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Inequality X^n<Y^n if x<y and n is odd

  1. Mar 16, 2015 #1
    Hi All,

    Question : Prove that xn<yn , given that x<y and n is odd .

    Attempt at solution :
    Assumptions:
    y-x>0
    y2>0
    x2>0

    So y2x<y3 & x3<x2y

    So i need to prove that x2y<y2x

    i.e need to prove

    then y2x-x2y>0

    then yx(y-x)>0, from assumptions y-x>0 so i need to prove that yx>0, so i have 3 cases

    Case 1 : y and x are both positive so the inequality is true and everything is true.
    Case 2: y and x are both negative so the inequality is true again and everything is true
    Case 3 : y is positive and x is negative (as y>x), then the above inequality is not true, but i go back knowing that y is positive and x is negative, then y>0 and x<0 so y3>0 and x3<0 so y3>x3

    At the end x3<y3 and by repetition xn <yn

    End.

    Is this proof true ? is there an easier way.

    Thanks.
     
  2. jcsd
  3. Mar 16, 2015 #2

    mfb

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    Staff: Mentor

    Looks fine.
     
  4. Mar 17, 2015 #3

    mathman

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    There may be an easier way, by considering 3 cases.
    1. both positive. [itex]y^n-x^n=(y-x)(x^{n-1}+yx^{n-2}+...y^{n-1}) \gt 0[/itex] Since y-x > 0 and series has all positive terms.
    2. y >0 and x < 0. Odd powers have same sign.
    3. both negative. Change sign for both and use case 1 to show [itex]-x^n \gt -y^n[/itex].
     
  5. Mar 18, 2015 #4
    Alternatively, show that f(x) = x2k+1 for all natural numbers k is strictly monotonically increasing.
     
    Last edited: Mar 18, 2015
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