For positive self adjoint $T$, show $|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$

Question

As in title, $T$ is a positive self adjoint, bounded linear operator on a Hilbert Space $X$ and I'd like to show

$$|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$$

Self adjoint means we have $|\langle Tx,y\rangle|=|\langle x,Ty\rangle|=|\langle Ty, x\rangle|$. So by CS-inequality we have

$$ |\langle Tx,y \rangle|^2\le \|Tx\|\cdot \|Ty\| \cdot \|x\| \cdot \|y\| $$

But I'm not sure if this is a right approach - is it true that for positive operators $\langle Tx,x\rangle=\|Tx\|\cdot \|x\|$? if so, I'm done but I'm rather skeptical about that - I know that equality holds iff $Tx$ is a scalar multiple of $x$, but I don't see any reason why that might be the case for every $x$.

Answer 1

Positive operator simply means that $\langle v,Tv\rangle\geq 0$, $\forall v$. For instance, given $t\in \mathbb{R}$, you have $$ 0\leq \langle y-tx,T(y-tx)\rangle = \langle y,Ty\rangle - 2t\langle x,Ty\rangle + t^2\langle x,Tx\rangle $$ and therefore...