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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?
Please help me, I have tried to prove it for many days but still have no idea about this...
One can use the same idea as in the proof of the Integral Chebyshev inequality (see for example "Theorem 3 (Chebyshev’s inequality)" in http://imar.ro/journals/Mathematical_Reports/Pdfs/2010/2/Niculescu.pdf):
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 $g([a, b])$.
