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I need this thing for my work, I think it is true but not sure, didn't manage to prove it.

$f, g, f', g': \mathbb{R} \to \mathbb{R}^+$ are all monotonic non-decreasing functions, continuous almost everywhere. We know that 1>$\int_{-\infty}^\infty f(x) dx > \int_{-\infty}^\infty f'(x) dx$ and $1>\int_{-\infty}^\infty g(x) dx > \int_{-\infty}^\infty g'(x) dx$. Prove that $\int_{-\infty}^\infty f(x)g(x) dx > \int_{-\infty}^\infty f'(x)g'(x) dx$.

Thanks!

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    How is $f$ integrable? It's domain is positive, and it's nondecreasing. – 2017-01-26

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I'll try a standard trick and see if it helps.

Spoiler: Didn't work. Oh, well. Maybe it will help someone else, so I'll submit it.

$\begin{array}\\ \int_{-\infty}^\infty (f(x)g(x)- f'(x)g'(x)) dx &=\int_{-\infty}^\infty (f(x)g(x)-f'(x)g(x)+f'(x)g(x)- f'(x)g'(x)) dx\\ &=\int_{-\infty}^\infty (f(x)g(x)-f'(x)g(x))dx+\int_{-\infty}^\infty(f'(x)g(x) -f'(x)g'(x)) dx\\ &=\int_{-\infty}^\infty (f(x)-f'(x))g(x)dx+\int_{-\infty}^\infty f'(x)(g(x)-g'(x)) dx\\ \end{array} $