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Math Help - Inequality question

  1. #1
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    Inequality question

    If x, y, z are real numbers, show that

    (y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2 >= yz + zx + xy.

    I simplified it to

    x^2 + y^2 + z^2 >= xy + xz + yz

    not really sure what Im supposed to do next though lol
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  2. #2
    MHF Contributor red_dog's Avatar
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    Multiply both sides by 2:

    2x^2+2y^2+2z^2\geq 2xy+2xz+2yz

    (x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2)\geq 0

    (x-y)^2+(x-z)^2+(y-z)^2\geq 0
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  3. #3
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    That's factorised, but I'm still not sure how I show that the Left Hand Side is >= to the Right Hand Side
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  4. #4
    Member SENTINEL4's Avatar
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    (x-y)^2+(x-z)^2+(y-z)^2\geq 0

    This is true because it is the sum of three positive numbers, which is >=0
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by kinetix2006 View Post
    If x, y, z are real numbers, show that

    (y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2 >= yz + zx + xy.

    I simplified it to

    x^2 + y^2 + z^2 >= xy + xz + yz

    not really sure what Im supposed to do next though lol
    You can't manipulate the inequality since that's you asked to prove.

    On red_dog's post, he wrote (x-y)^2+(x-z)^2+(y-z)^2\geq 0<br />
, which is an obvious fact, start from there to prove what you want.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    You can't manipulate the inequality since that's you asked to prove.

    On red_dog's post, he wrote (x-y)^2+(x-z)^2+(y-z)^2\geq 0<br />
, which is an obvious fact, start from there to prove what you want.
    Okay, if you can't manipulate the inequality though, then what would be the procedure for proving it from the starting point of

    (y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2 >= yz + zx + xy
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