Final step in proof of the Hellinger-Toeplitz Theorem Consider the following statement:

(Hellinger-Toeplitz Theorem) Let $H$ be a Hilbert space and $A : H \to H$ be linear and symmetric,
  i.e. $$\langle x,Ay \rangle = \langle Ax,y \rangle$$ holds for all
  $x,y \in H$. Then $A$ is bounded.

I prove this using the Banach-Steinhaus theorem. I defined $\varphi_y : H \to \mathbb{C}$ by $\varphi_y(x) := \langle A(y),x \rangle$ and $$\mathcal{F} := \{\varphi_y : y \in \partial B_1(0)\}$$ It is easy to show that $\mathcal{F}$ satisfies the prerequisites for the Banach-Steinhaus theorem and thus $$\sup_{T \in \mathcal{F}} \|T\| < \infty$$
Now I want to use this to show that $A$ is bounded. For $x \in H$ I compute $$\|A(x)\|^2 = \langle A(x),A(x) \rangle = \|x\|\langle A(x/\|x\|),A(x)\rangle = \|x\|\varphi_{x/\|x\|}(A(x))$$ But somehow I cannot get rid of the $A(x)$ in the argument of $\varphi$. Can anyone help me how to proceed?

Because $\varphi_{x/ \|x\|}(A(x)) \leq c \|A(x)\|$ where $c = \sup_{ T \in \mathcal{F}} \|T\|$ we have 
$$
\|A(x)\|^2 \leq c \|x \| \|A(x)\|.
$$
Then dividing by $\|A(x)\|$ gives you that $A$ is bounded.
