= Cauchy-Schwarz inequality for a positive operator
{c}
{title2=$|\langle Tx,y\rangle|^2\leq\langle Tx,x\rangle\langle Ty,y\rangle$}
For a bounded <positive operator> $T$, the form $B(x,y)=\langle Tx,y\rangle$ is positive semidefinite and Hermitian. Expanding $B(x+ty,x+ty)\geq0$ for complex $t$ and minimizing its quadratic polynomial gives the displayed inequality. If $B(y,y)=0$, varying $t$ along the negative direction of $B(x,y)$ forces $B(x,y)=0$, covering the degenerate case. In particular $\langle Tx,x\rangle=0$ implies $Tx=0$, by testing against arbitrary $y$.
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