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Thread: Pre-algebra: inequality

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    Pre-algebra: inequality

    Suppose $\displaystyle x,y,z$ are any positive real numbers such that $\displaystyle x+y+z=xyz.$ Prove that: $\displaystyle (x-1)(y-1)(z-1) \leq 6\sqrt{3}-10.$

    Source: JIPAM
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    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by NonCommAlg View Post
    Suppose $\displaystyle x,y,z$ are any positive real numbers such that $\displaystyle x+y+z=xyz.$ Prove that: $\displaystyle (x-1)(y-1)(z-1) \leq 6\sqrt{3}-10.$

    Source: JIPAM
    We are looking for $\displaystyle \max_{x>0,y>0,z>0}f_{0}(x,y,z),$ where $\displaystyle f_{0}(x,y,z):=(x-1)(y-1)(z-1)$ under the restriction $\displaystyle g(x,y,z):=x+y+z-xyz=0$.
    Hence, we set $\displaystyle f_{1}(x,y):=f_{0}(x,y,-(x+y)/(1-xy))$, therefore it suffices to find $\displaystyle \max_{x>0,y>0}f_{1}(x,y)$.
    Considering the symmetricity, and it suffices to examine the partial derivative of $\displaystyle f_{1}$ with respect to $\displaystyle x$.
    It follows that the critical points are $\displaystyle x_{0}=1$ and $\displaystyle x_{0}=2y/(y^{2}-1)$.
    Clearly, $\displaystyle f_{1}(1,y)=0$.
    On the other hand, let $\displaystyle f_{2}(y):=f_{1}(2y/(y^{2}-1),y).$
    We find that the critical point for $\displaystyle f_{2}$ is $\displaystyle y_{0}=\sqrt{3}$, and $\displaystyle f_{2}(\sqrt{3})=(\sqrt{3}-1)^{3}=6\sqrt{3}-10$, which is the max value of $\displaystyle f_{1}$ (at $\displaystyle (\sqrt{3},\sqrt{3})$) at the same time because the other critical point $\displaystyle (1,y)$ makes $\displaystyle f_{1}$ attain $\displaystyle 0$.
    Therefore, the given inequality holds.

    Not. I know this is a long solution and not so nice. :S
    Last edited by bkarpuz; Mar 31st 2009 at 02:23 PM.
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    Quote Originally Posted by bkarpuz View Post
    We are looking for $\displaystyle \max_{x>0,y>0,z>0}f_{0}(x,y,z),$ where $\displaystyle f_{0}(x,y,z):=(x-1)(y-1)(z-1)$ under the restriction $\displaystyle g(x,y,z):=x+y+z-xyz=0$.
    Hence, we set $\displaystyle f_{1}(x,y):=f_{0}(x,y,-(x+y)/(1-xy))$, therefore it suffices to find $\displaystyle \max_{x>0,y>0}f_{1}(x,y)$.
    Considering the symmetricity, and it suffices to examine the partial derivative of $\displaystyle f_{1}$ with respect to $\displaystyle x$.
    It follows that the critical points are $\displaystyle x_{0}=1$ and $\displaystyle x_{0}=2y/(y^{2}-1)$.
    Clearly, $\displaystyle f_{1}(1,y)=0$.
    On the other hand, let $\displaystyle f_{2}(y):=f_{1}(2y/(y^{2}-1),y).$
    We find that the critical point for $\displaystyle f_{2}$ is $\displaystyle y_{0}=\sqrt{3}$, and $\displaystyle f_{2}(\sqrt{3})=(\sqrt{3}-1)^{3}=6\sqrt{3}-10$, which is the max value of $\displaystyle f_{1}$ (at $\displaystyle (\sqrt{3},\sqrt{3})$) at the same time because the other critical point $\displaystyle (1,y)$ makes $\displaystyle f_{1}$ attain $\displaystyle 0$.
    Therefore, the given inequality holds.

    Not. I know this is a long solution and not so nice. :S
    well, the question is supposed to be a Pre-algebra (not calculus) problem. that's what the title of my post says.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Unhappy

    Quote Originally Posted by NonCommAlg View Post
    well, the question is supposed to be a Pre-algebra (not calculus) problem. that's what the title of my post says.
    I know, as I know in JIPAM there are so mant inequalities using convex functions, I guess u wish to see a solution in that direction. But I have no idea since I am not focused on this subject.
    I just wanted to share my long and poor solution.
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