If $T$ is a positive operator then $I+T$ is invertible Let $T$ be a positive operator on a Hilbert space $H$, prove that $I+T:H\to H$ is invertible and $(I+T)^{-1} \in B(H)$.
Now, If I prove $I+T$ is invertible, the bounded inverse theorem implies the second part. Now while proving that $I+T$ is invertible, I have proved that $I+T$ is one-one. But now I have to prove that $I+T$ is onto. In doing so my idea is to prove that $I+T$ is bounded below, so that $Range(I+T)$ becomes closed and then show that $Range(I+T)^{\perp}=\{\ 0 \}\ $, then by projection theorem we will have $Range(I+T)=H$.
But I couldn't execute this idea. Other ideas will also be appreciated
Thanks in advance!!

Note that 
$$
I+T-\lambda I=T-(\lambda -1)I.
$$
So $\lambda\in\sigma(I+T)\iff \lambda-1\in\sigma(T)$. In other words, 
$$
\sigma(I+T)=\{\lambda+1:\ \lambda\in\sigma(T)\}.
$$
As $T$ is positive, $\sigma(T)\subset[0,\infty)$. Thus $\sigma(I+T)\subset [1,\infty)$. 
It follows that $0\not\in\sigma(I+T)$, so $I+T$ is invertible. 
