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Integral given to SEEMOUS 2011.Need help!

Posted November 28th, 2011 by MSA
in

For a given integer $ n\ge1 $,let f:[0,1]--->R be a non-decresing function.Prove that:
$ \int^1_0 f(x)\,dx\le (n+1)\int^1_0 x^nf(x)\,dx $.
Find all non-decreasing continuous functions for wich equality holds.

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Answer from Romania

On December 8th, 2011 Structure says:

This is a nice problem.You can solve it like that:

Notice $ \int_0^1(n+1)x^n dx =1 $ You can see that $ u(x)=(n+1)x^n $ is increasing on [0,1] and $ u(x)=1 $ for

$$a=\frac{1}{\sqrt[n]{n+1}}$$

We consider now $ \int_0^1(1-u(x))dx=0 $ and $ \int_0^a(1-u(x))dx=\int_a^1 (u(t)-1) dt $ and for $ x\in [0,a] $ $ f(x)\le f(a) $ and for $ t\in [a,1] $

$$f(x)\le f(a)\le f(t)$$
$$\int_0^a(1-u(x))f(x)dx\le\int_0^a(1-u(x)) f(a) dx=\int_a^1 (u(t)-1)f(a) dt\le\int_a^1 (u(t)-1)f(t) dt$$

You have from here

$$\int_0^af(x)dx +\int_a^1 f(t) dt\le \int_0^au(x)f(x)+\int_a^1u(t) f(t) dt$$

From this solution you can see that equality can hold just for constant non decreasing function

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