Saturation of the Cauchy-Schwarz Inequality Going to as little details as possible, here is a statement from Wald's text on QFT in curved spacetimes(I am not quoting the book)
He considers two vector spaces ${\cal S}$ and ${\cal H}$.

Note - For more details about ${\cal S}$ and ${\cal H}$, read this box. I believe for most part of the question, the following details are irrelevant, but I will provide them nonetheless. Otherwise, skip below.
He starts by considering the solution space of a classical system ${\cal S}$ with symplectic structure $\Omega$. This has a natural vector space structure. He complexifies it to ${\cal S}^{\mathbb C}$ and extends $\Omega$ to ${\cal S}^{\mathbb C}$ by complex linearity on each variable. He then defines the map $(\cdot, \cdot): {\cal S}^{\mathbb C} \times {\cal S}^{\mathbb C} \to {\mathbb C}$ on ${\cal S}^{\mathbb C}$ as
  $$
(y_1, y_2) = - i \Omega( \overline{y_1}, y_2)
$$
  This satisfies all the properties of an inner product except positive-definiteness. He then considers the subspace ${\cal H}$ of ${\cal S}^{\mathbb C}$ on which the inner product above is positive-definite. (There are of course many such choices of ${\cal H}$. Any one of them will do.)

He then shows that there is a one-one onto map $K: {\cal S} \to {\cal H}$. He shows that one can define a real inner product $\mu: {\cal S} \times {\cal S} \to {\mathbb R}$ on ${\cal S}$. He then goes on to show that one can use this to define a complex inner product on $\cal H$ as
$$
\left( K y_1, K y_2 \right)_{\cal H} = \mu(y_1, y_2) - \frac{i}{2} \Omega(y_1, y_2)~\forall~y_1, y_2 \in {\cal S}
$$
where $\Omega: {\cal S} \times {\cal S} \to {\mathbb R}$ is an antisymmetric function on ${\cal S}$, i.e. $\Omega(y_1, y_2) = - \Omega(y_2, y_1)$.
He then uses the Cauchy-Schwarz Inequality for ${\cal H}$. This reads
$$
\left( K y_1, K y_1 \right)_{\cal H} \left( K y_2, K y_2 \right)_{\cal H} \geq \left| \left( K y_1, K y_2 \right)_{\cal H} \right|^2 \geq \left| \text{Im}  \left( K y_1, K y_2 \right)_{\cal H} \right|^2 
$$
Expanding it out, he writes
$$
\mu(y_1, y_1) \mu(y_2, y_2) \geq \mu(y_1, y_2)^2 + \frac{1}{4} \Omega(y_1, y_2)^2 \geq  \frac{1}{4} \Omega(y_1, y_2)^2
$$
More specifically
$$
\boxed{ \mu(y_1, y_1) \mu(y_2, y_2)\geq  \frac{1}{4} \Omega(y_1, y_2)^2 } 
$$
Now, here is the statement that confuses me

Indeed, since $K$ is one-to-one and onto and since the Schwarz inequality on ${\cal H}$ always can be ``saturated", we obtain the following stronger version of the last inequality: For each $y_1 \in {\cal S}$ we have
  $$
\mu(y_1, y_1) = \frac{1}{4} \max_{y_2 \neq 0} \frac{ \Omega(y_1, y_2)^2}{\mu(y_2, y_2)} 
$$
  Here's my question
  Q. Where did he get the above expression from?

He seems to be claiming that the boxed inequality is always saturated for some vector $y_2 \in {\cal S}$. Is that true? Why?
PS - I will understand if some people think that this question is more of a math question than a physics one. But, I thought that it might be possible that the answer relies on some of the assumptions we make in physics, so I asked it here. Any comments will be helpful.
 A: Let $y_1, y_2$ $2$ complex vectors and let $\langle ~,~\rangle$ be a complex inner product defined by $\langle y_1,y_2\rangle = \vec y_1^*.\vec y_2$.
Let $\vec a$ and $\vec b$ the real and imaginary part of $\vec y$ :
$\vec y = \vec a + i \vec b$
Then : 
$$\langle y_1,y_2\rangle  = \left(\vec a_1 \cdot \vec a_2 + \vec b_1 \cdot \vec b_2\right) + i \left(\vec a_1 \cdot \vec b_2 - \vec b_1 \cdot \vec a_2\right) = u(y_1,y_2) + iv(y_1,y_2)$$
The Cauchy-Schwartz inequality gives : 
$$\langle y_1,y_1\rangle \langle y_2,y_2\rangle   ~~\ge ~~|\langle y_1,y_2\rangle |^2$$
We note that : $\langle y_1,y_1\rangle  = u(y_1, y_1)$, so we have :
$$u(y_1,y_1)~~\ge ~~ \frac{u^2(y_1,y_2) + v^2(y_1,y_2)}{u(y_2,y_2)}$$
Now, fixing a particular $y_1$, we limit the set of $y_2$ to those which respect $u(y_1,y_2) =0$. So, we have now : 
$$u(y_1,y_1)~~\ge ~~ \frac{ v^2(y_1,y_2)}{u(y_2,y_2)}\tag 1$$
Now, take explicitly $y_2$ defined by $ \vec a_2 = - ~\vec b_1, \vec b_2 =\vec a_1,$, we see that $\vec a_1 \cdot \vec a_2 + \vec b_1 \cdot \vec b_2 = 0$, that is $u(y_1,y_2) = 0$, so this choice is coherent with our previous hypothesis.
Moreover, we have $v(y_1,y_2) = \vec a_1^2 + \vec b_1^2 $, and $u(y_2,y_2) =  \vec a_1^2 + \vec b_1^2 $, so we have, for this particular $y_2$.
$$u(y_1,y_1)~~= ~~ \frac{ v^2(y_1,y_2)}{u(y_2,y_2)}\tag 1$$
So, we see, that the inequality $(1)$ is effectively saturated  by our choice of this particular $y_2\,.$
