TITLE: General Cauchy-Schwarz for adjoint positive operators
QUESTION [5 upvotes]: I'm trying to prove the next inequality, like Cauchy-Schwarz standard inequality:
$$|\langle Tx,y\rangle |\leq\langle Tx,x\rangle ^{1/2}\langle Ty,y\rangle ^{1/2}\space\forall x,y\in\mathcal{H},$$ where $\mathcal{H}$ is a complex Hilbert space, $T$ bounded linear operator, $T\geq 0$ and $T=T^{*}.$
If we consider that, for $t\in\mathbb{R},$ 
$$0\leq\langle T(y-tx),y-tx\rangle =\langle Ty,y\rangle -2t\mathcal{Re}(\langle Ty,x\rangle )+t^{2}\langle Tx,x\rangle :=P(t),$$ then $P(t)$ is a polynomial of second grade, so its discrimante have to be 
$$\mathcal{Re}^{2}(\langle Ty,x\rangle )\leq\langle Tx,x\rangle \langle Ty,y\rangle ,$$
but I would like to conclude that $$|\langle Ty,x\rangle |\leq\langle Tx,x\rangle \langle Ty,y\rangle ,$$
How can we conclude the desire inequality? Is there something wrong?

REPLY [3 votes]: Replace $y$ in the inequality you have obtained by $ay$ where $a$ is a complex number. You get $\{Re (a\langle Ty,x \rangle \})^{2} \leq |a|^{2}\langle Tx, x\rangle  \langle Ty, y\rangle $. Now put $\bar a=\frac {\langle Ty,x \rangle} {|\langle Ty,x \rangle|}$.