For $F \int_\beta^1 G(x)|\ln x| dx$? I want to show for any two distribution functions $F<G$ defined on $[0,1]$, $$\int_\alpha^1 F(x)dx = \int_\beta^1 G(x)dx$$ implies $$\int_\alpha^1 F(x)|\ln x| dx \geq \int_\beta^1 G(x)|\ln x| dx.$$
It must be $\alpha<\beta$ as $F$ stochastically dominates $G$. My intuition is then, by multiplying by $|\ln x|$, $F(x)$ and $ G(x)$ get smaller for bigger $x$; while they get bigger for smaller $x$. So $\alpha<\beta$ would yield $\int_\alpha^1 F(x)|\ln x| dx > \int_\beta^1 G(x)|\ln x| dx$.
I find this is true at least for some cases by investigating numerically -- for instance, $F(x)=x^2$ and $G(x)=x$. But I am struggling to prove (or disprove) it. I would appreciate any hints or suggestions.

A simpler version of the question might be: For two monotone functions $F<G$ with $F(0)=G(0)=0$ and $F(1)=G(1)=1$,
$$\int_\alpha^1 F(x)dx = \int_\beta^1 G(x)dx \implies \int_\alpha^1 xF(x)dx \leq \int_\beta^1 xG(x) dx.$$
It looks simple, but I cannot prove (or disprove) it.
 A: I think I find a proof for a more general case. I would appreciate if anyone can confirm this. Please let me know if I miss something.
Let $F$ and $G$ be non-decreasing functions with $0\leq F(x)\leq G(x)$ for all $x$. For any $\alpha \leq \beta \leq B$ such that
$$\int_\alpha^B F(x)dx = \int_\beta^B G(x)dx,$$
and any non-decreasing function $M(x)$, it holds
$$\int_\alpha^B F(x)M(x)dx \leq \int_\beta^B G(x)M(x)dx.$$
Proof: Define $\beta(\alpha)$ as a solution to
$$\int_\alpha^B F(x)dx = \int_{\beta(\alpha)}^B G(x)dx.$$
Note $\beta(\alpha)$ uniquely exists for any $\alpha$; $\beta(\alpha)\geq \alpha$; $ \beta(B)=B$; and $\beta'(\alpha)G(\beta(\alpha))=F(\alpha)$.
Let $$K(\alpha)=\int_\alpha^B F(x)M(x)dx -\int_{\beta(\alpha)}^B G(x)M(x)dx$$ to show $K(\alpha)\leq 0$ for all $\alpha<B$. Note $K(B)=0$. For all $\alpha\leq B$, we have $K'(\alpha)\geq 0$, because
$$K'(\alpha)= -F(\alpha)M(\alpha)+G(\beta(\alpha))M(\beta(\alpha))\beta'(\alpha)= -F(\alpha)[M(\alpha)-M(\beta(\alpha))]\geq0.$$
Thus, $K(\alpha)\leq 0$ for all $\alpha<B$.
