Formula for the least element on the spectrum

Let $m=\inf\;\{ \lambda : \lambda\in\sigma(A) \}$. Then, for every positive integer $n$, $E_{A}[m,m+1/n] \ne 0$. So there exists a unit vector $x_n\in\mathcal{D}(A)$ such that $E_{A}[m,m+1/n]x_n = x_n$, which gives \begin{align} 0 & \le \langle (A-mI)x_n,x_n\rangle \\ & = \int_{m}^{m+1/n}(\lambda-m) d\langle E(\lambda)x_n,x_n\rangle \\ & \le \frac{1}{n}\langle E[m,m+1/n]x_n,x_n\rangle \\ & \le \frac{1}{n}\langle x_n,x_n\rangle = \frac{1}{n}. \end{align}

Therefore, $\lim_n \langle A x_n,x_n\rangle = m$.

For a selfadjoint operator, every element of the spectrum is an approximate eigenvalue. That shows your equality.