# An integral inequality for increasing continuous function

If $f$ is a increasing continuous real-valued function on $\mathbb{R}$ and $g$ is a continuous real-valued function on $[a,b]$. Then does the inequality $$\left(\int_a^b f(g(x))dx\right)\left(\int_a^b g(x)dx\right) \leq (b-a)\int_a^bf(g(x))g(x)dx$$holds ture?

Jan 7, 2016 at 20:29
• I tried to prove it by applying the mean value theorem for integration on $\int_a^b f(g(x))dx$ but I don't know how to show that $f(c)\int_a^bg(x)dx \leq \int_a^b f(g(x))g(x) dx$.
– CCC
Jan 7, 2016 at 21:13
• Actually, I don't know how to use the monotonicity of $f(x)$...
– CCC
Jan 7, 2016 at 21:14

The monotonicity of $f$ implies that $$\tag{*} 0 \le \bigl(f(g(x)) - f(g(y) \bigr) \cdot \bigl(g(x) - g(y) \bigr)$$ for all $x, y \in [a, b]$, and therefore $$0 \le \int_a^b \int_a^b \bigl(f(g(x)) - f(g(y) \bigr) \cdot \bigl(g(x) - g(y) \bigr) \, dx dy \\ = 2 (b-a) \int_a^b f(g(x)) g(x) \, dx - 2 \left(\int_a^b f(g(x))\,dx\right)\left(\int_a^b g(x)\,dx\right) \, .$$
One can also see that equality holds if and only if equality holds in $(*)$ for all $x, y \in [a, b]$ (since both function are assumed to be continuous), and that is the case if and only if $f$ is constant on $g([a, b])$.