Solution
= Solution
A separate expansion gives
$$
\|(I+T)x\|^2=\|x\|^2+\|Tx\|^2+2\operatorname{Re}\langle Tx,x\rangle
\geq\|x\|^2.
$$
Hence $\boxed{\|(I+T)x\|\geq\|x\|}$. In particular $I+T$ is injective and bounded below by one.
= Solution
A separate expansion gives
$$
\|(I+T)x\|^2=\|x\|^2+\|Tx\|^2+2\operatorname{Re}\langle Tx,x\rangle
\geq\|x\|^2.
$$
Hence $\boxed{\|(I+T)x\|\geq\|x\|}$. In particular $I+T$ is injective and bounded below by one.