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Math Help - (a^2+2)(b^2+2)(c^2+2)>= 3(a+b+c)^2

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    (a^2+2)(b^2+2)(c^2+2)>= 3(a+b+c)^2

    Prove that \forall a, b, c\in\mathbb{R}: (a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2.
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    Quote Originally Posted by james_bond View Post
    Prove that \forall a, b, c\in\mathbb{R}: (a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2.
    Let me take a stab at this. Let's define the function F as

    F(a,b,c) = (a^2+2)(b^2+2)(c^2+2) - 3(a+b+c)^2

    We will show that the minumum of this is zero. Taking partial derivatives and setting to zero gives

    F_a = 2a(b^2+2)(c^2+2) - 6(a+b+c) = 0
    F_b = 2b(a^2+2)(c^2+2) - 6(a+b+c) = 0
    F_c = 2c(a^2+2)(b^2+2) - 6(a+b+c) = 0,

    By substracting in pairs we obtain

     (a-b)(ab-2) = 0,\;\;\;(a-c)(ac-2) = 0,\;\;\;(b-c)(bc-2) = 0

    from which we deduce that a = b = c.

    Substituting into any of the partial derivatives gives

    2a(a-1)(a+1)(a^2+5) = 0

    giving rise to three cases:  a = 0, \;\;a = -1, \;\;a=1.

    The first gives F(0,0,0) = 8 whereas the remaining two gives F(- 1,- 1,- 1) = 0,\;\;\;F(1, 1, 1) = 0. Thus, the minimum of F is 0 giving

    F(a,b,c) \ge 0

    for which the inequality follows.
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  3. #3
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    Quote Originally Posted by james_bond View Post

    Prove that \forall a, b, c\in\mathbb{R}: (a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2.
    changing a,b,c to \sqrt{2}a, \sqrt{2}b, \sqrt{2}c respectively, we'll get the following equivalent inequality: 4(a^2 + 1)(b^2 + 1)(c^2 + 1) \geq 3(a+b+c)^2. \ \ \ \ \ (*)


    Lemma: \forall a,b,c \in \mathbb{R}: \ \ a^2 + b^2 + c^2 + 1 + 4a^2b^2c^2 - 2(ab+bc+ac) \geq 0.

    Proof: rewrite the inequality as: (1+4b^2c^2)a^2 - 2(b+c)a + (b-c)^2 + 1 \geq 0. now look at the LHS of the inequality as a quadratic function in a.

    the discriminant of this quadratic function is: -4[(2bc - 1)^2 + 4b^2c^2(b-c)^2] \leq 0. \ \ \Box


    Proof of (*): 4(a^2 + 1)(b^2+1)(c^2 + 1) - 3(a+b+c)^2 =

    (2ab \ - \ 1)^2 + (2bc \ - \ 1)^2 + (2ac \ - \ 1)^2 +a^2 + b^2 + c^2 + 1 + 4a^2b^2c^2 - 2(ab \ + \ bc \ + \ ac) \geq 0, by the Lemma.  \Box
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    Quote Originally Posted by danny arrigo View Post
     (a-b)(ab-2) = 0,\;\;\;(a-c)(ac-2) = 0,\;\;\;(b-c)(bc-2) = 0

    from which we deduce that a = b = c.
    What if a=\frac 2b=c?
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  5. #5
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    Quote Originally Posted by james_bond View Post
    What if a=\frac 2b=c?
    Yes, you are correct, they are solutions of my intermediate set of equations but will not satisfy the original set. Notice that

    a = 0,\;\;b = 0,\;\;c = 0

    is a solution of the original set but cannot be obtained from your result.
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