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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...

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  • $\begingroup$ "Please help me, I have tried to prove it for many days but still have no idea about this..."... Can you post some of the ideas you have tried? $\endgroup$
    – Adam
    Jan 7 '16 at 20:29
  • $\begingroup$ 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 $f(c)\int_a^bg(x)dx \leq \int_a^b f(g(x))g(x) dx$. $\endgroup$
    – CCC
    Jan 7 '16 at 21:13
  • $\begingroup$ Actually, I don't know how to use the monotonicity of $f(x)$... $\endgroup$
    – CCC
    Jan 7 '16 at 21:14
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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 $g([a, b])$.

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