Is $|\langle Ax, y\rangle|^4 \leq \langle |A|^{2}x, x\rangle\langle |A^*|^{2}y, y\rangle$?

Question

Let $A$ be a bounded linear operator on a complex Hilbert space $H$.

Why \begin{equation} |\langle Ax, y\rangle|^4 \leq \langle |A|^{2}x, x\rangle\langle |A^*|^{2}y, y\rangle, \end{equation} for every $x,y\in H$. Note that $|X|=(X^*X)^{1/2}$ is the absolute value of an operator $X$.

I think that the above inequality is used in the following proof

Answer 1

$$\begin{align}|\langle Tx,x\rangle|^2 + \|Tx\|^4 &= |\langle Tx,x\rangle|^2 + \left(\sqrt{\langle Tx, Tx \rangle}\right)^4\\&=|\langle Tx,x\rangle|^2 +|\langle Tx, Tx \rangle|^2\\&=|\langle Tx,x\rangle|^2 + |\langle T^*Tx, x \rangle|^2\end{align}$$

is the first line of the calculation. Though not directly mentioned, they continue it as: $$\begin{align}|\langle Tx,x\rangle|^2 + |\langle T^*Tx, x \rangle|^2 &= |\langle Tx,x\rangle|\,|\langle Tx,x\rangle| + |\langle T^*Tx, x \rangle|\,|\langle T^*Tx, x \rangle|\\ &=|\langle Tx,x\rangle|\,|\langle x,Tx\rangle| + |\langle T^*Tx, x \rangle|\,|\langle T^*Tx, x \rangle|\\ &=|\langle Tx,x\rangle|\,|\langle T^*x,x\rangle| + |\langle T^*Tx, x \rangle|\,|\langle T^*Tx, x \rangle|\end{align}$$ where the fact that $|\langle u,v\rangle| = |\langle v,u\rangle|$ was used in the 2nd line, and the definition of the adjoint was used in the third.

Now apply Cauchy-Schwarz (with square roots) to each of the 4 expressions.