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$\frac 12 \int_{0}^{1} x^4(1-x)^4 dx\leq \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx \leq \int_{0}^{1} x^4(1-x)^4dx$

This is my homework assignment whereby I have no idea of how should i get started. Could i have some hint on this question??

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Hint: if $0 \leq x \leq 1$, then $1/2 \leq 1/(1+x^2) \leq 1$. Now look at the areas that these integrals represent. – Old John Nov 15 '12 at 11:23
hmm... I still don't get it, @OldJohn – melyong Nov 15 '12 at 11:35

(**) I hope you know that $\,f(x)\geq 0\,\,\,\,\forall\,x\in [a,b]\Longrightarrow \int_a^bf(x)\,dx\geq 0\,$... If you don't then proving this by means of Riemann sums is really easy.

Well, taking now $\,[0,1]\,$, we have that

$$\frac{1}{2}\,x^4(1-x)^4\leq \frac{x^4(1-x)^4}{1+x^2}\leq x^4(1-x)^4$$

and now apply (**) to each inequality above

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Hint: For $x\in[0,1]$ $$\frac12\le\frac1{1+x^2}\le1$$ Multiply by $x^4(1-x)^4$ and integrate over $[0,1]$.

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I see that this is already given in a comment by OldJohn. – robjohn Nov 15 '12 at 16:08
ahhh. im getting what is @OldJohn was trying to say. Thanks – melyong Nov 15 '12 at 20:00

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