Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $f$ is a positive, monotone decreasing function, prove that

$\int_0^1xf(x)^2dx \int_0^1f(x)dx\le \int_0^1f(x)^2dx \int_0^1xf(x)dx$

share|improve this question

1 Answer 1

up vote 8 down vote accepted

Consider the function $g:(x,y)\mapsto\frac12(x-y)(f(x)-f(y))f(x)f(y)$ on $[0,1]^2$. On the one hand, $g\leqslant0$ on $[0,1]^2$ (why?). On the other hand, expanding $g(x,y)$ into the sum of four terms like $xf(x)^2f(y)$ or $xf(x)f(y)^2$, one gets the identity $$ \iint_{[0,1]^2} g(x,y)\mathrm dx\mathrm dy=\int_0^1xf(x)^2\mathrm dx\cdot\int_0^1f(x)\mathrm dx-\int_0^1xf(x)\mathrm dx\cdot\int_0^1f(x)^2\mathrm dx. $$ Since the LHS is $\leqslant0$, this proves the desired inequality.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.