Given two increasing continuous functions $f,g$ prove that $(b-a) \int^b_a f(x)g(x) dx > \int^b_a f(x) dx \int^b_a g(x) dx$

Given two monotonically increasing continuous functions $f,g$ prove that $$(b-a) \int^b_a f(x)g(x) dx > \int^b_a f(x) dx \int^b_a g(x) dx,\; b>a$$

what I have tried:

Let $$h(x) = (x-a)\int^x_a f(t)g(t) dt - \int^x_a f(t)dt\int^x_a g(t)dt $$ and trying to figure out that $h'>0$ using MVT to reform $h'(x)$ as

$$h'(x) = (x-a)(-f(x)g(\xi_1) - g(x)f(\xi_2) + f(x)g(x) + f(\xi_3)g(\xi_3))$$ where $\xi_{1,2,3} \in (a,x)$ (That method does not seem to work)

That is the Integral Chebyshev inequality. The following proof is from http://imar.ro/journals/Mathematical_Reports/Pdfs/2010/2/Niculescu.pdf (Theorem 3):

If $f$ and $g$ are both increasing (or both decreasing) then $$ \tag{*} 0 \le \bigl(f(x) - f(y) \bigr) \cdot \bigl(g(x) - g(y) \bigr) $$ for all $x, y \in [a, b]$. It follows that $$ 0 \le \int_a^b \int_a^b \bigl(f(x) - f(y) \bigr) \cdot \bigl(g(x) - g(y) \bigr) \, dx dy \\ = 2 (b-a) \int_a^b f(x) g(x) \, dx - 2 \left(\int_a^b f(x)\,dx\right)\left(\int_a^b g(x)\,dx\right) \, . $$

If $f$ is increasing and $g$ decreasing (or vice versa) then the reverse inequality holds.

If equality holds then equality holds in $(*)$ for all $x, y \in [a, b]$ (since $f$ and $g$ are assumed to be continuous). In particular $$ 0 = \bigl(f(a) - f(b) \bigr) \cdot \bigl(g(a) - g(b) \bigr) $$ which means that (at least one of) $f$ or $g$ is constant.

Here is a (somewhat) different proof I found, based on the integral mean value theorem.

The inequality is equivalent to: $$\int_a^b \left( f-\frac{1}{b-a}\int_a^b f\right) g\geq 0. $$

Let $\xi\in [a,b]$ be such that $f(\xi)=\frac{1}{b-a}\int_a^b f$. Then \begin{align*}\int_a^b \left( f-\frac{1}{b-a}\int_a^b f\right) g=\int_\xi^b (f-f(\xi))g+\int_a^{\xi}(f-f(\xi))g=:I_1+I_2. \end{align*} Since $f$ is increasing, we have $f-f(\xi)\geq 0$ on $[\xi,b]$ and $f-f(\xi)\leq 0$ on $[a,\xi]$. Moreover, since $g$ is increasing, we get $$I_1+I_2 \geq \int_{\xi}^b (f-f(\xi))g(\xi)+\int_a^{\xi}(f-f(\xi))g(\xi)=g(\xi)\int_a^b (f-f(\xi))=0.$$

EDIT: To have equality, we need $\int_{\xi}^b (f-f(\xi))(g-g(\xi))=\int_{a}^{\xi} (f-f(\xi))(g-g(\xi))=0$ which implies that $(f-f(\xi))(g-g(\xi))=0$ on $[a,b]$. From here, using the fact that $f,g$ are increasing it is not hard to deduce that either $f$ or $g$ must be constant on $[a,b]$.