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Proving inequality

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  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data

    this is the problem , if x and y are real positive numbers , I need to prove

    $$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 12xy$$



    2. Relevant equations

    $$x^2 + y^2 \ge 2xy$$ (Variation of AM GM Theorem)

    3. The attempt at a solution
    but $$x^2 + y^2 \ge 2xy $$, so $$6x^2 + 6y^2 \ge 12xy $$, from this if

    $$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 6x^2 + 6y^2$$ then the given inequality holds

    let's try this $$44x^4 + 4y^3 \ge 6x^2 + 6y^2$$

    $$4x^4 + 4y^3 / (6x^2 + 6y^2) \ge 1 $$

    $$ 4x^4 + 4y^3 /(6x^2 + 6y^2) - 1 \ge 0 $$

    $$4x^4 + 4y^3 - 6x^2 - 6y^2 /(6x^2 + 6y^2) \ge 0$$ , divide both sides by the numerator ( this step feels wrong)

    $$1/(6x^2 + 6y^2) \ge 0 $$

    since this is true the inequality $$4x^4 + 4y^3 \ge 6x^2 + 6y^2$$ holds , so from this we can say that

    $$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 4x^4 + 4y^3 $$

    $$5x^2 + y +1 \ge 0 $$

    , this is obviously true and thus the above inequality holds

    BUT , the inequality $$4x^4 + 4y^3 \ge 6x^2 + 6y^2$$ doesn't hold for x and y = 1 but it does seem to work for all values of greater than 2 . Where have I gone wrong ?
    There seems to be a better way to solve the problem because solving this in such a round about way seems weird . A better solution (this proof does not work for numbers from 0 - 2 ) to this would be helpful.
     
  2. jcsd
  3. Feb 22, 2015 #2

    LCKurtz

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    Not sure if it helps you or not, but you can find the critical point of ##4x^4 + 4y^3 + 5x^2 + y + 1 - 12xy## numerically and that it occurs at
    upload_2015-2-22_14-17-40.png
    and the minimum value of the function is
    1.036443430 in the first quadrant.
     
  4. Feb 22, 2015 #3

    Ray Vickson

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    Your above point P1 is is actually a saddle point. The min in the first quadrant is at P2 = {x = .5522496847, y = .6847746720}, giving f = 0.328132751. I found these by first solving ##f_y=0## for ##x##, then substituting into ##f_x = 0## to get a 6th degree polynomial in ##y## whose two positive roots are {y = .7508111365e-1} and {y = .6847746720}---Hoorray for Maple!. The Hessian of f at P1 is indefinite, while at P2 is positive definite. A 3d plot shows f > 0 in the first quadrant, and a contour plot of f near (0,0) shows clearly the saddle-point nature of P1.
     
  5. Feb 23, 2015 #4
    This really doesn't help , this question was asked in a regional math olympiad to 10th graders so this information doesn't mean much to me . I need a simple proof using the AM GM theorem and also maybe a way to prove this inequality works for x and y lesser than 2 but greater than 0
     
  6. Feb 23, 2015 #5

    Ray Vickson

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    Here is an elementary argument that comes close, but does not quite do what you want. First, note that if ##t_1, t_2, \ldots, t_6 > 0## are some positive numbers and ##a_1, a_2, \ldots, a_6 > 0, \; a_1 + a_2 + \cdots + a_6 = 1## are positive "weights" summing to 1, then
    [tex] a_1 t_1 + a_2 t_2 + \cdots + a_6 t_6 \geq t_1^{a_1} \cdot t_2^{a_2} \cdots t_6^{a_6} [/tex]
    This is a generalization of the AM/GM inequality, Apply it when the left-hand-side is written as
    [tex] \text{LHS} = \frac{1}{6} (24 x^4) + \frac{1}{4} (16 y^3) + \frac{1}{6} (30 x^2) + \frac{1}{4} (4y) + \frac{1}{6} (6) [/tex]
    That will give you
    [tex] \text{LHS} \geq (24 \cdot 30 \cdot 6)^{1/6} \, (16 \cdot 4)^{1/4} xy = 4320^{1/6} 64^{1/4} xy \doteq 11.41455411\, xy[/tex]
    This gives LHS >= 11.416 xy instead of LHS >= 12 xy. So, as I said: close, but not quite what you want. Again, I don't know if this helps you.

    Note added in edit: perhaps it is possible to group the terms on the left in another way and with different weights, allowing use of the more general AM/GM inequality
    [tex] \sum_{i=1}^m a_i t_i \geq \prod_{i=1}^m t_i^{a_i} [/tex]
    where ##t_i, a_i > 0, i=1, \ldots m## and ##\sum a_i = 1##.

    Edit note 2: we can get a constant >= 12 on the right by looking at the problem as follows: if
    [tex] t_1 = 4x^2, \: t_2 = 4 y^3, \: t_3 = 5x^2, \: t_4 = y, \: t_5 = 1 [/tex]
    we have
    [tex] f(x,y) = t_1 + t_2 + t_3 + t_4 + t_5 \\
    = a_1(t_1/a_1) + a_2 (t_2/a_2) + a_3 (t_3/a_3) + a_4 (t_4/a_4) + a_5 (t_5/a_5 )\\
    \geq (t_1/a_1)^{a_1} (t_2/a_2)^{a_2} (t_3/a_3)^{a_3} (t_4/a_4) ^{a_4} (t_5/a_5)^{a_5}\\
    = (4/a_1)^{a_1}(4/a_2)^{a_2} (5/a_3)^{a_3} (1/a_4)^{a_4} (1/a_5)_{a_5} x^{4a_1+ 2a_3} y^{3a_2+a_4} [/tex]
    Thus, ##f(x,y) \geq c xy##, if
    [tex] 4 a_1 + 2 a_3 = 1,\\
    3 a_2 + a_4 = 1\\
    a_1 + a_2 + a_3 + a_4 + a_5 = 1 [/tex]
    For given ##a_i## the constant ##c## on the right is
    [tex] c = \left( \frac{4}{a_1}\right)^{a_1} \left( \frac{4}{a_2} \right)^{a_2}
    \left( \frac{5}{a_3} \right)^{a_3} \left( \frac{1}{a_4} \right)^{a_4} \left( \frac{1}{a_5} \right)^{a_5} [/tex]
    There are many values of the ##a_i## that give ##c \geq 12##. (In particular, if we maximize ##c## subject to the constraints on the ##a_i## we can get ##c = 14.4902646505479994 > 12##. (In other words, we actually have ##f(x,y) \geq 14.49 xy##.) If you do not care to use an optimization package (or are not allowed to do so) you can just try some values manually until you stumble upon a value ##c \geq 12##.

    It would help to use the restrictions on the ##a_i## to solve for three of them in terms of the other two; for example, we can easily solve for ##a_3, a_4, a_5## in terms of ##a_1,a_2##. Then ##c## can be re-written in terms of ##a_1, a_2## only, and there are some restrictions on ##a_1, a_2## due to the requirements ##a_3,a_4,a_5 \geq 0##. One can then just try some trial values of ##(a_1,a_2)## until we get a value ##c \geq 12##. In principle, all you need is a good hand-held calculator.
     
    Last edited: Feb 23, 2015
  7. Feb 23, 2015 #6
    But @Vriska is saying that it was a regional Olympiad. In that Maple and calculators are not allowed.
    It is a bit of manipulation work and not calculative one by putting values for x and y( Olympiads not want that). At the moment I am not able to solve it but will try.
     
  8. Feb 23, 2015 #7

    LCKurtz

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    You're right of course. I let Maple solve the system and it only found the saddle point. And when I did a contour plot I didn't notice the min it showed was in a different place. And I didn't plot enough contours to see the saddle point. Just shows you have to be careful with Maple. As far as that being an Olympiad problem, well, good luck with that. I have other things to spend my time on...
     
  9. Feb 23, 2015 #8

    Ray Vickson

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    Post #5 (final part) presents a method that requires no more than a hand-held "scientific" calculator. Again, good luck to the OP!
     
  10. Feb 23, 2015 #9

    SammyS

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    Try this:

    ## 4x^4 + 4y^3 + 5x^2 + y + 1 ##

    ## =\left(4x^4 +1 \right) +y \left(4y^2 +1\right) + 5x^2##​

    Use your variation of AM/GM on each of the parenthetical quantities and proceed.
     
  11. Feb 23, 2015 #10
    Brilliant.
    Proceeding by your method we get this all expression is greater than 9x2 + 4y2.
    Now obviously as stated by OP 6x2+ 6y2 >= 12xy
    Now how we will prove that 9x2 + 4y2 is greater than 6x2+ 6y2?
     
  12. Feb 24, 2015 #11

    SammyS

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    OP may have been wrong in that regard, but the result doesn't depend upon that. It's not necessarily true that ##9x^2+4y^2\ge 6x^2+6y^2\ . ##

    I leave it to OP to complete the solution.
     
  13. Feb 24, 2015 #12
    I have an interest in the problem,
    But isn't
    6x2+ 6y2 is greater than or equal to 12xy by the AM-GM inequality?
     
  14. Feb 24, 2015 #13

    SammyS

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    Yes, but that is irrelevant to the problem at hand.

    ## 3x^2 + 12y^2 \ge 12xy \ ## also, as well as other combinations.
     
  15. Feb 24, 2015 #14
    How that's true? Initially we get x2 + y2 greater than or equal to 2xy
    Then how
    ## 3x^2 + 12y^2 \ge 12xy \ ## also, as well as other combinations.
     
  16. Feb 24, 2015 #15
    So this is wrong in reality.
    If taking x=1 and y=1
    $$4x^4 + 4y^3 - 6x^2 - 6y^2$$ will be negative and when we divide a thing by negative in inequalities the inequality direction changes. Same the case for taking reciprocals.
    For example,
    1> 0.
    When we divide by -1 on both sides,
    -1<0 so the sign changes.
     
  17. Feb 24, 2015 #16

    Ray Vickson

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    The AM/GM inequality say that for ##A,B > 0## we have ##A + B \geq 2 \sqrt{A B}##, so ##3 x^2 + 12y^2 \geq 2 \sqrt{3 x^2 \,12 y^2} = 2 \sqrt{36} xy = 12 xy##.
     
  18. Feb 24, 2015 #17
    Got it,
    And from here also the answer indirectly. Thanks.
    Now it's all for the OP to understand.
     
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