Prove that:$\int_a^bp(x)f(x)dx \int_a^bp(x)g(x)dx\leq\int_a^bp(x)dx\int_a^bp(x)f(x)g(x)dx$

Question

Problem:

For continuous, either both increasing or both decreasing functions $f, g$ on $[a, b]$, suppose that $p(x)$ is continuous and positive. Prove that $$\int_a^bp(x)f(x)dx \int_a^bp(x)g(x)dx\leq\int_a^bp(x)dx\int_a^bp(x)f(x)g(x)dx$$

I don't know what I can do. I think the monotonic property of $f, g$ will be used but I have no idea. Can anyone give me hints?

Answer 1

You can proceed similarly as in the proof of the Integral Chebyshev inequality, see for example

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

In your case, you use the fact that the monotonicity of $f$ and $g$ (and the positivity of $p$) implies that $$ 0 \le p(x) p(y) \bigl(f(x) - f(y) \bigr) \bigl(g(x) - g(y) \bigr) $$ for all $x, y \in [a, b]$. Integrating this relationship over $[a,b] \times [a, b]$ yields exactly the wanted inequality $$ \int_a^bp(x)f(x) \, dx \int_a^bp(x)g(x) \, dx \le \int_a^bp(x) \, dx \int_a^bp(x)f(x)g(x) \, dx \, . $$