Let $a$,$b$ be positive elements in a C* algebra with $\|a\|,\|b\|\leq 1$, does $\|a-b\|\leq 1$ hold?

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On Murphy's C*-algebra and operator theory page 89 it says

Now suppose that $a$ is positive and $\|a\| \leq 1$. Then $u_\lambda - a$ is hermitian and $\| u_\lambda -a\|\leq 1$.

Where $u_\lambda$ is an arbitrary approximate unit in a C*-algebra, which means $u_\lambda$ is a positive element satisfying $\|u_\lambda\|\leq 1$.

Could you tell me why is $\| u_\lambda -a\|\leq 1$ ? Thanks.

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Assertion: $\|a-b\|\leq \text{max}\{\|a\|,\|b\|\}$ holds for positive $a$,$b$.

Add a unit if there isn't one. Since $-\|b\|\leq -b\leq a-b \leq a\leq \|a\|$ we have $-\|b\|\leq a-b\leq \|a\|$.

By applying Gelfand transform to $C^*(a-b,1)$ we see $\|a-b\|\leq \text{max}\{\|a\|,\|b\|\}$.