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Math Help - Algebra

  1. #1
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    Algebra

    Given two real numbers x and y such that
    (x+y)^2\leq2xy
    find x+y.

    If x+y=6 and xy=3, find \displaystyle\frac{1}{x}+\frac{1}{y}.

    Given |x-y|=2 and |y-z|=7, find the minimum possible value of |x-z|.
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  2. #2
    o_O
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    1. (x+y)^{2} \leq 2xy
    x^{2} + 2xy + y^{2} \leq 2xy
    x^{2} + y^{2} \leq 0
    x^{2} \leq -y^{2}

    What values of x and y make this inequality even possible?

    2. \frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{x + y}{xy}

    3. |x - y| = 2 \quad \Rightarrow \quad x - y = \pm 2 \quad \Rightarrow \quad x = y \pm 2
    |y - z| = 7 \quad \Rightarrow \quad y - z = \pm 7 \quad \Rightarrow \quad z = y \pm 7

    Find x - z and see what the lowest value of |x - z| would be.
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  3. #3
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    Quote Originally Posted by o_O View Post
    1. (x+y)^{2} \leq 2xy
    x^{2} + 2xy + y^{2} \leq 2xy
    x^{2} + y^{2} \leq 0
    x^{2} \leq -y^{2}

    What values of x and y make this inequality even possible?
    Just skip your last line.
    x^2+y^2 \leq 0. Note x^2+y^2\geq 0 (since it is a sum of two squares). Thus, x^2+y^2 = 0 \implies x=0,y=0.
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