Formula for the least element on the spectrum Let $A$ be a self-adjoint operator defined on a dense subset of an Hilbert space $\mathcal{H}$. Assume that $A$ is bounded below in the sense there is $m \in \mathbb{R}$ such that
$$\langle Ax,x\rangle \geq m,~\forall x : \|x\| = 1.$$
I want to show that:
$$ m = \inf\{\lambda : \lambda \in \sigma(A)\} = \inf \{\langle Ax,x\rangle : \|x\| = 1\}.$$
I know that if $E_A$ denotes the unique spectral measure that represents $A$, then $\mathrm{supp}~E_A = \sigma(A),$ from which follows the first equality. So, it is only left to prove the last equality. Any hints?
Thanks in advance.

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