Show that if $A$ is a bounded linear operator on a Hilbert space then $||Ax||^2 \le ||A||\langle Ax,x\rangle$

Question

Im getting kinda stuck on this, so far I have this.

\begin{align}\|Ax\|^2 &= \langle Ax,Ax\rangle\\&=\langle A^{*}Ax,x\rangle\\&\le \|A^*A\|\cdot\|x\|^2 \\&= \|A^*A\|\langle x,x\rangle \\& = \|A^2\|\langle x,x\rangle \\&\le \|A\|\cdot||A|| \langle x,x\rangle \end{align}

Also,

$$\langle Ax,x\rangle \le \|A\|\cdot\|x\|^2=\|A\|\langle x,x\rangle$$

Can someone give me a hint on how I should move forward?

Answer 1

This is false. Suppose that your space is $\mathbb{R}^2$, endowed with the usual inner produrct, and that $A(x,y)=(-y,x)$. Then $\|A\|=1$ and $(\forall v\in\mathbb{R}^2):\langle Av,v\rangle=0$. However, $(\forall v\in\mathbb{R}^2):\|Av\|^2=\|v\|^2$.