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Math Help - Prove

  1. #1
    Super Member dhiab's Avatar
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    Prove

    Prove :
    If :  \left\{ \begin{array}{l}<br />
x,y \in \Re+ \\ <br />
n \in {\rm N} \\ <br />
x + y = 1 \\ <br />
\end{array} \right.<br />
    Then :  \left( {1 + \frac{1}{{x^n }}} \right)\left( {1 + \frac{1}{{y^n }}} \right) \ge \left( {1 + 2^n } \right)^2 <br />
    Last edited by dhiab; October 3rd 2009 at 07:23 AM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint: set x= \cos^2 \theta, y = \sin^2 \theta. Then

    \left(1+\frac{1}{x^n}\right)\left(1+\frac{1}{y^n}\  right) is minimized when \theta = \pi/4 so that x=y=\cos^2\theta = 1/2...
    Last edited by mr fantastic; October 4th 2009 at 02:35 AM.
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Haha... you're a helpful lad.

    Hint: set x= \cos^2 \theta, y = \sin^2 \theta. Then

    \left(1+\frac{1}{x^n}\right)\left(1+\frac{1}{y^n}\  right) is minimized when \theta = \pi/4 so that x=y=\cos^2\theta = 1/2...

    Hello : Thank you, Are you the details
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  4. #4
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    Hehehe, I try to help where I can


    That hint is more helpful ... did you use calculus to find the minimum?

    Cuz you could also use trig identities to get:

     \left(1+\frac{1}{x^n}\right)\left(1+\frac{1}{y^n}\  right) = 1+ \displaystyle{\frac{2^{2n}}{\sin^{2n}2 \phi}} + \displaystyle{\frac{2^{2n}}{\sin^{2n}2 \phi}} \geq 1 + 2^{n+1} + 2^{2n} = (1+2^n)^2
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