# Four Integrals in One Inequality

### Problem

### 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.

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