Let $A\colon \text{dom}(A) \to \mathcal{H}$ be a densely defined symmetric operator on a Hilbert space $\mathcal H$. The symmetry implies that $$ \|(A + i)f\|^2 = \|Af\|^2 + \|f\|^2 \quad \text{for all } f \in \text{dom}(A) $$ The operator $A + i \colon \text{dom}(A) \to \text{Ran}(A + i)$ therefore is a bijection, and its inverse $(A + i)^{-1}$ is a contraction. The latter should follow from the identity above by I struggle to verify this.
What I tried is the following: Substituting $y = (A + i)f$ and $f = (A + i)^{-1}y$, $Af = y - i(A+i)^{-1}f$ we see that for each $y \ne 0$,
\begin{align}
\|y\|^2 &= \|y - i(A + i)^{-1}y\|^2 + \|(A + i)^{-1}y\|^2 \\
&= \|y\|^2 - 2 \Re\big(\langle y,i(A + i)^{-1}y\rangle\big) + 2\|(A + i)^{-1}y\|^2
\end{align}
so that
$$
\|(A + i)^{-1}y\|^2 = \Re\big(\langle y,i(A + i)^{-1}y\rangle\big) \le |\langle y,i(A + i)^{-1}y\rangle| \le \|y\|\|(A + i)^{-1}y\|
$$
where $\Re(z)$ denotes the real part of $z$. Thus we obtain
$$
\|(A + i)^{-1}y\| \le \|y\|
$$
However we need a constant $<1$, is this going in the right direction?
Many thanks for your help!