Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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

  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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    No, you derived those inequalities in your post, but it's wrong. How does it have anything to do with the base case?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook