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FA NAKROTH

24/04/2018 at 09:27
Answers
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Let x , y , z be positive real numbers . Prove that \(\dfrac{x^2-z^2}{y+z}+\dfrac{y^2-x^2}{z+x}+\dfrac{z^2-y^2}{x+y}\ge0\)




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    Nguyễn Huy Thắng 24/04/2018 at 10:10

    \(\dfrac{x^2-z^2}{y+z}+\dfrac{y^2-x^2}{z+x}+\dfrac{z^2-y^2}{x+y}\ge0\)

    \(\Leftrightarrow\dfrac{x^4+y^4+z^4-x^2y^2-y^2z^2-x^2z^2}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\)

    \(\Leftrightarrow\dfrac{\left(x^4-2x^2y^2+y^4\right)+\left(y^4-2y^2z^2+z^4\right)+\left(z^4-2x^2z^2+x^4\right)}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\)

    \(\Leftrightarrow\dfrac{\left(x^2-y^2\right)^2+\left(y^2-z^2\right)^2+\left(z^2-x^2\right)^2}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\left(\text{*}\right)\)

    The last inequality is obvious

    \("="\Leftrightarrow x=y=z\)

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    Nguyễn Huy Thắng 24/04/2018 at 10:12

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