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.