Four Integrals in One Inequality

Solution

By Chebyshev's inequality (this is where we need function $f(x)\,$ to be increasing)

$\displaystyle \int_a^bxf(x)dx\ge\frac{1}{b-a}\left(\int_a^bxdx\right)\left(\int_a^bf(x)dx\right)=\frac{a+b}{2}\int_a^bf(x)dx.$

Similarly,

$\displaystyle \int_a^bx^3f(x)dx\ge\frac{a^3+a^2b+ab^2+b^3}{4}\int_a^bf(x)dx.$

By the Cauchy-Schwarz inequality,

$\displaystyle \int_a^bf^2(x)dx\ge\frac{1}{b-a}\left(\int_a^bf(x)dx\right)^2.$

By the AM-GM inequality, $\displaystyle \frac{a+b}{2}\cdot\frac{a^3+a^2b+ab^2+b^3}{4}\ge a^2b^2.\,$ Multiplying the three inequality yields the required one.

Acknowledgment

Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. Leo Giugiuc has commented with his solution.