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Polynomial inequality proof

  1. Feb 4, 2014 #1
    Prove that if x and y are not both , then x^4+x^3y+x^2y^2+xy^3+y^4 > 0

    I have no idea how to start this proof, can anyone give me an idea?
     
  2. jcsd
  3. Feb 4, 2014 #2

    Dick

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    Here's a hint for one way to do it. If x=0 it's clearly true. If x is not zero, then factor x^4 out and look at the other factor. Remind you of anything?
     
  4. Feb 5, 2014 #3
    sorry I meant to say that if x and y are both not 0
     
  5. Feb 5, 2014 #4
    Hint: This expression can be simplified.
     
  6. Feb 5, 2014 #5
    I factored out x^4 and got x^4 [(y/x)^4 + (y/x)^3 +(y/x)^2 + y/x +1] I see that all these terms have y/x so if i let t = y/x then it'll be x^4 [t^4 + t^3 +t^2 + t +1] (just so it is easier to visualize) but I am stuck, How do i simplify it further?
     
  7. Feb 5, 2014 #6
    Hint: think about a certain sequence.
     
  8. Feb 5, 2014 #7
    I am still lost
     
  9. Feb 5, 2014 #8
    What special sequences do you know of?
     
  10. Feb 5, 2014 #9
    x^n - y^n = (x - y) (x^n-1 + x^n-2y+ ....+ xy^n-2 + y^n-1 )??
     
  11. Feb 5, 2014 #10

    haruspex

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    That'll do it. So how can you use that to rewrite the original expression?
     
  12. Feb 5, 2014 #11
    (x^5 - y^5)/(x - y) = x^4 + x^3y + x^2y^2 + xy^3 + y^4 , is that correct?
     
  13. Feb 5, 2014 #12
    ahhhh I see it know, in both cases if x-y <0 or if x-y>0 the quotient will always be positive. Thanks everyone.
     
  14. Feb 5, 2014 #13

    haruspex

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    Right, but there is one special case you need to address separately.
     
  15. Feb 5, 2014 #14
    what would that be?
     
  16. Feb 5, 2014 #15
    x and y cannot be 0 by the way
     
  17. Feb 5, 2014 #16

    haruspex

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    Your expression involves 1/(x-y). What doubt should that create?
     
  18. Feb 5, 2014 #17

    Dick

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    But they can be nonzero and equal.
     
  19. Feb 6, 2014 #18
    But if you simplify the fraction then there is no problem.
     
  20. Feb 6, 2014 #19

    Dick

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    I didn't say there was a problem, but x=y is a special case.
     
  21. Feb 10, 2014 #20
    Assume that [tex]x,y>0[/tex].Let [tex]P(x,y)[/tex]be the polynomial, then [tex]P(x,y)=P(-x,-y)\ge P(x,-y)=x^4+y^4-x^3y-xy^3+x^2y^2\ge x^2y^2\ge 0[/tex] And the equality cannot hold.
     
    Last edited: Feb 10, 2014
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