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Thread: Real number problem

  1. #1
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    Real number problem

    About positive real numbers $\displaystyle a, b, c$ we know that $\displaystyle a+b+c=17$ and $\displaystyle a\cdot b\cdot d=64$.
    Show that $\displaystyle \max(a,b,c) \geq 8.$
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Arczi1984 View Post
    About positive real numbers $\displaystyle a, b, c$ we know that $\displaystyle a+b+c=17$ and $\displaystyle a\cdot b\cdot d=64$.
    Show that $\displaystyle \max(a,b,c) \geq 8.$
    $\displaystyle a+b=17-c $ and $\displaystyle ab=\frac{64}{c} \implies $ WLOG $\displaystyle a=\frac12\left(17-c+\sqrt{(17-c)^2-\frac{256}{c}}\right) $ and $\displaystyle b=\frac12\left(17-c-\sqrt{(17-c)^2-\frac{256}{c}}\right) $

    Now assume $\displaystyle a,b,c < 8 $

    Consider $\displaystyle f(x)=\frac12\left(17-c+\sqrt{(17-c)^2-\frac{256}{c}}\right)-8 $ on the interval $\displaystyle [1,8) $ and use calculus to find that the minimum is $\displaystyle 0 $ which implies $\displaystyle a\geq8 $ and that gives us a contradiction.
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  3. #3
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    Thanks.
    I've question about last statement.
    Is the function $\displaystyle f(x)$ constant? On the right side is only 'c' - positive real number (there is no 'x').
    And from where we know that $\displaystyle a\geq 8$?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Arczi1984 View Post
    Thanks.
    I've question about last statement.
    Is the function $\displaystyle f(x)$ constant? On the right side is only 'c' - positive real number (there is no 'x').
    And from where we know that $\displaystyle a\geq 8$?
    Whoops, it should be $\displaystyle f(x)=\frac12\left(17-x+\sqrt{(17-x)^2-\frac{256}{x}}\right)-8
    $.


    If the minimum of $\displaystyle f(x) $ is $\displaystyle 0 $, then that means for all $\displaystyle x\in[1,8),\;\;\frac12\left(17-x+\sqrt{(17-x)^2-\frac{256}{x}}\right)\geq8 $ which is a problem for us.
    Last edited by chiph588@; May 5th 2010 at 04:23 PM.
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  5. #5
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    Ok, now it is clear Once more thanks for help.
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