Estimate integral. (Lp Spaces, Holder)

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    Estimate Integral
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Homework Statement


Show that: \left(\int^{0}_{1}\frac{x^{\frac{1}{2}}dx}{(1-x)^{\frac{1}{3}}}\right)^{3}\leq\frac{8}{5}

Homework Equations


Holder inequality.

The Attempt at a Solution


First, I took the cube root of each side. This let me just deal with the 1-norm on the left. Then I broke the function into 2 parts: f=numerator. g=1/denominator. The 2-norm of each function is integrable. |f|2=(1/2)^(1/2). |g|2=3^(1/2). |f|2*|g|2=(3/2)^(1/2)>(8/5)^(1/3), which is counter the claim. So I obviously missed something.

Where is the flaw in this approach? Are there better approaches to this problem?
 
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You have used the inequality

\int_0^1 |fg|\leq \left(\int_0^1 |f|^p\right)^{1/p}\left(\int_0^1 |g|^q\right)^{1/q}

with the functions ##f(x)=x^{1/2}## and ##g(x) = (1-x)^{-1/3}##. This is correct.

However, you have used ##p=q=2##. This gives you an inequality, but not the correct one. Try to use other constants.
 
I was hoping that I was missing a more strategic approach. (I had already tried working through it with q=3 but ran into unboundedness of g. Working with 2 solved the unboundedness problem, but missed the mark.)

Problem does solve with p=3 and q=3/2. Thanks for the help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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