Let $(\Omega,\mathcal A,\mu)$ be a $\sigma$-finite measure space, $\mathbb P,\mathbb Q$ be probability measures on $(\Omega,\mathcal A)$ with density functions $f,g: \Omega \to (0,\infty)$ concerning $\mu$.
Prove that $$\int_\Omega\min(f,g)\text{d}\mu\ge\frac{1}{2}\left(\int_\Omega\sqrt{f\cdot g}\ \text{d}\mu\right)^2$$
I have already tried everything I can think of, like with the Jensen-Inequality:
$$\frac{1}{2}\left(\int_\Omega\sqrt{f\cdot g}\ \text{d}\mu\right)^2\le\int_\Omega\frac{f\cdot g}{2}\ \text{d}\mu$$ but it all leads nowhere. Can anyone give me a hint? I greatly appreciate any help!
This is just an application of Cauchy-Schwarz: \begin{align} \left( \int_\Omega \sqrt[]{fg} \, d\mu\right) ^2&=\left( \int_\Omega \sqrt[]{\min(f, g) \max(f, g) } \, d\mu\right) ^2 \\ &\leq \left( \int_\Omega \min(f, g) \, d\mu\right) \left( \int_\Omega \max(f, g) \, d\mu\right) \end{align} Now notice that $\max(f, g) \leq f+g$ to finish.