Prove that with the Cauchy Schwarz inequality, if $(f,g)=||f||||g||$ and $g \neq 0$ then $f=cg$ for a scalar $c$. $\textbf{Prompt:}$ Prove that with the Cauchy Schwarz inequality, if $(f,g)=||f||||g||$ and $g \neq 0$ then $f=cg$ for a scalar $c$.
I have an outline for suggestions for the proof, but I don't understand it. Here are the suggestions:
Assume $||f||=||g||=1$ and $(f,g)=1$. Then $f-g$ and $g$ are orthogonal, and $f=f-g+g$. Therefore, $||f||^{2}=||f-g||^{2}+||g||^{2}$.
Now, I don't understand why we assume that $||f||=||g||=1$. What if it's not 1? And...I'm trying to understand the orthogonal part...I understand the following algebra of equating $f$ to $f-g+g$, but how exactly does this conclusion imply that $f=cg$ for a scalar $c$?
 A: Since $\|f\|^2 = \|f-g\|^2 + \|g\|^2$ and $\|f\|^2 = \|g\|^2 = 1$ by assumption, we have $\|f-g\|^2 = 0$, so $f = g$ and the claim holds with $c=1$. To recover the general case from this one, suppose $(f,g) = \|f\|\|g\|$. If $f\ne 0$, put $\hat f = f/\|f\|$ and $\hat g = g/\|g\|$. By what we have just shown, since $\hat f$ and $\hat g$ are unit vectors whose inner product is $1$, we conclude that $\hat f = \hat g$. If $c = \|f\|/\|g\|$ then $f = cg$. If $f = 0$, then $f = 0g$.
A: Hint: It may behoove you to look at the quantity $\|f-g\|^2$. Is there a formula for this that implies $f$ and $g$ should be linearly dependent?
A: You can assume $||f||=||g||=1$ because if not, then you normalize your vectors by dividing each vector by its norm.
Once you prove the theorem for unit vectors, then you can generalize the result to arbitrary vectors because constants cancel out from both sides of $$  ||f||^{2}=||f-g||^{2}+||g||^{2}$$
For your second question, note that with the assumption of $||f||=||g||$, $$||f||^{2}=||f-g||^{2}+||g||^{2} \implies ||f-g||^{2} =0 $$ which implies $f=g.$
That is in this case the constant is $ c=1$ 
A: Suppose $g\ne 0$. Then the Cauchy-Schwarz inequality follows immediately from
$$
        0 \le \left\|\|g\|f-\frac{\langle g,f\rangle}{\|g\|}g\right\|^2
 = \|f\|^2\|g\|^2-|\langle g,f\rangle|^2.
$$
The Cauchy-Schwarz inequality is equality for this $g\ne 0$ iff
$$
              f = \frac{\langle g,f\rangle}{\|g\|^2}g.
$$
So, you don't need to normalize $f,g$.
