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.