How prove this integral inequality $\int_{0}^{1}(f(x)+g(x))dx\ge\int_{0}^{1}f(g(x))dx$?

Asked Dec 23 '14 at 16:43

Active Dec 23 '14 at 17:09

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let two function $f(x),g(x):[0,1]\to [0,1]$ are Continuous function,and $f(x)$ strictly monotone increasing,

show that $$\int_{0}^{1}(f(x)+g(x))dx\ge\int_{0}^{1}f(g(x))dx$$

I am not comfortable with compositions and don't have any idea on how to deal with the problem. Any help will be appreciated. Thanks!

This problem is from analysis problem book excises by Min hui xie (This book exsice problem it is very hard)see this

enter image description here

edited Jun 12 '20 at 10:38

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asked Dec 23 '14 at 16:43

math110

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maybe can use Geometric meaning to explain? – math110 Dec 23 '14 at 16:50
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Put your claim forth! – Swapnil Tripathi Dec 23 '14 at 16:51
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May we know the context behind this question ? It looks like homework/contest, and answering a homework/contest question which does not show the OP's work wouldn't feel right. "I don't have any idea" is surely not the way to go. – Hippalectryon Dec 23 '14 at 16:52
It would be great if your graph could be a bit more... readable.. – Hippalectryon Dec 23 '14 at 16:58
Is $f(x)$ strictly monotone increasing? – Nicholas Stull Dec 23 '14 at 17:08
@NicholasStull,Yes,Thank you,I have edit it – math110 Dec 23 '14 at 17:09
@NicholasStull "increasing" usually means strictly increasing (otherwise, it would have been non-decreasing). – Clement C. Dec 23 '14 at 17:10
@chinamath For $f(x)=x^n$ and $g(x) = x^{1/n}$ the right hand side is $1/2$, not $1$. – Winther Dec 23 '14 at 17:11
@Winther,Oh,I know ,But also such inequality – math110 Dec 23 '14 at 17:11
@ClementC. I was just confirming that. Some books differ in convention, which is why I asked. – Nicholas Stull Dec 23 '14 at 17:13

How prove this integral inequality $\int_{0}^{1}(f(x)+g(x))dx\ge\int_{0}^{1}f(g(x))dx$?

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