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

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.