# How prove this integral inequality $\int_{0}^{s}f(x)\,dx\le\int_{s}^{1}f(x)\,dx\le\dfrac{s}{1-s}\int_{0}^{s}f(x)\,dx$

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.

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Hey, math110, the problems in your questions are constantly challenging, and at competition level. Where did you find these problems? Could you please sharing some sources if possible? Nice problem btw+1. – Shuhao Cao May 7 '13 at 2:27
HaHa, in china, I am a competition teacher,so I have lot of some student ask some hard questions, and this problem is my good frend and ago is my teacher creat, and he is benjing universty teacher, He teache math,His profile:baike.baidu.com/view/1876355.htm That's all, Thank you – math110 May 7 '13 at 7:07

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 $$\int_s^1f(x)dx=1-F(s)\le\frac{s}{1-s}F(s)=\frac{s}{1-s}\int_0^sf(x)dx.\tag{8}$$

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nice, Is very nice. – math110 May 7 '13 at 11:48
@math110: Thank you. – 23rd May 7 '13 at 11:58

Because $x\in[0,1],xf(x)\le f(x)$, then $\int_0^1 xf(x)dx < \int_0^1 f(x)dx,s<1$

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Thank you,@Genming Wang, Then ? – math110 May 4 '13 at 17:48
How come this is an answer? – Shuhao Cao May 7 '13 at 2:25
No, I think this problem not easy. Thank you – math110 May 7 '13 at 7:03