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Absolute value...

  1. Jul 17, 2016 #1
    1. The problem statement, all variables and given/known data
    X and Y 2 real numbers / |x| <1 and |y|<1
    Prove that |x+y|<|xy+1|


    2. Relevant equations


    3. The attempt at a solution
    |x+y|<2
    I couldn't prove that |xy+1| >2
    And couldn't find a way to solve the problem
    Please help
     
  2. jcsd
  3. Jul 17, 2016 #2

    phinds

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    That's good because
    (1) It's not true
    (2) It's not at all what you have been asked to prove.
     
  4. Jul 17, 2016 #3

    PeroK

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    Note that you have, essentially, three cases:

    1) ##x, y \ge 0##
    2) ##x, y \le 0##
    3) ##x \le 0, \ y \ge 0##

    You could try those cases separately and see what you can do.
     
  5. Jul 17, 2016 #4

    Filip Larsen

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    Since, by definition, |x| = x when x ≥ 0 and |x| = -x when x < 0, a general method for "removing" absolute values is to divide each instance of an absolute value into two cases. Perhaps you can find a way to divide the relation you are trying to prove into such different cases? For instance, how must x and y relate to each other if you are to "remove" the absolute value on |x+y| ?

    Edit: I type way too slow :rolleyes:
     
  6. Jul 17, 2016 #5

    Mark44

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    And this one?
    4) ##x \ge 0, \ y \le 0##?
     
  7. Jul 17, 2016 #6

    PeroK

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    It's symmetric in ##x## and ##y##. That's just the same as case 3).
     
  8. Jul 17, 2016 #7

    phinds

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    You're thinking one step ahead and realizing that it works out that way, but for slow people like me it's beneficial to state ALL the cases up front so I think Mark's statement is appropriate.

    I MIGHT have thought to make one of the conditions just "x and y have different signs" and then it does become just one condition.
     
  9. Jul 17, 2016 #8

    PeroK

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    That would be fair enough if I hadn't said "essentially". Formally, I would have said wlog (without loss of generality).
     
  10. Jul 17, 2016 #9

    phinds

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    A fair point. I hadn't notice that.
     
  11. Jul 19, 2016 #10
    I think I got it?
    If x>0 and y>0
    Then |x+y |= x+ y and | xy+1|=xy +1
    We have to prove that x+y < xy +1
    X-1 < xy -y
    X-1 < y(x-1) / x-1<0
    1>y which is true tgeb x+y <xy+1
    --if x and y are negative
    |x+y |= -x-y and xy +1>0 then |xy+1 |=xy +1
    Let us prove again thet -x-y <xy +1
    -x-xy<1+y
    -x(1+y)< 1+y / -1<y<0 => y+1>0
    -x<1< = > x>-1 which is true .
    ----case 3 if xis positif and y is negatif
    0<x<1 , -1<y<0
    -1<x+y<1 ,1+ xy<1
    If x+y >0
    We have to prove : x+y < xy +1
    X-1< y(x-1) /x-1 <0
    1> y true.
    If x++y <0 ..we have to prove that -x-y <xy +1 ..
    -x(1+y) <1+y ../ 1+y >0
    So -x<1
    X>-1 which is true
     
  12. Jul 19, 2016 #11

    haruspex

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    Probably - I haven't checked the details - but there is a much easier way.
    Note that |a|>|b| if and only if a2>b2.
     
  13. Jul 19, 2016 #12
    Yes I know i tried it out first and didn't knowif its alright to use the same method bec I couldn't remember how inequalities work at this point
    Thats said it is shorter probably if I didn't make any mistake...I though I absolutely did
     
  14. Jul 19, 2016 #13

    haruspex

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    Would you mind posting that attempt?
     
  15. Jul 21, 2016 #14

    haruspex

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    There being no response, I'll post my full solution.
    |x|<1, |y|<1
    x2<1, y2<1
    (1-x2)(1-y2)>0
    1+x2y2>x2+y2
    1+2xy+x2y2>x2+2xy+y2
    (1+xy)2>(x+y)2
    |1+xy|>|x+y|

    Note that |x|>1, |y|>1 leads to the same result.
     
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