Let $A$ be a unbounded, self-adjoint operator on some Hilbert space $H$. Suppose that $\lVert (\lambda - A) x \rVert \geq \lambda \lVert x \rVert$ for all $x \in D(A)$ and $\lambda > 0$ and let $y \in H$ such that $(A y, y) = 0$. I want to prove that $A y = 0$.
I know two proofs for that statement: One involves the spectral theorem and the second one uses another not so trivial result about positive operators. Both proofs work just fine but I wonder if one can use just the norm inequality and the properties of the inner product of $H$ (maybe like polarization identity, etc.) to estimate $\lVert A y \rVert^2 \leq 0$ to get the desired result. I tried it but I couldn't find an argument with that flavour.
If it helps, you can also assume that e.g. $1 \in \rho(A)$ and use $\lVert R(1, A) \rVert \leq 1$.
The assumption $(\lambda I-A)x\|\ge \lambda \|x\|$ for $\lambda>0$ implies that $-A$ is a nonnegative operator. Thus the antilinear form $(x,y)\to -\langle x,Ay\rangle $ defined on $D(A)\times D(A)$ satisfies the Cauchy-Schwarz inequality $$|\langle x,Ay\rangle|^2\le \langle x,Ax\rangle\langle y,Ay\rangle $$ Let $\langle y,Ay\rangle=0$ for some $y\in D(A).$ Then $\langle x,Ay\rangle =0$ for any $x\in D(A).$ Since $D(A)$ is dense, we get $Ay=0.$