let $f(x)>0$ is continuous and is increasing on $[0,1]$,and $s=\dfrac{\int_{0}^{1}xf(x)dx}{\int_{0}^{1}f(x)\,dx}$
show that $$\int_{0}^{s}f(x)\,dx\le\int_{s}^{1}f(x)\,dx\le\dfrac{s}{1-s}\int_{0}^{s}f(x)\,dx$$
I have see this problem:
prove the following inequality of Steffensen: if $g$ is Riemann integrable on [a,b] and $0\le g(x)\le 1$ for every $x\in [a,b]$,and $ f$ decreases on that interval,then $$\int_{b-c}^{b}f(x)\,dx\le\int_{a}^{b}f(x)g(x)\,dx\le\int_{a}^{a+c}f(x)\,dx$$ where $$c=\int_{a}^{b}g(x)\,dx$$
poof:since $0\le c\le b-a$,we see that $a+c,b-c\in [a,b]$, now we prove the left inequality, we have \begin{align} &\int_{a}^{b}f(x)g(x)\,dx-\int_{b-c}^{b}f(x)\,dx\\ &=\int_{a}^{b-c}f(x)g(x)\,dx+\int_{b-c}^{b}f(x)(g(x)-1)\,dx\\ &\ge\int_{a}^{b-c}f(x)g(x)\,dx+f(b-c)\left(\int_{b-c}^{b}g(x)dx-c\right)\\ &=\int_{a}^{b-c}f(x)g(x)\,dx-f(b-c)\int_{a}^{b-c}g(x)\,dx\\ &=\int_{a}^{b-c}g(x)(f(x)-f(b-c))\,dx\ge 0 \end{align} The other inequality can be proved analogously.
Without loss of generality, we may assume that $\int_0^1f(x)dx=1$.
Define $F(x)=\int_0^x f(t)dt$. By definition, $F(0)=0$ and $F(1)=1$. Moreover, since $F'=f$ is positive and increasing, $F$ is increasing and convex. Therefore, by Jensen's inequality, $$F(s)=F\big(\int_0^1xf(x)dx\big)\le \int_0^1F(x)f(x)dx=\int_0^1F(x)F'(x)dx=\frac{1}{2}.\tag{1}$$
It follows that $$\int_0^sf(x)dx=F(s)\le 1-F(s)=\int_s^1f(x)dx.\tag{2}$$ By the convexity of $F$, when $0\le t\le 1$, $$F(ts)\le tF(s)+(1-t)F(0)=tF(s)\tag{3}$$ and $$F(ts+1-t)\le tF(s)+(1-t)F(1)=tF(s)+1-t.\tag{4}$$ Due to $(3)$ and $(4)$, we have $$\int_0^s F(x)dx=s\int_0^1F(ts)dt\le\frac{s}{2}F(s)\tag{5}$$ and $$\int_s^1 F(x)dx=(1-s)\int_0^1F(ts+1-t)dt\le\frac{1-s}{2}(F(s)+1).\tag{6}$$ $(5)+(6)$ implies that $$\frac{1}{2}(F(s)+1-s)\ge\int_0^1 F(x)dx= xF(x)\big|_0^1-\int_0^1xf(x)dx =1-s.\tag{7}$$ It follows that $$\int_s^1f(x)dx=1-F(s)\le\frac{s}{1-s}F(s)=\frac{s}{1-s}\int_0^sf(x)dx.\tag{8}$$