Show that $\Vert T\Vert = \sup\{\vert \langle Tx, y\rangle\vert : x,y\in E, \,\Vert x\Vert \leq 1, \,\Vert y\Vert \leq 1\}$

Question

I have the following task where I'm blocked:

Show that if $E$ is an inner product space and $T: E\rightarrow E$ is a bounded linear operator, that is

$$\underset{\Vert x\Vert \leq 1}{\sup} \Vert Tx\Vert < \infty,\quad x\in E$$

then $$\Vert T\Vert = \sup\{\vert \langle Tx, y\rangle\vert : x, y\in E,\, \Vert x\Vert \leq 1,\, \Vert y\Vert \leq 1\}.$$

In the problem $\langle, \rangle$ denotes inner product. I'm basically stuck here and I don't know how to proceed. I know that generally the norm of a linear operator is defined as:

$$\Vert T\Vert =\underset{\Vert x\Vert \leq 1}{\sup} \Vert Tx\Vert,\quad x\in E.$$

So I guess I need to show that:

$$\Vert T\Vert =\sup \{\Vert Tx\Vert : x\in E,\, \Vert x\Vert \leq 1\} = \sup\{\vert \langle Tx, y\rangle\vert : x, y\in E,\, \Vert x\Vert \leq 1,\, \Vert y\Vert \leq 1\}.$$

Am I in the correct trail? Any hints what I should notice here?