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Math Help - Prove

  1. #1
    Super Member dhiab's Avatar
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    Prove

    x,y,z >= 0 .. Prove :
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by dhiab View Post
    x,y,z >= 0 .. Prove :
    My two cents on this problem:

    If at least one of x, y, z is zero, the statement is obviously true:

    Let's say x=0, we get:
    y^2 + z^2 + 1 \geq 2yz
    which, after factoring gives:

    (y-z)^2 + 1 \geq 0

    So we consider the case where all three x,y,z are greater than zero. Assume further that x < y < z and that y = l x and z = kx for some l, k > 0. Note that k > l.

    Now we write the inequality in terms of x, l, and k:

    x^2 + (lx)^2 + (kx)^2 + 2 x (lx)(kx) + 1 \geq 2(xlx + xkx + lxkx)

    and move all terms to the left:

    x^2 + l^2 x^2 + k^2 x^2 + 2 klx^3 + 1 \geq 2(x^2 l + x^2 k + lk x^2)

    Divide both sides by x^2, since x^2 > 0 this will not change the direction of the inequality we are trying to prove:

    1 + l^2 + k^2 + 2klx + \frac{1}{x^2} \geq 2l + 2lk + 2k

    Note that l^2 + k^2 \geq 2lk (Cauchy inequality?)

    The above result comes from 'arithmetic mean' is greater than or equal to the geometric mean, formally:

    \frac{x+y}{2} \geq \sqrt{xy}


    So it suffices to show that:

    1 + 2klx + \frac{1}{x^2} \geq 2l + 2k

    Moving all terms to the left hand side:

    1+ 2klx + \frac{1}{x^2} - 2 (l-k) \geq 0

    Because l < k , all the terms on the left hand side are positive.

    I hope this helps.
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