We define the norm of an operator $A:X \to Y$, with $X$ a pre Hilbert space, as $$ \lVert A \lVert := \sup\{\lVert Ax \lVert \text{ for } x \in X, \lVert x \rVert \leq 1 \} $$
When $A$ is symmetric, i.e. $Ax.y=x.Ay$, then we have that $$\lVert A \lVert = \sup\{\lvert Ax.x\rvert \text{ for } x \in X, \lVert x \rVert \leq 1 \}=:M $$
The proof I was given easily shows that $M \leq \lVert A \lVert$. But to show that $M \geq \lVert A \lVert$, it uses polarization formulas of the inner product and uses a vector $v:= \frac {Ax}{b}$ out of nowhere and then finds the right $b$ so that everything works.
I was wondering if someone has another proof to show that $M \geq \lVert A \lVert$.