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Math Help - Inequalities Question

  1. #1
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    Inequalities Question

    If a,b,c are positive real numbers prove that  \Bigl[ (1+a)(1+b)(1+c) \Bigr]^{7}>7^{7} a^{4}b^{4}c^{4}
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  2. #2
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    Use AM  \geq GM .  a^7 + b^7 + c^7 + (ab)^7 + (ac)^7 + (bc)^7 + (abc)^7  \geq  7 (abc)^4

    Replace a,b,c by  a^{\frac{1}{7}}, b^{\frac{1}{7}}, c^{\frac{1}{7}}
    it becomes :

     a + b + c + ab + ac + bc + abc  \geq  7 (abc)^{\frac{4}{7}}

    Therefore ,

     1 + a + b + c + ab + ac + bc + abc  \geq  1 + 7 (abc)^{\frac{4}{7}} >  7 (abc)^{\frac{4}{7}}

    Then ,

     (1+a)(1+b)(1+c) > 7 (abc)^{\frac{4}{7}}

     [(1+a)(1+b)(1+c)]^7 > 7^7 (abc)^4
    Last edited by simplependulum; June 20th 2009 at 10:21 PM.
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  3. #3
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    Quote Originally Posted by simplependulum View Post
    Use AM  \geq GM .  a^7 + b^7 + c^7 + (ab)^7 + (ac)^7 + (bc)^7 + (abc)^7 \geq 7 (abc)^4

    Replace a,b,c by  a^{\frac{1}{7}}, b^{\frac{1}{7}}, c^{\frac{1}{7}}
    it becomes :

     a + b + c + ab + ac + bc + abc \geq 7 (abc)^{\frac{4}{7}}

    Therefore ,

     1 + a + b + c + ab + ac + bc + abc \geq 1 + 7 (abc)^{\frac{4}{7}} > 7 (abc)^{\frac{4}{7}}

    Then ,

     (1+a)(1+b)(1+c) > 7 (abc)^{\frac{4}{7}}

     [(1+a)(1+b)(1+c)]^7 > 7^7 (abc)^4
    similarly, for any positive real numbers x_1,x_2, \cdots , x_n we have: (1+x_1)(1+x_2) \cdots (1+x_n) \geq 1 + (2^n - 1) \sqrt[2^n - 1]{(x_1x_2 \cdots x_n)^{2^{n-1}}}. as a result:

    [(1+x_1)(1+x_2) \cdots (1+x_n)]^{2^n - 1} > (2^n - 1)^{2^n - 1} (x_1x_2 \cdots x_n)^{2^{n-1}}.
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