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Math Help - Tricky Inequality

  1. #1
    Super Member craig's Avatar
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    Tricky Inequality

    We got given this problem in a tutorial a few weeks ago and I can't for the life of me figure out how to solve it, probably something simple but would love it if someone could tell me how to go about this.

    Prove that:

    \frac{[x^2(3-x)(2+x)]^\frac{1}{3}}{3+x} \leq 1 if |x| \leq 1

    Cheers
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  2. #2
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    Quote Originally Posted by craig View Post
    We got given this problem in a tutorial a few weeks ago and I can't for the life of me figure out how to solve it, probably something simple but would love it if someone could tell me how to go about this.

    Prove that:

    \frac{[x^2(3-x)(2+x)]^\frac{1}{3}}{3+x} \leq 1 if |x| \leq 1

    Cheers

    As |x|\leq 1 , we get: x^2\leq 1\,,\,3-x\leq 4\,,\,2+x\leq 3\,,\,3+x\geq 2 , so now input this in your expression and you'll get it is less than or equal to \frac{8^{1/3}}{2}\leq 1 .

    Tonio
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  3. #3
    Super Member craig's Avatar
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    Ahh thank you, you have no idea how long that was bugging me for!
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