Showing operator norm is bounded above and below (from Lax-Milgram Theorem)

Question

Let $H$ be a Hilbert space, $B(x,y)$ a bilinear form $H$ (so that $B:H\times H \to \mathbb{R}$ and $B(x,y)$ is linear in $x$ and $y$). Suppose that there exist constants $a,b>0$ such that $|B(x,y)|\le b\|x\|\|y\|$ and $B(x,x)\ge a\|x\|^2$ for all $x,y\in H$.

Let $T:H\to H$ be a linear operator satisfying the condition $B(x,y)=\langle x, Ty\rangle$ for all $x\in H$. Show that $a\|x\|\le \|Tx\|\le b\|x\|$.

The first part is straightforward. We have that $|B(x,x)|\le b\|x\|^2$, $B(Tx,x)=\langle Tx, Tx \rangle=\|Tx\|^2\le b\|Tx\|\|x\|\iff \|Tx\| \le b\|x\|$.

With the second part, however, I'm stuck. $|B(Tx,Tx)|=|\langle Tx, TTx \rangle|\ge a\|Tx\|^2$.

I'd appreciate some help with this. I think there is some property related to bilinear forms and/or bounded linear operators that I do not recognize here.

Answer 1

$a\|x\|^{2}\leq B(x,x)=\left<x,Tx\right>\leq\|x\|\|Tx\|$, so $\|Tx\|\geq a\|x\|$.

The second inequality is due to Cauchy-Schwartz.