Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 That's all, Thank you – math110 May 7 '13 at 7:07
up vote 5 down vote accepted

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

share|cite|improve this answer
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 $\int_0^1 xf(x)dx < \int_0^1 f(x)dx,s<1$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.