Let $ f(x) = \frac{1}{\sqrt{2\pi}} e ^{-x^2/2} $. In a paper the author claims that $ \int_{\mathbb{R}} |w| f(w) dw \leq \left( \int_{\mathbb{R}} |w|^2 f(w) dw \right)^{1/2} $. I don't see a good argument for the step. I would use Cauchy-Schwarz inquality (with $ L^2 $- norm), Minkowsky inquality or Hölder inquality but they don't seem to fit.
Do you have any clue?
Since $f(x)$ is the probability density function of a normal distribution, it reduces to $$ \mathbb{E}\lvert Z\rvert\leq[\mathbb{E}\lvert Z\rvert^2]^{1/2} $$ which is immediate from Cauchy-Schwarz.