Problem about proving the uncertainty principle $f$ and $g$ are two quadratially integrable functions. It is shown from Schwarz inequality, def of variance and an identity of complex numbers that
$$(\Delta A)^2(\Delta B)^2=\langle f|f\rangle \langle g|g\rangle~\geq~|\langle f|g\rangle|^2~\geq~[\frac{(\langle f|g\rangle-\langle g|f \rangle)}{2i}]^2$$
Then the proof says: Use the definition of $f$ and $g$ and the normalization of $\Psi$ to verify that 
$$\left\langle f|g> = <g|f \right\rangle^*=\langle\Psi|\hat A \hat B|\Psi\rangle-\langle A \rangle \langle B \rangle \tag{I}$$   
Which leads to the formula:
$$(\Delta A)^2(\Delta B)^2~\geq~\frac{1}{4}|\langle \Psi| [\hat A, \hat B] | \Psi \rangle|^2 $$
I don't get the equality in (I). I also don't get the notation on the left side. Can anyone help out?
The proof is taken from problem 7.60 in Ira Levine quantum mechanics. 
 A: If $\langle A\rangle=\langle \Psi| \hat A| \Psi \rangle$ exists and is finite. And $\langle B\rangle=\langle \Psi| \hat B| \Psi \rangle$ exists and is finite. Then we can define $|f\rangle=\hat A |\Psi\rangle-\langle A\rangle|\Psi\rangle$ and define $|g\rangle=\hat B |\Psi\rangle-\langle B\rangle|\Psi\rangle.$
If $f$ and $g$ are square integrable then you can compute the integrals of their squares (which are $\langle f|f\rangle$ and $\langle g|g\rangle$ respectively). By simplifying the results you'll see they equal the variance of $A$ and $B$ respectively. Then you can combine with Cauchy-Schwarz to get:
$$(\Delta A)^2(\Delta B)^2=\langle f|f\rangle\langle g|g\rangle~\geq~|\langle f|g\rangle |^2.$$
Now $\langle f|g\rangle = \langle g|f\rangle^*$ is just the generalization of $\vec a \cdot \vec b=\vec b\cdot \vec a$ to complex vectors where instead of being symmetric it is conjugate symmetric. For example, if $\langle f|g\rangle=\int f^*g$ and $\langle g|f\rangle=\int g^*f$ then $\langle f|g\rangle = \langle g|f\rangle^*$ since:
$$\begin{array} & \int f^*g &=\left(\int f^*g\right)^{**}\\&=\left(\int (f^{*}g)^*\right)^*\\&=\left(\int f^{**}g^*\right)^*\\&=\left(\int g^*f\right)^*.\end{array}$$
And to get the right hand side you do the same tricks as showing they equaled the variance, bring constants outside of the integrals.
