Two quick notes before starting:
1. f and g must be either both increasing or both decreasing (presumably that's what you mean by increasing or decreasing simultaneously). A counterexample where f is decreasing and g is increasing is $\displaystyle f(x) = 1-x$ and $\displaystyle g(x)=x$.
2. If f and g are both decreasing, then $\displaystyle -f$ and $\displaystyle -g$ are both increasing, so it suffices to do the proof with the condition that f and g are both increasing.
So we assume that f and g are both increasing on [a,b], and we want to prove $\displaystyle \int_a^b f(x) g(x)\,dx \ge \frac{1}{b-a}\int_a^b f(x)\,dx\int_a^b g(x)\,dx$.
Let $\displaystyle f_{ave}$ be the average value of f: $\displaystyle f_{ave}=\frac{1}{b-a}\int_a^b f(x)\,dx$. Since f is increasing, there is a c, $\displaystyle a\le c\le b$, such that $\displaystyle f(x)\le f_{ave}$ if $\displaystyle x<c$ and $\displaystyle f(x)\ge f_{ave}$ if $\displaystyle x>c$.
We have $\displaystyle \int_a^c f_{ave}-f(x)\,dx = \int_c^b f(x)-f_{ave}\,dx$ and the integrands are always nonnegative. Since g is increasing, $\displaystyle g(x_1) \le g(x_2)$ for $\displaystyle x_1$ in [a,c] and $\displaystyle x_2$ in [c,b]. So $\displaystyle \int_a^c (f_{ave}-f(x))g(x)\,dx \le \int_c^b (f(x)-f_{ave})g(x)\,dx$. Rearranging gives $\displaystyle \int_a^b f_{ave}\,g(x)\,dx \le \int_a^b f(x)g(x)\,dx$, and the conclusion follows.
- Hollywood
$\displaystyle \displaystyle \int_a^c (f_{ave}-f(x))g(x)\,dx \le \int_c^b (f(x)-f_{ave})g(x)\,dx$
could you please explain how you got this.
$g$ is not necessarily a positive function so what properties of integrals are being used here?
Let $p = sup_{x<c}g(x) \text{ and } q = inf_{x>c} g(x)$. So $p \le q$, and if $g$ is continuous then $p=q=g(c)$. Start with:
$\int_a^c \bar f - f(x)~dx = \int_c^b f(x) -\bar f ~dx$
Put $p$ and $q$ in the left and right side of that to get the middle inequality below to get his inequality:
$\int_a^c (\bar f - f(x))g(x)~dx \le \int_a^c (\bar f - f(x))p~dx \le \int_c^b (f(x) -\bar f)q~dx \le \int_c^b (f(x) -\bar f)g(x)~dx $