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Math Help - Math problem

  1. #1
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    Math problem

    let a,b are positiv real such abc=1
    prove that a+b+c>=sqrt(a)+sqrt(b)+ sqrt(c)
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  2. #2
    Senior Member Shanks's Avatar
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    because a+1\geq 2\sqrt{a},b+1\geq 2\sqrt{b},c+1\geq 2\sqrt{c}, we obtain
    a+b+c \geq \sqrt{a}+\sqrt{b}+\sqrt{c}+(\sqrt{a}+\sqrt{b}+\sqr  t{c}-3)<br />
\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+(3\sqrt[3]{\sqrt{abc}}-3)<br />
=\sqrt{a}+\sqrt{b}+\sqrt{c}
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  3. #3
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    Inequalite and square root

    a , b are positiv real such a+b=1
    justify that:
     \sqrt{1+4b}+2\sqrt{1+4a}\geq 4
    Last edited by Jhevon; January 9th 2010 at 03:15 PM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Lamalif View Post
    a , b are positiv real such a+b=1
    justify that:
    [tex] \sqrt{1+4b}+2\sqrt{1+4a}\geq 4
    [\math]
    I meant, what do you want? Do you just want an answer? This is in the CALCULUS section. Do you want us to use Lagrange Multipliers? Do you want us to just use basic IMO like inequality stuff? You tell me
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  5. #5
    Senior Member Shanks's Avatar
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    Notice the fact that for any t in [0,1], we have
    \sqrt{1+4t}\geq 1+(\sqrt{5}-1)t.
    Applying the fact, we get the inequality immediately!
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  6. #6
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    Exact inequalite

    a,b are positiv real such a+b=1
    justify this
    <br />
\sqrt{1+4a}+\sqrt{1+4b}\geq 4<br />
    thank you
    Last edited by Jhevon; January 9th 2010 at 03:19 PM.
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  7. #7
    Senior Member Shanks's Avatar
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    Is there a digit 2 in the problem?
    If not, the inequality does not hold for some positive a,b, a+b=1.
    If there is a digit 2, then applying the inequlity mentioned above, substitute t with a and b to get two inequality, add them up, the result follows immediately.
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  8. #8
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    Excuse me (difficulty to use Latex)
    the exact expression is this
    a,b are positiv real such a+b=1
    justify this

    \sqrt{1+4a}+\sqrt{1+4b}\le 4

    thank you again
    Last edited by Jhevon; January 9th 2010 at 03:16 PM.
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  9. #9
    Senior Member Shanks's Avatar
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    since positive a and b satisify a+b=1,
    \Rightarrow a<1, b<1
    \Rightarrow \sqrt{1+4a} <\sqrt{5},\sqrt{1+4b}<\sqrt{5}
    \Rightarrow (\sqrt{1+4a})^2+2\sqrt{1+4a}\sqrt{1+4b}+(\sqrt{1+4  b})^2<br />
<1+4a+2\sqrt{5}\sqrt{5}+1+4b=16
    \Rightarrow \sqrt{1+4a}+\sqrt{1+4b}\le 4
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  10. #10
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    Another answer

    for any positive number n
    (1+2n)\le(1+n)^2

    \sqrt{1+4a}\le\sqrt{(1+2a)^2},\sqrt{1+4b}\le\sqrt{  (1+2b)^2}

    \sqrt{1+4a}+\sqrt{1+4b}\le(1+2a+1+2b)
    Last edited by Jhevon; January 11th 2010 at 05:32 PM.
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