Cauchy-Schwarz Hermitian Inner Product Remainder
Instead of posting more and more comments (I've deleted most of them) it's better to formulate sort of an answer.
Let's first get straight the derivation of Schwarz inequality for the complex inner product.
The most concise proof I could find on the internet is in the (Dutch) Wikipedia page
"Ongelijkheid van Cauchy-Schwarz".
The inequality reads, with $a$ and $b$ complex vectors:
$$
|\langle a,b \rangle|^2 \le \langle a,a \rangle \langle b,b \rangle
$$
Equality is when $b = \lambda a$ ($\lambda$ complex). Then trivialiter:
$$
|\langle a,b \rangle|^2 = |\lambda|^2 \langle a,a \rangle =
\langle a,a \rangle \langle \lambda a, \lambda a \rangle =
\langle a,a \rangle \langle b,b \rangle
$$
Assume trivial cases done and assume that $b \ne 0$ . Then for each
complex number $\lambda$ :
$$
0 \le \langle a-\lambda b, a-\lambda b \rangle =
\langle a,a \rangle - \lambda \langle b,a \rangle
- \overline{\lambda} \langle a,b \rangle + |\lambda|^2 \langle b,b \rangle
$$
Where $\overline{\lambda}$ is the complex conjugate of $\lambda$. Now take:
$$
\lambda = \frac{\langle a,b \rangle}{\langle b,b \rangle}
$$
Then the Schwarz inequality easily follows from:
$$
0 \le \langle a,a \rangle - \frac{\langle a,b \rangle}{\langle b,b \rangle}\langle b,a \rangle
- \frac{\langle b,a \rangle}{\langle b,b \rangle} \langle a,b \rangle + \left|\frac{\langle a,b \rangle}
{\langle b,b \rangle}\right|^2 \langle b,b \rangle\\ \Longleftrightarrow \qquad
0 \le \langle a,a \rangle - \frac{|\langle a,b \rangle|^2}{\langle b,b \rangle}
$$
Of course $f(a,b)$ must be a positive function if we write instead:
$$
|\langle a,b \rangle|^2 = \langle a,a \rangle \langle b,b \rangle - f(a,b)
\quad \Longrightarrow \quad f(a,b) = \langle a,a \rangle \langle b,b \rangle - |\langle a,b \rangle|^2
$$
If that is what's meant by the OP ..