How is $\|T^\ast T\| \ge \sup_{\|x\| = 1} (x, T^\ast Tx) = \sup_{\|x\|=1} \|Tx\|^2 = \|T\|^2$

Question

As part of a proof that $\|T^\ast T\| = \|T\|^2$ where $T$ is a bounded linear transformation from a Hilbert space to itself, the following is stated: $$\|T^\ast T\| \ge \sup_{\|x\| = 1} (x, T^\ast Tx) = \sup_{\|x\|=1} \|Tx\|^2 = \|T\|^2$$

Where is the first inequality coming from?

Answer 1

It's Cauchy-Schwartz: $(x,T^*Tx)\leq\Vert T^*Tx\Vert\Vert x\Vert ,\forall x\in H$ then $\forall x\in H,\Vert x\Vert=1, (x,T^*x)\leq\Vert T^*Tx\Vert$

Answer 2

You have that $|(x,T^*Tx)|≤\|x\|\cdot\|T^*Tx\|$, so: $$\|T^*T\|:=\sup_{\|x\|≤1}\|T^*Tx\|≥\sup_{\|x\|≤1}|(x,T^*Tx)|.$$ On the other hand $(x,T^*Tx)=(Tx,Tx)$ is positive so you can drop the absolute values.