Lower bound for the norm of the resolvent

Question

I need to prove next statement (I want to do it for general case) $\|R_A(z)\| = \lVert \frac{1}{A-zI} \rVert \ge \text{dist}(z,\sigma(A))^{-1}$ I think it could be like this let $a\in \sigma(A) z \notin \sigma (A)$ then $(A-a)^{-1}$ is unbounded. $(A-z-(a-z))=(A-z)(I-(a-z)(A-z)^{-1}) ||(a-z)(A-z)^{-1}||>1$ then we get statement. but my proof doesn't cover eigenvalue.

Answer 1

See Corollary 6.9(1), p242 in Invitation To Operator Theory (here).