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Help with proofs

  1. Sep 20, 2011 #1
    1. prove that if 0<a<b, then
    a<[itex]\sqrt{}ab<a+b/2<b[/itex]




    2. [itex]\sqrt{}ab\leq(a+b)/2[/itex] holds for all a,b [itex]\geq 0[/itex]



    3. Where do I begin? I have no clue! Thank you to anyone who can help!
     
  2. jcsd
  3. Sep 20, 2011 #2

    dynamicsolo

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    Try squaring the various expressions and compare those: since a and b are both positive, the order of the inequalities will not be changed.
     
  4. Sep 20, 2011 #3
    Thanks for the quick response, would you be able to elaborate a little bit more on what you said? I'm not looking for the answer, just a little more elaboration on what you posted.

    Any help is appreciated!
     
    Last edited: Sep 20, 2011
  5. Sep 20, 2011 #4

    dynamicsolo

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    #2 is probably easier to start with: what do you get when you square both sides of the inequality? In light of the fact that a and b are positive numbers, is it clear that that inequality works? If so, since the two sides of the inequality are the squares of the original sides of the inequality, the original inequality will also hold, for the specified condition.

    This is not "all there is to it": you will have to manipulate the inequality in some way to arrive at an inequality you know must be true.
     
  6. Sep 20, 2011 #5
    can you explain that in layman's terms?
     
  7. Sep 20, 2011 #6

    dynamicsolo

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    If 0 < X ≤ Y , then X2 ≤ Y2 . Squaring does not change the direction of the inequality and the squared expressions may be easier to work with.
     
  8. Sep 20, 2011 #7
    cool. that's really helpful, thanks
     
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