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Math Help - (2p)^2013+(2q)^2013>2(pq)^2013

  1. #1
    Junior Member darence's Avatar
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    (2p)^2013+(2q)^2013>2(pq)^2013

    Proove that for real numbers p ang q \therefore \frac{1}{p}+\frac{1}{q}=1 ,
    (2p)^{2013}+(2q)^{2013}>2(pq)^{2013}
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: (2p)^2013+(2q)^2013>2(pq)^2013

    It's not true for all real numbers p and q. For example if p = 1/2 and q = -1 the left hand side is less than the right hand side for odd values of the exponent.

    I suspect the problem should have been written "for positive real values of p and q..." In this case you can show that p = q/(q-1) and that p and q are both > 1. The equation boils down to  (q-1)^n > -1 }, where 'n' is the exponent (2013 in this case).
    Last edited by ebaines; February 6th 2013 at 11:45 AM.
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  3. #3
    Junior Member darence's Avatar
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    Re: (2p)^2013+(2q)^2013>2(pq)^2013

    Omg
    This was on my examination and I don't pass because I could not proove.
    Thank you very much.
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    Re: (2p)^2013+(2q)^2013>2(pq)^2013

    Quote Originally Posted by darence View Post
    Proove that for real numbers p ang q \therefore \frac{1}{p}+\frac{1}{q}=1 ,
    (2p)^{2013}+(2q)^{2013}>2(pq)^{2013}
    like ebaines, I guess the real question was: prove that for positive reals p and q, \frac{1}{p}+\frac{1}{q}=1 ,
    (2p)^{2013}+(2q)^{2013}>2(pq)^{2013}

    then, I think its pretty obvious because \frac{1}{p}+\frac{1}{q}=1 \implies p+q=pq (now for the given relation to hold remember that p and q must be both positive and also greater than 1)
    p+q=pq \implies (p+q)^{2013}=pq^{2013} \implies p^{2013}+q^{2013}+\underbrace{........}_{\mathclap  {\text{other positive terms of expansion}}}=(pq)^{2013}
    \implies p^{2013}+q^{2013}>(pq)^{2013} \implies 2^{2013}(p^{2013}+q^{2013})>2^{2013}(pq)^{2013}>2(  pq)^{2013}
    hence (2p)^{2013}+(2q)^{2013}>2(pq)^{2013}
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  5. #5
    Junior Member darence's Avatar
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    Re: (2p)^2013+(2q)^2013>2(pq)^2013

    It's OK, Thank you very much. But problem was as I wrote and I failed exam I will see what to do now...
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