Let $f,g,$ be integrable on $[a,b]$. Prove that
$$\int_a^b(fg)^2\le\int_a^bf^2\int_a^bg^2$$
I know that from Cauchy-Schwarz we have
$$\left(\int_a^bfg\right)^2\le\int_a^bf^2\int_a^bg^2$$
so if we showed that
$$\int_a^b(fg)^2\le\left(\int_a^bfg\right)^2$$
we would be done. But I don't think this is even true in general, so this method doesn't seem to lead anywhere. Is there another approach to this problem that I'm missing?
Edit: The inequality seems to be false. Perhaps the inequality was given incorrectly.
Best Answer
This doesn't seem to be true in general. For example:
Take $f(x)=1$ for $x \in [0,\frac{1}{2}]$
$g(x)=x$ (same interval.)
Then, your LHS= $\frac{1}{24}$, RHS is $\frac{1}{48}$