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Spivak Calculus 4th ed (Ch1 problem 7)

  1. Feb 8, 2016 #1
    Prove that if 0<a<b, then [itex]a < \sqrt{ab} < \frac {a+b} {2} < b [/itex]

    Please excuse if format is messy, this is my first time writing in Latex.

    Suppose 0<a<b then 0<a+O =a (1). Since a<b we can can rewrite this using (1) which is a+0<b (2).
    Adding a to both sides of (2) (closed under addition) we get 2a< a+b (3). Using the multiplicative inverse of 2, (3) now becomes ## a < \frac {a + b} {2} ## (4). Using the fact that a<b, ## a < \frac {a + b} {2} ##
    < ## a < \frac {a + b} {2} ## < ## a < \frac {b + b} {2} ## =b. Therefore a<b .

    Here is were I am confused. I am missing the step were ## a < \sqrt {ab} ##. Can I say that a*b<a*a=a^2 ?
    Then take the square root? If this is true how would I connect it with ## a < \frac {a + b} {2} ## ?

    Strong criticism welcomed.
     
  2. jcsd
  3. Feb 8, 2016 #2

    Samy_A

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    Here something went wrong.

    For ##\sqrt {ab} \lt \frac {a+b}{2}##, you could use the following:
    ##(a+b)²>0,\ (a-b)²>0##.
     
  4. Feb 8, 2016 #3
    I am thinking that 0<a<b

    a<b
    ## a * a < a * b ##
    ## \sqrt{a*a} < \sqrt{a*b} ##
    ## a + b < \sqrt{a*b} + b < \sqrt{b*b} +b= 2*b ##
    (a+b)/2 < b ?

    I am thinking this is correct.
     
  5. Feb 8, 2016 #4
    this come from the fact that if a<b then b^2-a^2= (b-a)(b+a)?
     
  6. Feb 8, 2016 #5

    Samy_A

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    This is needlessly complicated:
    ##\frac{a+b}{2} \lt \frac{b+b}{2}=b## is all you need for the last inequality.

    What I meant with "something went wrong" was about the following:
    ##\frac {a + b} {2} ## < ## a## is obviously wrong, probably a typo.
    I don't see how that helps you in proving the missing piece: ##\sqrt {ab}< \frac {a + b} {2}##
    You could square both sides and calculate the difference. That's where ##(a-b)²>0## will help.
     
  7. Feb 8, 2016 #6
    Ahh very clever would have never thought about (a-b)^2 >0 part. By using this fact, it we just play with addition and subtraction of these terms. Thanks alot.
     
  8. Feb 8, 2016 #7

    Mark44

    Staff: Mentor

    @MidgetDwarf, in future posts, please don't delete the homework template. Its use is required in problems posted in the Homework sections.
     
  9. Feb 8, 2016 #8
    Thanks Mark for informing.
     
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