TITLE: Prove that $A\geq I$ implies that $A$ is invertible.
QUESTION [5 upvotes]: Here's the question:

Let $A$ be a positive operator on a (possibly infinite dimensional) Hilbert space.  Let $I$ denote the identity operator.  Suppose that $A \geq I$, which is to say that $A - I$ is a positive operator.  Prove that $A$ is invertible.

I think that this is true, but I haven't been able to find a proof one way or the other.  I would like to avoid invoking any heavy machinery (like the spectral theorem) if possible.  I would also be interested in a proof that carries over to more general $C^*$ algebras.
Of course, the proof in the case of finite dimensional spaces is fairly obvious, since it suffices to show that the operator has a trivial kernel.  Since that does not suffice here, I really have no clue what my next move should be.
Any guidance here would be greatly appreciated.

REPLY [2 votes]: Since $A$ is positive, its spectrum is contained in $[0,\infty)$. Then the spectrum of $A+I$ is contained in $[1,\infty)$; thus $0$ is not in the spectrum of $A+I$ and $A+I$ is invertible. 
To justify that the spectrum translates note that $A+I-\lambda I=A-(\lambda-1)I$. So $A+I-\lambda$ is not invertible precisely when $A-(\lambda -1)I$ is not invertible, i.e. when $\lambda-1$ is in the spectrum of $A$. So 
$$
\sigma(A+I)=\{\lambda+I:\ \lambda\in\sigma(A)\} 
$$