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Math Help - Nice Inequality !!

  1. #1
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    Nice Inequality !!

    Hii !


    a, b and c the réels posifives numbers , prove That :

    a^2+b^2+c^2 \geq a+b+c+2( \frac{a-1}{a+1} + \frac{b-1}{b+1} + \frac{c-1}{c+1} )

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  2. #2
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    Quote Originally Posted by Perelman View Post
    Hii !


    a, b and c the réels posifives numbers , prove That :

    a^2+b^2+c^2 \geq a+b+c+2( \frac{a-1}{a+1} + \frac{b-1}{b+1} + \frac{c-1}{c+1} )

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    Spoiler:
    Take all the terms over to the left side:

    a^2 - a -2\Bigl(\frac{a-1}{a+1}\Bigr) = \frac{(a-1)^2(a+2)}{a+1}, which is obviously positive if a is! – and similarly for the b and c terms.
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  3. #3
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    Hii!!

    Perfect Mr Opalg !

    So My Solution :
    a^2+b^2+c^2 \geq a+b+c+2( \frac{a-1}{a+1} + \frac{b-1}{b+1} + \frac{c-1}{c+1} )
    \Leftrightarrow a^2+ b^2+ c^2 + 4(\frac{1}{a+1} + \frac{1}{b+1}+\frac{1}{c+1}) \geq a +b +c + 6

    Put : a+1=x , b+1=y and c+1=z :
    x^2+y^2+z^2+4(\frac{1}{x} + \frac{1}{y}+\frac{1}{z}) \geq 3(x+y+z)
    By Cauchy Schwartz :
    \frac{1}{x} + \frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z}

    And We Have : x^2+ y^2+ z^2 \geq \frac{(x+y+z)2}{3}

    So, We need prove : \frac{(x+y+z)^2}{3} + \frac{36}{x+y+z} \geq 3(x+y+z)

    \Leftrightarrow (x+y+z)^2 -9(x+y+z)^2+108 \geq 0
    \Leftrightarrow (x+y+z-6)^2 (x+y+z+3) \geq 0

    Cqfd .
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  4. #4
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