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Prove a^3+b^3+c^3>=3abc

  1. Sep 18, 2005 #1
    So I've been thinking about this for a while for an analysis class. I proved that a^4 + b^4 + c^4 + d^4 = 4abcd. Now I'm supposed to prove the inequality above using w=(abc)^1/3/. I'm not asking anyone to do my homework for me, but maybe someone could point me in the right direction?

    Nice to be back here.
     
  2. jcsd
  3. Sep 18, 2005 #2
    i realize i might have posted this in the wrong place. i'm sorry. please move it, although i do think that it is a nice topic for this section.
     
  4. Sep 18, 2005 #3
    Although this is slightly off-topic I don't how:

    a^4 + b^4 + c^4 + d^4 === 4abcd

    Let:
    a = 1
    b = 2
    c = 4
    d = 7

    1^4 + 2^4 + 4^4 + 7^4 = 2674

    However:
    4*1*2*4*7 = 224

    I'm confused :S
     
  5. Sep 18, 2005 #4
    This inequality dosen't hold for a=-1,b=-1,c=1 so I assume that it must be proven for positive numbers only? About the other response, I assume that you meant to type ">= 4abcd"; is this correct? My suggestion is to consider separately the situation where the numbers are less than 1 from what happens as the numbers a,b,c,d become greater than 1. Also what if you rewrote the equation so that one side would be a more definite quantity and still managed to use w somehow. Just some thoughts.
     
  6. Sep 19, 2005 #5
    I meant prove a^3+b^3+c^3>=3abc using w=(xyz)^(1/3)

    and yes i did mean earlier >=
     
  7. Sep 19, 2005 #6
    argh i meant using a^4 + b^4 + c^4 + d^4 = 4abcd and d=(abc)^(1/3) prove a^3 + b^3 + c^3 + d^3 = 3abc for positive numbers a,b,c
     
    Last edited: Sep 20, 2005
  8. Sep 20, 2005 #7
    i got it, will post proof tomorrow.
     
  9. Sep 20, 2005 #8
    4xyzw<=x^4+y^4+z^4+w^4
    (((( w=(xyz)^(1/3) ))))
    4(xyz)^(4/3)=x^4+y^4+z^4+(xyz)^(4/3)
    3(xyz)^(4/3)=x^4+y^4+z^4
    ((((a,b,c=x^4,y^4,z^4 respetively))))
    3(abc)^(1/3)<= a + b + c
    and now use a,b,c = j^3,k^3,l^3
     
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