Is this true?:
$$\frac{\displaystyle\left|\int_a^bf\left(x\right)\cdot g\left(x\right)\space\text{d}x\right|^2}{\displaystyle\left|\int_a^bf\left(x\right)\space\text{d}x\right|^2}\le\int_a^b\left|g\left(x\right)\right|^2\space\text{d}x$$
(I think it is by the use of the Cauchy-Schwarz inequality)
No, it is not true for any set up where the bottom integral evaluates to $0$.