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Math Help - ~~~hard problem---inequalities~~

  1. #1
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    Exclamation ~~~hard problem---inequalities~~

    hey guys!!! could you please help me with this question that iv been stuck on for ages.....!!!!! ok here i go...

    Let a and b be positive real numbers such that
    (a^2) +(b^2)=1
    Prove that (a^4)+(b^4) is greater than or equal to 1/2

    plz help!!! i reeli dun understand the q
    thanx guys!!!!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fring
    hey guys!!! could you please help me with this question that iv been stuck on for ages.....!!!!! ok here i go...

    Let a and b be positive real numbers such that
    (a^2) +(b^2)=1
    Prove that (a^4)+(b^4) is greater than or equal to 1/2

    plz help!!! i reeli dun understand the q
    thanx guys!!!!!
    Since a^2+b^2=1

    <br />
a^4+b^4=a^4+(1-a^2)^2<br />

    The minimum of a^4+(1-a^2)^2, if it exists, occurs either at an end point of the valid interval for a or at a stationary point of a^4+(1-a^2)^2. A quick check shows the minimum does not occur at a=\pm 1, so it must occur at a stationary point.

    The stationary points of this occur when:

    <br />
\frac{d}{da}a^4+(1-a^2)^2=0<br />

    which simplifies to:

    <br />
8a^3-2a=0<br />

    so either a=0, or a=\pm 1/\sqrt{2}.

    Now we need to classifty these stationary points. Examining the second derivative of a^4+(1-a^2)^2 at the relevant points shows that a=\pm 1/\sqrt{2} corresponds to a local minimum and that minimum is 1/2, and the other point corresponds to a local maximum. Comparison between all the maxima, and minima shows that these are all global maxima and minima, so:

    if a^2+b^2=1

    <br />
a^4+b^4 \ge 1/2<br />

    RonL
    Attached Thumbnails Attached Thumbnails ~~~hard problem---inequalities~~-gash.jpg  
    Last edited by CaptainBlack; July 25th 2006 at 06:18 AM.
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  3. #3
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    Quote Originally Posted by fring
    hey guys!!! could you please help me with this question that iv been stuck on for ages.....!!!!! ok here i go...

    Let a and b be positive real numbers such that
    (a^2) +(b^2)=1
    Prove that (a^4)+(b^4) is greater than or equal to 1/2

    plz help!!! i reeli dun understand the q
    thanx guys!!!!!
    You have,
    a^2+b^2=1
    Then,
    b^2=1-a^2
    Thus,
    a^4+b^4=a^4+(b^2)^2
    Substitute,
    a^4+(1-a^2)^2
    Open sesame,
    a^4+1-2a^2+a^4=2a^4-2a^2+1
    Let,
    x=a^2 then,
    2x^2-2x+1
    This is parabola with minimum point cuz, (2>0) which occurs at,
    x=-\frac{-2}{2\cdot 2}=1/2
    Its minimum value is,
    2\left( \frac{1}{2} \right)^2-2\left( \frac{1}{2} \right)+1=\frac{1}{2}-1+1=\frac{1}{2}
    Thus, you have that,
    a^4+b^4\geq 1/2
    (Since, the smallest possible value of this expression is 1/2 then any other value [which is more] must be greater than 1/2)
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