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Proove number inequality

  1. Oct 17, 2011 #1
    [tex]a,b,c,d\in\mathbb{R^{+}}\;\;,a+b+c+d=1.[/tex]

    Then prove that

    [tex]\left( a+\dfrac{1}{b}\right).\left(b+\dfrac{1}{c}\right).\left(c+\dfrac{1}{a}\right)\geq \left(\dfrac{10}{3}\right)^3[/tex]

    Anyone an idea on how to start with this exercise?
     
  2. jcsd
  3. Oct 17, 2011 #2
    implies 0<a+b+c<1
     
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