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Thread: algebra inequality

  1. #1
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    algebra inequality

    if a>0,b>0,c>0 then prove that a/(b+c)+b/(a+c)+c/(a+b)>=3/2
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let $\displaystyle b+c=x, \ a+c=y, \ a+b=z$

    Then $\displaystyle a=\dfrac{-x+y+z}{2}, \ b=\dfrac{x-y+z}{2}, \ c=\dfrac{x+y-z}{2}$

    The inequality becomes

    $\displaystyle \dfrac{-x+y+z}{2x}+\dfrac{x-y+z}{2y}+\dfrac{x+y-z}{2z}\geq \dfrac{3}{2}$

    or $\displaystyle \dfrac{1}{2}\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfra c{x}{z}+\dfrac{z}{x}\right)+\left(\dfrac{y}{z}+\df rac{z}{y}\right)-3\right]\geq\dfrac{3}{2}$

    But it is well known that if $\displaystyle a>0$ then $\displaystyle a+\dfrac{1}{a}\geq 2$

    Can you finish?
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  3. #3
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