Hi I am trying to solve the problem below and somehow I do not have the right idea. A particle is located between x and x + dx with probability P(x)dx. If
$\displaystyle \langle |x|^n \rangle \equiv \displaystyle \displaystyle\int_{-\infty}^{\infty} |x|^nP(x)\ dx$
show, using Schwarz's inequality, that $\langle |x| \rangle^2 \leqslant \langle |x|^2 \rangle $ .
This is a special case of Jensen's inequality. The proof of this special case can be obtained from Cauchy-Schwarz's inequality.
Note that \begin{align} \langle |x|\rangle &=\int_{-\infty}^{\infty}|x|P(x)dx \le \left(\int_{-\infty}^{\infty}|x|^2P(x)dx\right)^{1/2}\left(\int_{-\infty}^{\infty} 1^2P(x)dx\right)^{1/2}\\ &=\left(\int_{-\infty}^{\infty}|x|^2P(x)dx\right)^{1/2} = \langle |x|^2\rangle^{1/2}. \end{align}
Squaring both sides gives you the desired inequality. The inequality in the first line uses Cauchy-Schwarz and the $\int_{-\infty}^{\infty} 1^2P(x)dx=1$ because $P(x)$ is a probability density.
As an aside, one can prove a general version of this result known as Jensen's inequality. For any positive convex function $\phi,$ we have $\phi(\langle |x|\rangle)\le \langle \phi(|x|)\rangle.$
