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.
Best Answer
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.