TITLE: spectral radius and numerical radius of matrix
QUESTION [5 upvotes]: Let $\mathcal{H}$ be a complex finite dimensional Hilbert space ($\dim\mathcal{H}=d$). Let $A\in M_d(\mathbb{C})$. 

How can I show without using spectral considerations that
  $$\lim_{n\to\infty}\|A^n\|^{\frac{1}{n}}
\le\displaystyle\sup_{\|x\|=1}|\langle Ax,x\rangle|??
$$

Thank you for your help!!
This question is a motivation to the following one:
It is well known that the spectral radius of a bounded linear operator $T$ acting on a complex infinite dimensional Hilbert space is given by
$$r(T)=\lim_{n\to\infty}\|T^n\|^{\frac{1}{n}}.$$
I want to prove using the above formula (rather than the definition of the spectral radius) that $$r(T)=\lim_{n\to\infty}\|T^n\|^{\frac{1}{n}}\le \sup_{\|x\|=1}|\langle Tx.x\rangle|.$$

REPLY [2 votes]: 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.