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An inequality

  1. Aug 16, 2011 #1
    Suppose that x1,x2,x3,x4 are four positive real numbers such that x1x2x3x4=1.

    Show that x13+x23+x33+x43 ≥ x1+x2+x3+x4.

    How to solve this? (Not a homework!)

    Any hint is appreciated.
     
  2. jcsd
  3. Aug 16, 2011 #2
    Looks like it should be pretty easy with induction if you can prove the base case of:

    if x1,x2 are positive real numbers such that x1x2=1 then x13 + x23≥ x1 + x2

    For the base case you know that x1x2=1 ⇒ x2= 1/x1

    i.e. show x3 + 1/x3 - x - 1/x ≥ 0, for x ≥ 0 which shouldn't be to bad with some calculus

    At least that's the route I would go.
     
  4. Aug 16, 2011 #3
    MO Question?
     
  5. Aug 16, 2011 #4

    disregardthat

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    Use lagrange. It will pop right out.
     
  6. Aug 19, 2011 #5
    Thanks for the replies.

    That's a part of a problem of a national mathematical olympiad. The full problem is as follows:

    Suppose that x1,x2,x3,x4 are four positive real numbers such that x1x2x3x4=1. Show that

    Cgeq%20%5Cmax%5Cleft%20%5C{%20%5Csum_{i=1}^{4}x_i,%5Csum_{i=1}^{4}%5Cfrac{1}{x_i}%5Cright%20%5C}.gif
     
  7. Aug 19, 2011 #6
    Base case could just be n=1, i.e. one x. I guess you could also argue for n=0 too.
     
  8. Aug 19, 2011 #7
    good point, that makes it even easier
     
  9. Aug 19, 2011 #8
    Well, you gotta watch out for which natural number corresponds to the base case when carrying out inductive proofs.

    Check this out: We will show [itex]n^3 \leq n^2[/itex] for all n > 0 by induction. If n = 1, we get [itex]1^3 \leq 1^2[/itex] which is certainly true. Now assume [itex]k^3 \leq k^2[/itex]. Then

    [tex](k + 1)^3 = k^3 + 3k^2 + 3k + 1 \leq k^2 + 3k^2 + 3k + 1 \leq k^2 + 2k + 1 = (k + 1)^2[/tex]

    which was the desired result. But wait, [itex]2^3 = 8 > 4 = 2^2[/itex]. Contradiction? The problem comes from "choice" of base case.
     
  10. Aug 19, 2011 #9

    disregardthat

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    Analytical solution:

    Let [tex]f(x_1,x_2,x_3,x_4) = x_1^3+x_2^3+x_3^3+x_4^3-(x_1+x_2+x_3+x_4)[/tex], and [tex]g(x_1,x_2,x_3,x_4) = x_1x_2x_3x_4-1[/tex].

    we will find the minimum of f under the condition g = 0.

    We find the solutions to the lagrange equations

    [tex]\frac{\delta f}{\delta x_i} = \lambda \frac{\delta g}{\delta x_i}[/tex]

    and g = 0, or

    [tex]3x_i^2 -1= \lambda x_1x_2x_3 \Leftrightarrow 3x_i^3-x_i=\lambda[/tex], so [tex]3x_i^3-x_i = 3x_j^3-x_j[/tex], or [tex](x_i-x_j)(3(x_i^2+x_ix_j+x_j^2)-1) = 0[/tex]. Pick the largest one, say x_k. It will be larger or equal than 1. And thus [tex]3(x_k^2+x_ix_k+x_k^2)-1 \geq 2 > 0[/tex]. Hence [tex]x_i = x_k[/tex] for all i, so they are all equal. Hence x_i = 1 for all i.

    Now this is the minimum on some compact set containing (1,1,1,1).

    The set determined by [tex]x_i > 0[/tex], and [tex]x_1x_2x_3x_4 = 1[/tex], and [tex]x_1+x_2+x_3+x_4 <=5[/tex] contains (1,1,1,1) and is compact.

    Furthermore, by the general QM-AM [tex]\sqrt[3]{\frac{x_1^3+x_2^3+x_3^3+x_4^3}{4}} \geq \frac{x_1+x_2+x_3+x_4}{4}[/tex]

    So [tex]x_1^3+x_2^3+x_3^3+x_4^3 \geq \frac{(x_1+x_2+x_3+x_4)^2}{16} (x_1+x_2+x_3+x_4)[/tex]

    Hence outside our set [tex]f(x_1,x_2,x_3,x_4) > 0[/tex]. So (1,1,1,1) is actually the value for which f is minimized, and the value is 0. Hence [tex]x_1^3+x_2^3+x_2^3+x_4^3 \geq x_1+x_2+x_3+x_4[/tex].

    (EDIT: I assumed we were working in the subset x_i > 0 originally, so just assume I said that before I did :) )
     
    Last edited: Aug 19, 2011
  11. Aug 19, 2011 #10

    disregardthat

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    k^2 + 3k^2 + 3k + 1 \leq k^2 + 2k + 1 that does follow from the induction hypothesis and is certainly not true.
     
  12. Aug 19, 2011 #11
    That's what I said.
     
  13. Aug 19, 2011 #12

    disregardthat

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    No, you derived those inequalities in your post, but it's wrong. How does it have anything to do with the base case?
     
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