Estimate integral. (Lp Spaces, Holder)

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SUMMARY

The discussion centers on estimating the integral \(\left(\int^{0}_{1}\frac{x^{\frac{1}{2}}dx}{(1-x)^{\frac{1}{3}}}\right)^{3}\) and demonstrating that it is less than or equal to \(\frac{8}{5}\). The participants utilized Hölder's inequality, initially attempting with \(p=q=2\), which led to an incorrect conclusion. The correct approach involves using \(p=3\) and \(q=\frac{3}{2}\), successfully resolving the integral estimation problem.

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Homework Statement


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

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

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

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.
 

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