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Math Help - Inequalities

  1. #1
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    Inequalities

    X2+3+x-2>_7
    is the above statement true for all real x>0?

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  2. #2
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    Re: Inequalities

    What do you get when you expand \left(x+\frac{1}{x}\right)^2?
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  3. #3
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    Re: Inequalities

    Quote Originally Posted by Trianagt View Post
    X2+3+x-2>_7
    is the above statement true for all real x>0?

    Thanks
    \displaystyle \begin{align*} x^2 + 3 + x^{-2} &\geq 7 \\ x^4 + 3x^2 + 1 &\geq 7x^2 \\ x^4 - 4x^2 + 1 &\geq 0 \\ x^4 - 4x^2 + (-2)^2 - (-2)^2 + 1 &\geq 0 \\ \left(x^2 - 2\right)^2 - 3 &\geq 0 \\ \left(x^2 - 2\right)^2 &\geq 3 \\ \left|x^2 - 2\right| &\geq \sqrt{3} \\ x^2 - 2 \leq -\sqrt{3} \textrm{ or } x^2 - 2 &\geq \sqrt{3} \\ x^2 \leq 2 - \sqrt{3} \textrm{ or } x^2 &\geq \sqrt{2 + \sqrt{3}} \\ |x| \leq \sqrt{2 - \sqrt{3}} \textrm{ or } |x| &\geq \sqrt{2 + \sqrt{3}} \\ -\sqrt{2 - \sqrt{3}} \leq x \leq \sqrt{2 - \sqrt{3}} \textrm{ or } x \leq -\sqrt{2 + \sqrt{3}} \textrm{ or } x &\geq \sqrt{2 + \sqrt{3}} \end{align*}
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  4. #4
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    Re: Inequalities

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} x^2 + 3 + x^{-2} &\geq 7 \\ x^4 + 3x^2 + 1 &\geq 7x^2 \\ x^4 - 4x^2 + 1 &\geq 0 \\ x^4 - 4x^2 + (-2)^2 - (-2)^2 + 1 &\geq 0 \\ \left(x^2 - 2\right)^2 - 3 &\geq 0 \\ \left(x^2 - 2\right)^2 &\geq 3 \\ \left|x^2 - 2\right| &\geq \sqrt{3} \\ x^2 - 2 \leq -\sqrt{3} \textrm{ or } x^2 - 2 &\geq \sqrt{3} \\ x^2 \leq 2 - \sqrt{3} \textrm{ or } x^2 &\geq \sqrt{2 + \sqrt{3}} \\ |x| \leq \sqrt{2 - \sqrt{3}} \textrm{ or } |x| &\geq \sqrt{2 + \sqrt{3}} \\ -\sqrt{2 - \sqrt{3}} \leq x \leq \sqrt{2 - \sqrt{3}} \textrm{ or } x \leq -\sqrt{2 + \sqrt{3}} \textrm{ or } x &\geq \sqrt{2 + \sqrt{3}} \end{align*}
    Is that necessary?

    x^2+3+x^{-2}=(x+x^{-1})^2+1
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  5. #5
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    Re: Inequalities

    Quote Originally Posted by a tutor View Post
    Is that necessary?

    x^2+3+x^{-2}=(x+x^{-1})^2+1
    It would be if you actually wanted to solve for x...
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  6. #6
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    Re: Inequalities

    Quote Originally Posted by Prove It View Post
    It would be if you actually wanted to solve for x...
    Yes.

    You're short of things to do today aren't you.
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