If $f$ is a positive, monotone decreasing function, prove that $\int_0^1xf(x)^2dx \int_0^1f(x)dx\le \int_0^1f(x)^2dx \int_0^1xf(x)dx$

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

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

Solution 1:

Consider the function $g$ defined on $[0,1]^2$ by $$g(x,y)=\tfrac12(x-y)(f(x)-f(y))f(x)f(y)$$ Then, 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.