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Proving inequalities - Does induction work?

  1. Nov 12, 2008 #1

    ibc

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    Proving inequalities - Does induction work??

    1. The problem statement, all variables and given/known data
    prove that, for a,b,c>0, a+b+c=1, 1/a+1/b+1/c≥9


    2. Relevant equations
    it says that i might want to use the fact that for all X=/=0, X+1/X ≥ 2


    3. The attempt at a solution
    using the tip I could make it:
    a+1/a+b+1/b+c+1/c ≥ 10
    but that's as far as I got.





    1. The problem statement, all variables and given/known data
    prove using induction (or anything else):
    |sin(nx)|≤n|sin(x)|
    for natural n

    2. Relevant equations



    3. The attempt at a solution
    well it's true for n=1
    after using the induction assumption I made it so I have to prove that:
    |sin(x(n+1))|≤|sin(nx)| + |sin(x)|
    now i'm suck, I don't see how I could use trigonometry equivalences since they all give me cosins and sins*cosins and stuff like that, and these absolute values are a pain aswell, I could square it here and there, to get rid of them, but I don't see where it's going again.
     
  2. jcsd
  3. Nov 12, 2008 #2
    Re: Inequalities

    I only have time for the first one right now. Use a+b+c=1 !! :smile: and did you prove this inequality a+1/a+b+1/b+c+1/c >=10 ?
     
  4. Nov 12, 2008 #3

    Mark44

    Staff: Mentor

    Re: Inequalities

    For the first problem, pick some numbers and see how things work, and that might help you understand what is going on.

    Given that a, b, and c > 0 and that a + b + c = 1, these numbers are necessarily less than 1.

    One choice for all three variables is 1/3, from which the sum of the reciprocals is 9. If you choose a value for a that is larger than 1/3, then the other two variables will have to each be less than 1/3. For example, choose a = 1/2, b= 1/4, c = 1/4. What is the sum of the reciprocals? Is it larger or smaller than what you got before?
     
  5. Nov 12, 2008 #4

    ibc

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    Re: Inequalities

    ya I get that, but I need to prove it, in a mathematical way... which I can't figure out how, if I take the case when a>1/3 then b or c could be >1/3 aswell, and then the other one must be smaller, of course the bigger I take the numbers, the smaller the third one will be, and thus the whole thing will be larger and larger, but I can't find a mathematical procedure to show it, I can't find the right connection between a,b,c and the fact that the sum of all equals 1 doesn't help me since I still have 1 inequality with 2 variables.

    and I couldn't prove that a+1/a+b+1/b+c+1/c >=10 ?
     
  6. Nov 13, 2008 #5
    Re: Inequalities

    You said you obtained that equation from the hint and than were stuck....
     
  7. Nov 13, 2008 #6

    Mark44

    Staff: Mentor

    Re: Inequalities

    I was able to prove it, although it took a bit of doing. The high points of my proof are as follows:
    We want to find the minimum value of 1/a + 1/b + 1/c, subject to a, b, and c > 0 and a + b + c = 1.

    Let f(x, y) = 1/x + 1/y + 1/(1 - x - y)
    (The 3rd term uses the fact that the three numbers add up to 1.)

    Take partials wrt x and y and set them to 0.
    [tex]f_x = \frac{-1}{x^2} + \frac{1}{(1 - x - y)^2}[/tex]
    [tex]f_y = \frac{-1}{y^2} + \frac{1}{(1 - x - y)^2}[/tex]

    Setting both partials to 0 results eventually in y = [tex]\pm[/tex] x. Since x and y must be positive, we take y = x, with both positive.

    Now, f(x, y) = f(x, x) = g(x) = 1/x + 1/x + 1/(1 - 2x) = 2/x + 1/(1 - 2x)
    = (3x - 2)/(x^2 - x)

    g'(x) = -2(3x^2 -4x + 1) / (2x^2 - x)^2
    = -2(3x - 1)(x - 1) / (2x^2 - x)

    [tex]g'(x) = \frac{-2(3x^2 -4x + 1)}{(2x^2 - x)^2}[/tex]
    [tex] = \frac{-2(3x - 1)(x - 1)}{(2x^2 - x)}[/tex]

    From the above we see that g'(x) = 0 when x = 1/3 or when x = 1.
    The graph of g has a local minimum when x = 1/3 and a local maximum when x = 1. The latter value is of no interest in this problem, since it implies that y = 1 and 1 - x - y = -1.

    Putting everything together, we see that x = 1/3 and y = 1/3 (from previous work), so 1 - x - y = 1/3 as well.

    Going back to the a, b, and c of the original problem, the value of 1/a + 1/b + 1/c is smallest when a = 1/3, b = 1/3, and c = 1/3.

    What is that smallest value?
    1/a + 1/b + 1/c = 3 + 3 + 3 = 9.
    For any other values of a, b, and c that are positive and sum to 1,
    1/a + 1/b + 1/c will be larger than 9, which shows that under the conditions of this problem,
    [tex]\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9 [/tex]
    as was required.

    I didn't use the suggestion that x + 1/x >= 2 for all x not equal to 0, so it's possible that a proof that uses this fact is simpler.

    Mark
     
  8. Nov 13, 2008 #7

    ibc

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    Re: Inequalities

    Thanks
    though I didn't get that part,
    what does the partial derivative of this equation means, and what does it mean that you set them to zero?
     
  9. Nov 13, 2008 #8

    Mark44

    Staff: Mentor

    Re: Inequalities

    If y = f(x) and f is differentiable, you can talk about dy/dx ( = f'(x)), the derivative of y with respect to x. Here f is a function of one variable, x.

    If z = g(x, y), g is a function of two variables, so things get a bit more complicated. There is no longer the concept of "the derivative", but you can find partial derivatives with respect to x or with respect to y. When you take the partial of a function with respect to one of its variables, you treat the other variable as if it were a constant.
     
  10. Nov 13, 2008 #9
  11. Nov 13, 2008 #10

    ibc

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    Re: Inequalities

    ya, but why do we set them both to be zero?
    when partial derivative for y = partial derivative for x = 0, then? it's the function's minimum?
    if so, then by saying x=y is enough to know that x=y=z=1/3
     
  12. Nov 13, 2008 #11

    Mark44

    Staff: Mentor

    Re: Inequalities

    If both partials are zero at a given point, you can't tell if you're at a local minimum, a local maximum, or neither. The function I was looking at, f(x, y) = 1/x + 1/y + 1/(1 - x - y) is such that I was reasonably certain that there was a minimum at (1/3, 1/3).

    Just knowing that x = y wasn't enough to know that x, y, and z all had to be 1/3. There was some additional work that I did to establish that, which I described in my earlier post.
     
  13. Nov 15, 2008 #12

    ibc

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    Re: Inequalities

    but how can you assume there is a minimum at 1/3, 1/3? that's what we're trying to prove...
     
  14. Nov 15, 2008 #13
    Re: Inequalities

    One could always check the Hessian to make sure that the critical point at (1/3,1/3) is truly a local minimum.
     
  15. Nov 15, 2008 #14

    ibc

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    Re: Inequalities

    now I have no idea of what you are talking about
    I think there should be a simpler solution, it's just the beginning of calculus course
     
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