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Prove this inequality

  1. Nov 20, 2012 #1


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    1. The problem statement, all variables and given/known data
    If a,b,c are the positive real numbers, prove that [itex]a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2) \geq 6abc[/itex]

    2. Relevant equations

    3. The attempt at a solution
    With a little simplification L.H.S = [itex](a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2)[/itex]
    Using A.M>=G.M
    [itex]\dfrac{a^2+b^2+c^2}{3} \geq (a^2b^2c^2)^{\frac{1}{3}} \\
    a^2+b^2+c^2 \geq 3a^{2/3}b^{2/3}c^{2/3} \\
    [itex] \dfrac{a^2b^2+b^2c^2+c^2a^2}{3} \geq (a^2b^2.b^2c^2.c^2a^2)^{1/3} \\
    a^2b^2+b^2c^2+c^2a^2 \geq 3a^{4/3}b^{4/3}c^{4/3}
    Adding the two inequalities
    (a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2) \geq 3[a^{2/3}b^{2/3}c^{2/3}+a^{4/3}b^{4/3}c^{4/3}]

    Now how do I simplify next?
  2. jcsd
  3. Nov 20, 2012 #2


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    Staff: Mentor

    Your last inequality can be written as
    LHS >= 3 (x+x^2)
    with an appropriate x.

    And you have to show that
    LHS >= 6x3/2

    You can just use the same trick again at your new sum.
  4. Nov 20, 2012 #3


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    Gold Member

    Thanks. I got it.
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