Solution
= Solution
Expanding both squared <norms> gives
$$
\|(I+T)x\|^2-\|(I-T)x\|^2
=4\operatorname{Re}\langle Tx,x\rangle\geq0.
$$
Both <norms> are nonnegative, so taking square roots proves $\boxed{\|(I+T)x\|\geq\|(I-T)x\|}$.
= Solution
Expanding both squared <norms> gives
$$
\|(I+T)x\|^2-\|(I-T)x\|^2
=4\operatorname{Re}\langle Tx,x\rangle\geq0.
$$
Both <norms> are nonnegative, so taking square roots proves $\boxed{\|(I+T)x\|\geq\|(I-T)x\|}$.