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Math Help - Real number problem

  1. #1
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    Real number problem

    About positive real numbers a, b, c we know that a+b+c=17 and a\cdot b\cdot d=64.
    Show that \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 a, b, c we know that a+b+c=17 and a\cdot b\cdot d=64.
    Show that \max(a,b,c) \geq 8.
     a+b=17-c and  ab=\frac{64}{c} \implies WLOG  a=\frac12\left(17-c+\sqrt{(17-c)^2-\frac{256}{c}}\right) and  b=\frac12\left(17-c-\sqrt{(17-c)^2-\frac{256}{c}}\right)

    Now assume  a,b,c < 8

    Consider  f(x)=\frac12\left(17-c+\sqrt{(17-c)^2-\frac{256}{c}}\right)-8 on the interval  [1,8) and use calculus to find that the minimum is  0 which implies  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 f(x) constant? On the right side is only 'c' - positive real number (there is no 'x').
    And from where we know that 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 f(x) constant? On the right side is only 'c' - positive real number (there is no 'x').
    And from where we know that a\geq 8?
    Whoops, it should be  f(x)=\frac12\left(17-x+\sqrt{(17-x)^2-\frac{256}{x}}\right)-8<br />
 .


    If the minimum of  f(x) is  0 , then that means for all  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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