Mean of product vs. product of means
Using the Cauchy-Schwarz inequality we can prove that\begin{equation} \frac{1}{b-a} \int \limits_a^b f(x)^2 \, \mathrm{d}x \geq \left( \frac{1}{b-a} \int \limits_a^b f(x) \, \mathrm{d}x \right)^2. \end{equation}This is true because\begin{eqnarray} \left( \frac{1}{b-a} \int \limits_a^b f(x) \, \mathrm{d}x \right)^2 &=& \frac{1}{(b-a)^2} \left( \int \limits_a^b f(x) \cdot 1 \, \mathrm{d}x \right)^2\\ &\stackrel{\textrm{C-S}}{\leq}& \frac{1}{(b-a)^2} \int \limits_a^b f(x)^2 \, \mathrm{d}x \underbrace{\int \limits_a^b 1^2 \, \mathrm{d}x }_{b-a} = \frac{1}{b-a} \int \limits_a^b f(x)^2 \, \mathrm{d}x. \end{eqnarray}Is the following more general statement also true:\begin{equation} \frac{1}{b-a} \int \limits_a^b f(x) g(x) \, \mathrm{d}x \geq \frac{1}{b-a} \int \limits_a^b f(x) \, \mathrm{d}x \frac{1}{b-a} \int \limits_a^b g(x) \, \mathrm{d}x \end{equation}(say for non-negative functions to avoid problems)?
In words we would state this as "The mean of the product of two functions is greater or equal than the product of the means of the two functions", which is true if both functions are the same.
Any hint on how to prove or disprove this would be appreciated.
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$\begingroup$The statement\begin{equation} \frac{1}{b-a} \int \limits_a^b f(x) g(x) \, \mathrm{d}x \geq \frac{1}{b-a} \int \limits_a^b f(x) \, \mathrm{d}x \frac{1}{b-a} \int \limits_a^b g(x) \, \mathrm{d}x \end{equation}is false in general, even for strictly positive functions $f$ and $g$. To see this we first write the difference in the form\begin{eqnarray} && \frac{1}{b-a} \int \limits_a^b f(x) g(x) \, \mathrm{d}x - \frac{1}{b-a} \int \limits_a^b f(x) \, \mathrm{d}x \frac{1}{b-a} \int \limits_a^b g(x) \, \mathrm{d}x\\ && = \frac{1}{b-a} \int \limits_a^b \left( f(x) - \frac{1}{b-a} \int \limits_a^b f(x) \, \mathrm{d}x \right) \left( g(x) - \frac{1}{b-a} \int \limits_a^b g(x) \, \mathrm{d}x \right) \, \mathrm{d}x. \end{eqnarray}The last expression may be negative in general, for example with\begin{equation} f(x) := 1+\frac{x-a}{b-a}, \quad g(x) := 2-\frac{x-a}{b-a} \end{equation}(strictly positive on $[a,b]$). We obtain\begin{equation} \frac{1}{b-a} \int \limits_a^b f(x) \, \mathrm{d}x = \frac{1}{b-a} \int \limits_a^b g(x) \, \mathrm{d}x = \frac{3}{2}, \end{equation}and\begin{equation} \frac{1}{b-a} \int \limits_a^b \left( f(x) - \frac{1}{b-a} \int \limits_a^b f(x) \, \mathrm{d}x \right) \left( g(x) - \frac{1}{b-a} \int \limits_a^b g(x) \, \mathrm{d}x \right) \, \mathrm{d}x = -\frac{1}{12} < 0, \end{equation}so this is a counterexample.
However, if $f, g$ are monotonic functions of the same monotonicity, then the statement holds and is called the integral Chebyshev inequality.
Thanks for the helpful comments!
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