# $\int_\Omega\min(f,g)\text{d}\mu\ge\frac{1}{2}\big(\int_\Omega\sqrt{f\cdot g}\ \text{d}\mu\big)^2$

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!

• In this case f,g wouldnt be density functions. $\int f\ \text{d}\mu=5\ne 1$ – newbie Oct 26 '18 at 14:23
• This question has a poor choice of fonts for probability measures $\Bbb{R}$ and $\Bbb{Q}$. I was very confused at first because $\Bbb{R}$ and $\Bbb{Q}$ usually denote the set of real numbers and the set of rational numbers. Can you perhaps change $\mathbb{R}$ and $\mathbb{Q}$ to just plain $R$ and $Q$, @newbie? – user593746 Oct 26 '18 at 16:34

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.
• @newbie did you look at the Wikipedia page? You have a product of two functions which I have chosen to be $\sqrt[]{\min(f, g)}$ and $\sqrt[] {\max(f, g)}$. – Shashi Oct 26 '18 at 16:13