I have a question where I am given that
$$\langle|x|^n\rangle=\int_{-\infty}^{\infty}|x|^nP(x)\,dx$$
And must show that $\langle|x|\rangle^2\leq \langle|x|^2\rangle$ using Schwarz's inequality.
EDIT: $P(x)\,dx$ is probability of finding a particle between $x$ and $x+dx$ so these are probabilities and positive!
I have come to the expression of
$$\langle |x|^n\rangle \leq \int_{-\infty}^\infty |x|^2\int_{-\infty}^\infty P^2(x)\,dx$$
Now I feel like my lack of familiarity with integral manipulation is limiting me. I think Scwarz's inequality woyld imply that:
$$\int_{-\infty}^\infty P^2(x) \, dx = \left(\int_{-\infty}^\infty P(x) \, dx\right)^2$$
So then perhaps I could say that
$$\langle|x|^n\rangle \leq \left(\int_{-\infty}^\infty |x|^2 \int_{-\infty}^\infty P(x) \, dx \right) \int_{-\infty}^\infty P(x) \, dx$$
Then if I could combine $\int_{-\infty}^\infty |x|^2 \int_{-\infty}^\infty P(x) \, dx$
to
$$\int_{-\infty}^\infty |x|^2P(x) \, dx$$
then the problem is complete! I am just not sure if this is allowed and would appreciate any extra information on combining/manipulating integrals!