Let $H$ be a $\mathbb R$-Hilbert space, $A\in\mathfrak L(H)$ be self-adjoint, $r(A)$ and $\sigma(A)$ denote the spectral radius and spectrum of $A$, respectively, and \begin{align}\lambda_{\text{min}}&:=\inf\sigma(A),\\\lambda_{\text{max}}&:=\sup\sigma(A),\\m&:=\inf_{\left\|x\right\|_H=1}\langle Ax,x\rangle_H,\\M&:=\sup_{\left\|x\right\|_H=1}\langle Ax,x\rangle_H.\end{align}
I know that $$r(A)=\left\|A\right\|_{\mathfrak L(H)}=\max_{\lambda\in\sigma(A)}|\lambda|=\max(-\lambda_{\text{min}},\lambda_{\text{max}})\tag1.$$ On the other hand, $$\left\|A\right\|_{\mathfrak L(H)}=\sup_{\left\|x\right\|_H=1}|\langle Ax,x\rangle_H|=\max(-m,M)\tag2.$$
Question 1: From $(1)$ it should follow that at least one of $\lambda_{\text{min}}$ and $\lambda_{\text{max}}$ is contained in $\sigma(A)$. In the same way, by $(2)$ it should follow that at least one of $m$ and $M$ is contained in $\sigma(A)$. Or am I missing something?
Question 2: How are $\lambda_{\text{min}}$, $\lambda_{\text{max}}$, $m$ and $M$ related? It's tempting to expect that $\lambda_{\text{min}}=m$ and $\lambda_{\text{max}}=M$, but can we actually show that?