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Homework Help: Trig Inequality (partial solutions included)

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Given 0 <= a <= b

    show that,

    a <= sqrt(ab) <= (a+ b / 2) <= b

    2. Relevant equations

    a * b <= a^2 / a*b <= a* a



    3. The attempt at a solution

    I think i know where im going but i wanted to make sure if its correct so far.

    So we know that the order of least to greatest, 0 -> a -> b

    and the first part of inequality states that a <= (Sqrt)ab
    so i take the sqrt and squre the left side so it makes a ^2.

    The inequality is now a^2 <= ab
    and this is true because a < b. and a^2 = a * a
    there fore a* a is smaller than a * b.

    The second part is [sqrt(ab) <= a+ b/ 2]
    Sqrt(ab) smaller or equal to (a + b) / 2.

    Solution: square both sides, ab <= [ (a+b)/ 2 ]^2
    and this gives 4ab <= (a+b)^2
    4ab <= a^2 + 2ab + b^2

    and last part

    a + b / 2 <= b
    multiply by 2 to both sides.
    a+b <= 2b

    and sinec a < b and
    a + b <= b * b

    is this correct? am i allowed to do what i just did ? please help me
     
    Last edited: Sep 21, 2010
  2. jcsd
  3. Sep 21, 2010 #2

    Mark44

    Staff: Mentor

    What does this mean -- 0 <\ a <\ b?

    Why are the slashes there?

    For less than, use <
    For less than or equal, use <=
    Similar for greater than and greater than or equal.
     
  4. Sep 21, 2010 #3
    Sorry about that, fixed to <=
     
  5. Sep 21, 2010 #4

    Mark44

    Staff: Mentor

    Can you show that (a + b)/2 is between a and b?
    Can you show that sqrt(ab) <= (a + b)/2?
     
  6. Sep 21, 2010 #5
    a) (a+b)/2 between a and b

    yes, setting up an equality so that a <= (a+b)/2 <= b
    which is same as 2a <= a+b <= 2b

    this prooves that (a+b)/2 is between a and b


    2) sqrt(ab) <= (a+b)/2

    square both sides to remove sqrt from ab.

    (ab) <= [(a+b)/2]^2
    (ab) <= [ (a*b)^2/4]

    4ab <= (a*b)^2

    is it ok to leave it as 4ab <= (a*b)^2
    or should i be expanding so its 4ab <= a^2 + 2ab + b^2 or even go beyond this??
     
  7. Sep 21, 2010 #6

    Mark44

    Staff: Mentor

    I wouldn't do it that way.
    You're given that 0 <= a <= b.
    a = (a + a)/2 <= (a + b)/2, since a/2 <= b/2 (which is because a <= b).

    It's about as easy to show that (a + b)/2 < b.
    The step above is not obvious (to me). I would expand the (a + b)/2 part.
     
  8. Sep 21, 2010 #7
    im really confused, did you just want me to make it 4ab <= a^2 + 2ab + b^2
    and subtract 2ab from both sides to make 2ab <= a^2 + b^2 ?
     
  9. Sep 21, 2010 #8

    Mark44

    Staff: Mentor

    No, subtract 4ab from both sides.
     
  10. Sep 21, 2010 #9
    oh, just make " 0 " on the otherside? (Didnt know this was allowed)
    this indefinetly proves that it is greater than left side right?

    so i would end up with 0 <= a^2 - 2ab + b^2
    making the right side greater.
     
  11. Sep 21, 2010 #10

    Mark44

    Staff: Mentor

    Sure, you can always add or subtract the same amount from both sides of an equation or inequality.

    For 0 <= a^2 - 2ab + b^2, for what values of a and b? That's crucial.
     
  12. Sep 21, 2010 #11
    do i have to find what values a and b have to be ?
    a^2 - 2ab + b^2 = (a-b)^2
    or just leave it at 0 <= a^2 - 2ab + b^2 to prove sqrt(ab) <= (a+b)/2

    from the question it seems like their only asking for the 0 <= a^2 - 2ab + b^2 but also i think i should prove further to see if they are indeed greater than 0.

    (a-b)^2 how would this be interpreted as number or equality?
    confused.
     
  13. Sep 21, 2010 #12

    Mark44

    Staff: Mentor

    What you have is a sequence of equivalent inequalities, starting with
    sqrt(ab) <= (a + b)/2
    and going to 0 <= a^2 - 2ab + b^2

    If you know which values the last inequality is true for, you know the values the first one is true for.
     
  14. Sep 21, 2010 #13
    how do i find the values?
    im slowly getting lost.

    i have to prove that
    a^2 - 2ab + b^2 is greater or equal to 0.
    but how would i do that?

    (a-b)^2 >= 0
    any number of A that is smaller than B would give a positive real number.

    sry im lost here
     
  15. Sep 21, 2010 #14

    Mark44

    Staff: Mentor

    For this problem a - b can be negative or zero, but when you square a - b, you get a number that is always at least zero.

    So (a - b)^2 >= 0 for all real numbers a and b. This means that (a + b)/2 >= sqrt(ab) for all nonnegative real numbers a and b. The square root isn't a real number if a and b are opposite in sign, but you're given that both a and b are >= 0.
     
  16. Sep 21, 2010 #15
    ah.. i see the problem.
    a * b cannot be a negative number because of the square root.

    not sure how it would work out..
    ab would always have to > 0
     
  17. Sep 21, 2010 #16

    Mark44

    Staff: Mentor

    Not a problem. You're given that 0 <= a <= b, so neither one of them can be negative.
     
  18. Sep 21, 2010 #17
    oh right, so it does work out afterall.
    0 <= a^2 - 2ab + b^2

    this would be the correct final stage then?
    given 0 < a < b
     
  19. Sep 21, 2010 #18

    Mark44

    Staff: Mentor

    No, you don't want to end up with 0 <= a^2 - 2ab + b^2. You want to show from this being true for all values of a and b such that 0 <= a <= b, that sqrt(ab) <= (a + b)/2. That's what you want to conclude in this part of the problem.

    Part of doing that is to explain why the inequalities sqrt(ab) <= (a + b)/2 and 0 <= a^2 - 2ab + b^2 have the same solution set, at least for 0 <= a <= b.
     
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