I've read in a few places that Cauchy-Schwarz's inequality for the inner product of $L^2$ is just Hölder's inequality for the special case $p=q=2$ but I think there is missing some (maybe trivial) step.
Cauchy-Schwarz's inequality for the inner product of $L^2$ gives $$\left\vert{\int_X f \overline{g} }\right\vert \leq \int_X|f|^2\int_X|g|^2$$ And triangle inequality for integrals gives $$\left\vert{\int_X f \bar{g}}\right\vert \le \int_X \left\vert{f g}\right\vert $$ I don't see how to arrive to $$\int_X \left\vert{f g}\right\vert \leq \int_X|f|^2\int_X|g|^2$$ from Cauchy-Schwarz's, I mean I would understand it if the triangle inequality were the other way round
Hint: Consider C-S inequality for $|f|$ and $|g|$.