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!!

Let $A=I+T$. Recall $\ker A^* = (\operatorname{ran} A)^\perp$. For injective self-adjoint operators, this already implies that the range is dense. But your operator is not just injective but is bounded from below, meaning  there is $c>0$ such that 
$$\|Ax\|\ge c\|x\| \quad \forall x \tag{1}$$
Property (1) implies $\operatorname{ran} A$ is closed.  To summarize: a self-adjoint operator that is bounded from below is invertible. 
The proof of (1) is an application of monotonicity:
$$
\langle x+Tx, x+Tx\rangle  \ge \langle x, x\rangle
$$ 
so $c=1$ works.
