I only post this answer since you asked how to prove this using the Cauchy–Schwarz inequality. Here's the best argument I could come up with (and there's not much difference to percusse's argument, of course).
Write $B = \varepsilon A$ and multiply the inequality by $\varepsilon \gt 0$ to get $\DeclareMathOperator{\eps}{\varepsilon}$
$$\langle Bx,x \rangle \leq \langle (B+B^2)\xi,\xi \rangle + (1 + \|B\|)\, \|x - \xi\|^2.$$
Now estimate using the symmetry condition $\langle Sy,z\rangle = \langle y, Sz\rangle = \langle Sz,y\rangle$ several times
$$\begin{align*}
\langle Bx, x\rangle &\leq
\langle Bx, x\rangle + \|(1+B)\xi - x\|^2 \\
%&=
%\langle Bx, x\rangle+
%\|(1+B)\xi\|^2-2\langle(1+B)\xi,x\rangle +\|x\|^2\\
&=
\color{green}{\langle Bx, x\rangle}+
\color{red}{\langle (1+B)\xi,\xi\rangle}+
\color{blue}{\langle(1+B)\xi,B\xi\rangle}-
\color{red}{2\langle(1+B)\xi,x\rangle}+\color{green}{\langle x,x\rangle} \\
&=\color{blue}{\langle(B+B^2)\xi,\xi\rangle}+
\color{red}{\langle(1+B)\xi,\xi-x\rangle}-
\color{red}{\langle(1+B)\xi,x\rangle}+
\color{green}{\langle(1+B)x,x\rangle} \\
&=\langle(B+B^2)\xi,\xi\rangle+
\langle(1+B)(\xi-x),\xi\rangle
-\langle(1+B)(\xi-x),x\rangle \\
&=\langle(B+B^2)\xi,\xi\rangle + \langle (1+B)(\xi-x),\xi-x\rangle.
\end{align*}$$
Now we are in position to apply the Cauchy–Schwarz inequality:
$$\langle(1+B)(\xi-x),\xi-x\rangle \leq \|(1+B)(\xi-x)\|\,\|\xi-x\|.$$
Using $\|(1+B)(\xi-x)\| \leq \|1+B\|\,\|\xi-x\| \leq (1+\|B\|)\|\xi-x\|$ by definition of the operator norm, we get
$$\langle Bx,x\rangle \leq \langle(B+B^2)\xi,\xi\rangle + (1+\|B\|)\,\|\xi-x\|^2,$$
as desired.