If $A$ is normal: why $\sigma(A)=\sigma_\pi(A)$ implies that $\omega(A)\geq r(A)$?

Question

I see in this paper the following theorem:

The proof is not difficult. However, I don't understant the first sentence:

I don't understant why the fact that $\sigma(A)=\sigma_\pi(A)$ implies that $\omega(A)\geq r(A)$? If $n=1$, it is well known that for every normal operators $A$, we have $$\sigma(A)=\sigma_\pi(A).$$ Even if $n=1$, I don't see why $\sigma(A)=\sigma_\pi(A)$ implies that $\omega(A)\geq r(A)$?

Here,

and $\sigma(A)$ is defined as

and

Answer 1

Let $z_1,\dots,z_n\in \sigma(A)$ and $\epsilon>0$. Since $\sigma(A)=\sigma_\pi(A)$, there exist a unit vector $x$ such that $\|(A_k-z_k)x\|<\epsilon$. Thus \begin{align*} \sum_{k=1}^n|z_k|^2&=\sum_{k=1}^n|\langle z_k x,x\rangle|^2\\ &=\sum_{k=1}^n|\langle(z_k-A_k)x,x\rangle+\langle A_k x,x\rangle|^2\\ &\leq\sum_{k=1}^n\left((1+\delta^{-1}) |\langle (z_k-A_k)x,x\rangle|^2+(1+\delta)|\langle A_k x,x\rangle|^2\right)\\ &\leq(1+\delta^{-1})n\epsilon^2+(1+\delta)\sum_{k=1}^n|\langle A_k x,x\rangle|^2\\ &\leq (1+\delta^{-1})n\epsilon^2+(1+\delta) w(A)^2 \end{align*} Letting first $\epsilon\to 0$ and then $\delta\to 0$, we arrive at the desired inequality.