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Thread: Bigger or equal to

  1. #1
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    Bigger or equal to

    Prove that 1/(a+b)+1/(b+c)+1/(a+c) is bigger or equal to 4.5 where a, b and c are positive and their sum 1.
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  2. #2
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    Re: Bigger or equal to

    codecogs
    Last edited by Idea; Dec 15th 2017 at 05:05 AM.
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  3. #3
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    Re: Bigger or equal to

    $a<=b<=c$
    Last edited by Idea; Dec 15th 2017 at 05:53 AM.
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  4. #4
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    Re: Bigger or equal to

    What do you mean?
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  5. #5
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    Re: Bigger or equal to

    minimum=4.5 when a=b=c=1/3
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    Re: Bigger or equal to

    Quote Originally Posted by DenisB View Post
    minimum=4.5 when a=b=c=1/3
    While I agree with your answer Sir Denis I think you should demonstrate it is in fact a minimum!
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  7. #7
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    Re: Bigger or equal to

    Was leaving that for you Sir Roamsect...
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  8. #8
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    Re: Bigger or equal to

    Yeah, it has to be demonstrated.
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    Re: Bigger or equal to

    Quote Originally Posted by louis33 View Post
    What do you mean?
    I couldn't get my equations to display properly
    I typed it all up but codecogs swallowed it
    what is codecogs anyway?

    now a+b is less than or equal to 2c and so on
    next use the AM-GM inequality

    try again

    a\leq b\leq c
    Last edited by Idea; Dec 16th 2017 at 06:15 AM.
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  10. #10
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    Re: Bigger or equal to

    Quote Originally Posted by Idea View Post
    I couldn't get my equations to display properly
    I typed it all up but codecogs swallowed it
    what is codecogs anyway?

    now a+b is less than or equal to 2c and so on
    next use the AM-GM inequality

    try again

    a\leq b\leq c
    Do not use the [tex /tex] tags.

    Use dollar \$ signs at the start and end: $a\leq b\leq c$.

    Click the {Reply with quote} tab to see.
    Thanks from skeeter
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  11. #11
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    Re: Bigger or equal to

    $a\leq b\leq c$
    Last edited by Idea; Dec 16th 2017 at 08:03 AM.
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  12. #12
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    Re: Bigger or equal to

    If x,y,z are positive numbers then by AM-GM we have

    $x+y+z\geq 3 (x y z)^{1/3}$

    and

    $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 3(x y z)^{-1/3}$

    Multiply these two equations to get

    $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 9$

    Now replace $x=a+b$, $y=b+c$, $z=c+a$
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