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Math Help - Inequalities

  1. #1
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    Inequalities

    a) Prove that the inequality x(x+1)(x+2)(x+3) >= -1 holds for all real numbers x.

    b) Prove that (ab+bc+ca)^2 >= 3abc(a+b+c) holds for all real numbers a, b and c
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    a) Prove that the inequality x(x+1)(x+2)(x+3) >= -1 holds for all real numbers x.

    obviously \left(t^2 - \frac{5}{4} \right)^2 \geq 0, \ \forall t \in \mathbb{R}, which can be rewritten as: \left(t^2 - \frac{9}{4} \right) \left(t^2 - \frac{1}{4} \right) \geq -1.

    thus: \left(t-\frac{3}{2} \right) \left(t + \frac{3}{2} \right) \left(t - \frac{1}{2}\right) \left(t + \frac{1}{2} \right) \geq -1, \ \forall t \in \mathbb{R}. now put t=x+\frac{3}{2}. \ \ \ \square


    b) Prove that (ab+bc+ca)^2 >= 3abc(a+b+c) holds for all real numbers a, b and c
    Fact: for any real numbers x,y,z: \ (x+y+z)^2 \geq 3(xy+yz+zx).

    Proof: (x+y+z)^2-3(xy+yz+zx)=\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2) \geq 0. \ \ \ \square

    now your inequality follows quickly if you put x=ab, \ y=bc, \ z=ca.
    Last edited by NonCommAlg; July 31st 2008 at 05:11 AM.
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