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Another inequality proof

  1. Apr 16, 2007 #1
    The problem

    For all a>0 and b>0 if a^3+b^3 = a^5+b^5 prove that

    I have no idea on even how to start... i have tried using trigonometry (like in previous post) but come to a dead end... i am looking for the simplest method...
  2. jcsd
  3. Apr 16, 2007 #2


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    The only proof I've been able to come up with is kind of weird. Maybe you can do better. You should easily be able to convince yourself that if both a and b are >1 the equality can't hold. Same for <1. So we can take a>=1 and b<=1. Now consider f(n)=a^n+b^n (n>0). Now show f(n) has one extermum and it's a minimum. Your premise says f(3)=f(5). This says the extremal n is between 3 and 5. So f(1)>=f(3). Or a+b>=a^3+b^3. Factor the LHS and divide by (a+b) and you have your result. Funny, huh?
  4. Apr 17, 2007 #3
    I tried starting with a different approach. Since LHS = RHS, then RHS/LHS = 1. Doing a little creative long division, I obtained:

    [tex]\frac{a^5+b^5}{a^3+b^3} = 1 = a^2 + b^2 - \frac{a^3 b^2 + a^2 b^3}{a^3 + b^3}[/tex]

    From there,
    [tex] 1 + \frac{a^3 b^2 + a^2 b^3}{a^3 + b^3} = a^2 + b^2 [/tex]

    Then, perhaps on the LHS, factor out an ab:
    [tex] 1 + ab( \frac{a^2 b + a b^2}{a^3 + b^3}) = a^2 + b^2 [/tex]

    I'd think you can turn it around to [tex]a^2 + b^2 = 1 + ab( \frac{a^2 b + a b^2}{a^3 + b^3}) = (re-written) <= re-written with a term dropped out. [/tex]

    I'm drawing a momentary blank (and have to get home!) but I can't "see" the next step from here; but maybe it's enough that you can continue.
    Last edited: Apr 17, 2007
  5. Apr 17, 2007 #4


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    Keep trying, drpizza. I'd love to see the non-calc proof. I tried stuff like that for quite a while.
  6. Apr 18, 2007 #5


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    It's equivalent to show that:

    [tex](a^3 + b^3)(a^2 - ab + b^2) \leq a^5 + b^5[/tex]

    This line is true iff:

    [tex]a^5 + a^2b^3 - a^4b - ab^4 + a^3b^2 + b^5 \leq a^5 + b^5[/tex]


    [tex]a^2b^3 - a^4b - ab^4 + a^3b^2 \leq 0[/tex]


    [tex]ab^2 - a^3 - b^3 + a^2b \leq 0[/tex]


    [tex]a(b^2 - a^2) - b(b^2 - a^2) \leq 0[/tex]


    [tex](a-b)(b^2 - a^2) \leq 0[/tex]


    [tex](a-b)(b-a)(b+a) \leq 0[/tex]


    [tex]-(a-b)(a-b)(b+a) \leq 0[/tex]


    [tex](a-b)^2(b+a) \geq 0[/tex]

    Well (a-b)2 is a square, hence non-negative. b and a are both positive, so (b+a) is positive. So the product on the left is indeed non-negative, so the desired result holds.
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