Let $A$ be a self-adjoint operator on a Hilbert space $H$. Let $\lambda$ and $\mu$ be real values such that $\lambda + i \mu$ is in the resolvent of $A$. Then it can be shown that if $\mu \neq 0$ that $$\|(A - (\lambda + i \mu)I)x\| \geq \mu \|x\| \tag{1}$$ for all $x \in H$. I would like to show that we may bound the resolvent of $A$ at $\lambda + i \mu$ as follows: $$\|(A - (\lambda + i \mu)I)^{-1}\| \leq \frac1\mu.$$
Since $\lambda + i \mu$ is in the resolvent of $A$, we know that $(A - (\lambda + i \mu)I)^{-1}$ exists and is bounded. To achieve the desired bound, I have to use (1) to conclude that $$\|(A - (\lambda + i \mu)I)^{-1}x\| \leq \frac1\mu \|x\| \tag{2}$$ but I am having some trouble with this. I think there are some properties of the inverse that I am missing. How may I use (1) to obtain (2)?
In $$\|(A - (\lambda + i \mu)I)x\| \geq \mu \|x\| \tag{1}$$ take $x=(A - (\lambda + i \mu)I)^{-1}y$. You get $$\|(A - (\lambda + i \mu)I)^{-1}y\| \leq \frac1\mu\|y\|.$$ Rename $y$ as $x$. You get (2).