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Math Help - classic inequality

  1. #1
    Senior Member DivideBy0's Avatar
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    classic inequality

    Prove that if x, y, and z are positive real numbers, then

    \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geq \frac{3}{2}

    I haven't done much with inequalities so please explain your steps, thanks!
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by DivideBy0 View Post
    Prove that if x, y, and z are positive real numbers, then

    \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geq \frac{3}{2}

    I haven't done much with inequalities so please explain your steps, thanks!
    This is Nesbitts inequality, I know many proofs for this. Nesbitt's inequality - Wikipedia, the free encyclopedia
    The above link proves it using AM-GM and Rearrangement inequality
    I know a few more too
    I will post them if you want
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  3. #3
    Super Member PaulRS's Avatar
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    Define x+y+z=\lambda>0

    THus: y+z=\lambda-x; x+z=\lambda-y; x+y=\lambda-z

    So we have to prove: \frac{x}{\lambda-x}+\frac{y}{\lambda-y}+\frac{z}{\lambda-z}\geq{\frac{3}{2}}

    Let: f(x)=\frac{x}{\lambda-x}=\frac{\lambda}{\lambda-x}-1 which is clearly convex in (0; \lambda) (use the second derivative)

    It follows from Jensen's Inequality that: \frac{1}{3}\cdot{\left(f(x)+f(y)+f(z)\right)}\geq{  f\left(\frac{x+y+z}{3}\right)}

    Therefore: \frac{1}{3}\cdot{\left(\frac{x}{\lambda-x}+\frac{y}{\lambda-y}+\frac{z}{\lambda-z}\right)}\geq{\frac{\left(\frac{x+y+z}{3}\right)}  {\lambda-\left(\frac{x+y+z}{3}\right)}}

    Now remember that: x+y+z=\lambda and the assertion follows

    Jensen's inequality - Wikipedia, the free encyclopedia
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  4. #4
    Senior Member DivideBy0's Avatar
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    Thanks!
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  5. #5
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    It can also be done with AM-HM, that is how I learned it.
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