= Solution
Set $B(x,y)=\langle Tx,y\rangle$. It is a positive semidefinite <Hermitian form>. Write $A=B(x,x)$, $C=B(y,y)$ and $c=B(x,y)$. For every complex $t$,
$$
0\leq B(x+ty,x+ty)=A+\bar t c+t\bar c+|t|^2C.
$$
If $C>0$, choosing $t=-c/C$ gives $A-|c|^2/C\geq0$. If $C=0$ and $c\ne0$, choosing $t=-sc$ for large positive $s$ makes the expression negative, a contradiction. Thus $c=0$ in that case. In all cases,
$$
\boxed{|\langle Tx,y\rangle|^2\leq\langle Tx,x\rangle\langle Ty,y\rangle.}
$$
This is the <Cauchy-Schwarz inequality for a positive operator>.
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