spectral radius and numerical radius of matrix

Let me denote $\rho:=\lim \|A^n\|^{1/n}$. Then $|\lambda|\le\rho$ for all eigenvalues. Moreover, there is an eigenvalue $\lambda$ of $A$ such that $|\lambda|=\rho$. Using the corresponding unit eigenvector, $$ \rho \le \sup_{\|x\|=1}| \langle x,Ax\rangle| $$ follows.

I am not sure what the question is about. For instance, the inequality $$ \|A^n\|^{1/n} \le \sup_{\|x\|=1}| \langle x,Ax\rangle| $$ is not true for all $n$ in general. To see this, take the nilpotent matrix $$ A =\pmatrix{ 0 & 1 & 0 & 0 \\ \vdots & \ddots & \ddots &0\\ \vdots&\ddots&\ddots& 1 \\ 0 & \dots &\dots &0}. $$ Then $\|A^n\|=1$ for all $n<d$ and $\sup_{\|x\|=1} |\langle x,Ax\rangle|<1$.

In Michel Crouzeix - Numerical range and functional calculus in Hilbert space, we see that

Using the inequality $2$ with $P(z)=z^n$ we get the desired result.