Functional analysis: Why $|[x,y]|\leq [x,x]^{1/2}[y,y]^{1/2}$ Let $\varepsilon>0$. We denote $[x,y]:=\langle \varepsilon x+Ax,y\rangle$ where $A$ is symmetric and $A\geq 0$. Why do we have that $$\big|[x,y]\big|\leq [x,x]^{1/2}[y,y]^{1/2}\ \ \ ?$$
The hint is to use Cauchy-Schwarz on $\langle \varepsilon x+Ax,y\rangle$, but when I do it I get
$$\big|[x,y]\big|^2\leq \langle\varepsilon x+Ax,\varepsilon x+Ax\rangle\langle y,y\rangle$$ 
whereas I would like to have $$\langle\varepsilon x+Ax,x\rangle\langle\varepsilon y+Ay,y\rangle.$$
Question : How can I prove my inequality ?
 A: I've corrected your notation fully. For any fixed $\varepsilon > 0$, the form $[x,y]$ defines an inner product on $X$. The properties are easy to verify. Therefore, by the Cauchy-Schwarz inequality
$$
                 |[x,y]| \le [x,x]^{1/2}[y,y]^{1/2},\;\;\; x,y \in X.
$$
In terms of $A$, the above becomes
$$
           |(\varepsilon x + Ax,y)| \le (\varepsilon x + Ax,x)^{1/2}(\varepsilon y + Ay,y)^{1/2}
$$
The above hols for all $\varepsilon > 0$; so you may let $\varepsilon \rightarrow 0$ to also obtain
$$
                     |(Ax,y)| \le (Ax,x)^{1/2}(Ay,y)^{1/2}.
$$
I suspect that's the end result they're steering you toward.
A: If $A$ is positive definite then we could define a new inner product (or at least a bilinear form) $\langle x,h\rangle' = \langle x | \color{red}{A} + \color{red}{\epsilon \, I} | y \rangle =\langle ( \color{red}{A} + \color{red}{\epsilon \, I})x | y \rangle = [x,y]$. Here I use bra-ket notation for to make things look symmetric.
If $A$ is symmetric and positive definite, all the eigenvalues $\lambda > 0$ are positive so $A + \epsilon I$ is also symmetric and positive definite with eigenvalues $\lambda + \epsilon > 0$.  Therefore $[x,y]$ is also an inner product and satisfies Cauchy-Schwarz inequality.

Here is standard proof for $\big|[x,y]\big|\leq [x,x]^{1/2}[y,y]^{1/2}$:

We can verify that everywhere along the line $x + ty$ with $t \in \mathbb{R}$: 
$$\langle x+ty | A + \epsilon \, I | x+ty \rangle
= t^2 \langle y |\dots | y \rangle
+ 2t  \langle x | \dots | y \rangle
+ \langle x | \dots | x \rangle \geq 0$$
and since the discriminant is positive, we must have Cauchy-Schwarz here also.


Did I understand your notation correctly?  
