I was told:
$$ \max_{i} | \lambda(A) | \leq \| A \|_P $$
I tried thinking through it. So the operator norm is defined as:
$$ \| A \|_P = \sup_{y \neq 0} \frac{ \| A y \|_P }{ \| y\|_P } = \sup_{ \| y \| = 1 } \| A y \|_P$$
So I recalled the definition of eigenvalue/vector:
$$ A y_i = \lambda_i y_i $$
so for (unit) eigenvectors we have:
$$ \| A y_i \| = |\lambda_i | \| y_i \| = |\lambda_i(A) | $$
since we are getting the largest number $\| A y \|$ then that set also include the ones with eigenvalues and is a bigger set. Therefore, the operator norm can't be small (since it must be at least the size of the eigenvalues) and since the other vectors might be bigger it might be larger. So we get:
$$ \max_i |\lambda_i(A) | \leq \sup_{ \| y \| = 1 } \| A y \| = \| A \|_P$$
not sure if this is rigorous enough, but it seems correct top me.