Let $f(x),g(x)$ be $\mu$ measureable functions with $g(x) > 0$ almost surely.
Does $$\frac{\left(\int f(x)\mu(x)dx\right)^2}{\int g(x)\mu(x)dx} \leq \int \frac{(f(x))^2}{g(x)}\mu(x)dx$$ hold?
I think I have a proof for the discrete case, but the continuous one eludes me so far. Discrete case: $$\frac{(\sum a_iw_i)^2}{\sum b_i w_i} \leq \sum \frac{a_i^2}{b_i}$$ where $\sum w_i = 1$ and $b_i > 0$
Proof:
$$2a_ia_jw_iw_j = \frac{2a_ia_jw_iw_jb_ib_j}{b_ib_j} = \frac{2w_iw_j\sqrt{a_i^2a_j^2b_i^2b_j^2}}{b_ib_j} \leq \frac{a_i^2w_iw_jb_j^2}{b_ib_j} + \frac{a_j^2w_iw_jb_i^2}{b_ib_j} = \frac{b_i}{b_j}a_j^2w_iw_j + \frac{b_j}{b_i}a_i^2w_iw_j$$ and $$a_i^2w_i^2 = \frac{b_i^2}{b_i^2}a_i^2w_i^2$$ $$\implies \sum\sum a_ia_jw_iw_j \leq \sum\sum \frac{b_i}{b_j}a_j^2w_iw_j \\ \implies \left(\sum a_iw_i\right)^2 \leq \left(\sum w_ib_i\right)\left(\sum w_i \frac{a_i^2}{b_i}\right) \\ \implies \frac{\left(\sum w_ia_i\right)^2}{\sum w_ib_i} \leq \sum \frac{a_i^2}{b_i}$$
This is Cauchy-Schwarz inequality in the space $L^2(\mu)$ applied to the functions $\frac f{\sqrt g}$ and $\sqrt g$.