How prove $\int_{0}^{1}F(x)dx\le \int_{0}^{a}G(x)dx$

Question

if $G$ and $F$ are integrable,$a>0,G(x)\ge F(x)\ge 0$,and $$\int_{0}^{1}xF(x)dx=\int_{0}^{a}xG(x)dx$$ show that $$\int_{0}^{1}F(x)dx\le \int_{0}^{a}G(x)dx$$ let$$F(x)=\int_{0}^{x}f(t)dt,G(x)=\int_{0}^{x}g(t)dt$$ by parts:we have $$\int_{0}^{1}F(x)dx=xF(x)|_{0}^{1}-\int_{0}^{1}xdF(x)=F(1)-\int_{0}^{1}xf(x)dx$$ and $$\int_{0}^{a}G(x)dx=xG(x)|_{0}^{a}-\int_{0}^{a}xg(x)dx=G(a)-\int_{0}^{a}xg(x)dx$$ then follwing how can works? This problem from $MM$ (Mathematics Magazine) problem $622$,Thank you everone

Answer 1

Let $J(x)=\int_0^xF(t)dt$ and $K(x)=\int_0^xG(t)dt$. Since $0\le F\le G$, $J$ and $K$ are increasing, and $$J(x)\le K(y),\quad\forall\ 0\le x\le y.\tag{1}$$ We want to show $J(1)\le K(a)$. From $(1)$ we know that when $a\ge 1$, there is nothing to prove, so let us assume that $0<a<1$. On the one hand, due to integration by parts, $$J(1)-\int_0^1 J(x)dx=\int_0^1xF(x)dx=\int_0^axG(x)dx=aK(a)-\int_0^a K(x)dx.\tag{2}$$ On the other hand, by $(1)$ and noting that $J$ is increasing, $$\int_0^1 J(x)dx-\int_0^a K(x)dx\le \int_a^1 J(x)dx\le (1-a)J(1). \tag{3}$$ Combining $(2)$ and $(3)$, it follows that $J(1)\le K(a)$.