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Math Help - Inequality

  1. #1
    Member great_math's Avatar
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    Inequality

    Prove that for all x,y,z>0

    \frac{x}{(x+y)(x+z)}+\frac{y}{(y+z)(y+x)}+\frac{z}  {(z+x)(z+y)}\le \frac{9}{4(x+y+z)}<br />
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  2. #2
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    Quote Originally Posted by great_math View Post
    Prove that for all x,y,z>0

    \frac{x}{(x+y)(x+z)}+\frac{y}{(y+z)(y+x)}+\frac{z}  {(z+x)(z+y)}\le \frac{9}{4(x+y+z)}
    so we want to prove that: 8(x+y+z)(xy+yz+zx) \leq 9(x+y)(y+z)(z+x), which after simplifying becomes: 6xyz \leq x^2y+yz^2+x^2z +y^2z + xy^2+ xz^2 . \ \ \ \ (1)

    dividing both sides of (1) by xyz gives us: 6 \leq \frac{x}{z} + \frac{z}{x} + \frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y}, which is clearly true by AM-GM, or (if you don't like AM-GM) because: \forall t > 0 : \ t + \frac{1}{t} \geq 2. \ \ \Box
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