0$. Since $h(a-b)\in\sigma(a-b)$, we get $a-b\not\in\mathcal{A}_+$.

Answer 1

The claim is false, as is the proof you give for the commutative case (the fallacy lies in $\|b\|-h(a)<0$). Consider $\mathcal A=\mathbb C$, $a=4$, and $b=7$.

EDIT

Note that now the claim is equivalent to "If $0\leq a\leq b$, then $\|a\|\leq\|b\|$." This can be proven as follows: By functional calculus, we have $b\leq\|b\|$, and thus $a\leq\|b\|$. Applying functional calculus again gives $\|a\|\leq\|b\|$.