Given a Banach space.
Consider a closed operator: $$T:\mathcal{D}(T)\to E:\quad T=\overline{T}$$
Due to the Neumann series it holds: $$R(\lambda):=(\lambda- T)^{-1}:\quad\|R(\lambda)\|\geq\frac{1}{d(\lambda,\sigma(T))}\quad(\lambda\in\rho(T))$$ Can you give an example for strict inequality, please?
Example
Consider a nilpotent operator: $$N\in\mathcal{B}(E):\quad N^K\neq0\quad N^{K+1}=0$$
Then its resolvent has a singularity at zero only: $$\sigma(N)=(0):\quad(\lambda-N)^{-1}=\lambda\sum_{k=0}^K\lambda^{-k}N^k\quad(\lambda\neq0)$$
So the inequality necessarily becomes strict!
Explanation
Nilpotent operators have singularities of higher order at zero.
Normal operators have singularities of first order at zero.
Bounded quasinilpotent operators have essential singularities at zero ($\dim E=\infty$).
Note that a matrix whose resolvent has only poles of order one is diagonalizable.