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A,b,c are given:

  1. Jul 8, 2003 #1
    and a+b+c>0 and abc>0
    proove that abc>=a+b+c

    can someone please help?
     
  2. jcsd
  3. Jul 8, 2003 #2
    I'm afraid it's not correct...
    if a=0.1,b=0.1,c=0.1 then abc=0.001 and a+b+c=0.3, 0.001<0.3...
    [?]
     
  4. Jul 8, 2003 #3
    ****, damn.
    btw, bogdan, did these numbers were the first to pop up your head?
     
    Last edited: Jul 8, 2003
  5. Jul 11, 2003 #4
    It is not really hard to get those numbers.
    All you need to know is think for a moment ..
    If you have a positive number (call it X), and you multiply it but another positive number (say Y), you can get one of the following results :
    XY > X
    XY < X
    XY = X
    Now, if XY = X then Y = 1, obviously this is the changing point between XY > X and XY < X
    Try numbers bigger than 1 for Y, say, Y = 2, then XY = 2X, therefore XY > X (remember that X is positive).
    Now, try numbers smaller than 1 (and more than 0, remember Y is positive), say, Y = 0.1, then you get XY = X/10, and therefore XY < X
    bogdan was trying to disproove the original inequality, logically if he gets a sinlge case that gives a wrong answer in the inquality, then the inquality is wrong.
    The inequality says that abc will be bigger or equal to a+b+c,
    So, to disproove this try to get smaller value of abc than expected (since this MIGHT turn the inequality wrong, by making left side smaller).
    Since values smaller than 1 for a,b,c will make abc smaller and smaller, bogdan have chosen 0.1 for all of them.
    I see infinite number of cases that proove the inequality wrong.
    Try those :
    1- a=0.01, b=0.1, c=0.1
    2- a=1, b=1, c=1
    3- a=1.1, b=0.0001, c=0.0002
     
  6. Jul 13, 2003 #5
    I wonder if maybe this was supposed to be in the domain of all the positive integers?
     
  7. Jul 13, 2003 #6
    let's say it is, does it change the answer?
     
  8. Jul 13, 2003 #7
    Yes...for positive integers it is true...
    let's say 2<=a<=b<=c...it doesn't "particularize"...
    we have abc>=2*(bc)>=4*c=c+c+c+c>a+b+c...
     
  9. Jul 13, 2003 #8
    do you have proof or what you have shown was your proof?
     
  10. Jul 13, 2003 #9
    That's the proof...read it carefuly...:smile:
    (that's not true for integers equal to 1...in specific cases...)
     
  11. Jul 13, 2003 #10
    thanks.
     
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