Proving an inequality with Cauchy-Schwarz In the "User's guide to viscosity solutions" by Crandall, Ishii and Lions (link), they make the following claim (inequality (A.4) p. 58) :

Given $x$, $\xi$ $\in \mathbb{R}^n$, $A \in \cal{S}(n)$ (space of symmetric $n \times n$ matrices) , for all $\varepsilon >0$, the Cauchy-Schwarz inequality yields
$$\langle Ax,x \rangle \leq \langle (A+\varepsilon A^2) \xi,\xi \rangle+\left(\frac{1}{\varepsilon} + \|A\|\right)|x-\xi|^2,$$

where I guess $\|A\|$ is the spectral radius of $A$.
I have tried without success to prove this inequality and would appreciate some help.
 A: So first of all I will use $y$ instead of $\xi$ because I love my eyes and I will use $e$ for $\varepsilon$ because it is simpler to type. I hope you don't get offended by that. 
I think they use some sort of a triangle inequality of the form, $\|x-y\| + \|y\| \geq \|x\|$ and $x^TAx\leq\lambda_{max}x^Tx$. But I couldn't see it and I doubt that C-S is needed anyway. Let's do the dirty work, take the LHS to the right and denote that expression with $(\star)$, then: 
$$\begin{align}
(\star) &\geq -x^TAx +y^T(A+e A^2)y+(x-y)^T\left(\frac{1}{e}I + A \right)(x-y) \\
&=y^T(A+e A^2)y+(x-y)^T\frac{1}{e}I(x-y) + y^TAy - 2x^TAy\\
&=y^TA(I+e A)y+y^TAy + \frac{1}{e}y^Ty-2\frac{1}{e}x^Ty+\frac{1}{e}x^Tx-2x^TAy\\
&=y^TA(I+e A)y + y^T(\frac{1}{e}I+A)y -2x^T(\frac{1}{e}I+A)y + \frac{1}{e}x^Tx\\
&= \frac{1}{e}y^T(I+e A)^2y -2\frac{1}{e}x^T(I+eA)y + \frac{1}{e}x^Tx\\
&=\frac{1}{e}\|x-(I+eA)y\|^2\\
&\geq 0
\end{align}
$$
To be honest, this type of math snobbery makes me sick. Maybe there is a direct argument using C-S inequality. Then it would be shorter than this so why not including it in the document. But anyway, hope it helps.
EDIT: by the way the norm definition is given on page 17.
A: 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.
