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Math Help - Solving an inequality

  1. #1
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    Solving an inequality

    Let x and y be positive real numbers. Prove that:

    (p + 2)(q + 2)(p + q) \geq 16pq

    I'm not sure how to approach this. Do I simplify/expand somehow or is there a special thing I should do with problems like these.

    Please help, thanks, BG
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  2. #2
    MHF Contributor red_dog's Avatar
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    Use AM-GM inequality:

    p+2\geq 2\sqrt{2p}

    q+2\geq 2\sqrt{2q}

    p+q\geq 2\sqrt{pq}

    Now multiply the inequalities.
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by BG5965 View Post
    Let x and y be positive real numbers. Prove that:

    (p + 2)(q + 2)(p + q) \geq 16pq

    I'm not sure how to approach this. Do I simplify/expand somehow or is there a special thing I should do with problems like these.

    Please help, thanks, BG
    Hello there is the solution:
    I'have :

    \left( {p + 2} \right)\left( {q + 2} \right)\left( {p + q} \right) = p^2 q + q^2 p + 4\left( {p + q} \right) + 2\left( {p^2 + q^2 } \right) + 4pq<br />
    but :

    \frac{{p^2 q + q^2 p + 4p + 4q}}{4} \ge \sqrt[4]{{16p^4 <br />
q^4 }} = 2pq<br />
    p^2 q + q^2 p + 4p + 4q \ge 8pq.....(1)


    \frac{{2p^2 + 2q^2 }}{2} \ge \sqrt {4p^2 q^2 } = 2pq

    2\left( {p^2 + q^2 } \right) \ge 4pq.....(2)


    ^2 q + q^2 p + 4\left( {p + q} \right) + 2\left( {p^2 + q^2 } \right) \ge 8pq + 4pq" alt="(1) + (2)^2 q + q^2 p + 4\left( {p + q} \right) + 2\left( {p^2 + q^2 } \right) \ge 8pq + 4pq" />
    Conclusion :

    <br />
p^2 q + q^2 p + 4\left( {p + q} \right) + 2\left( {p^2 + q^2 } \right) + 4pq \ge 8pq + 4pq + 4pq = 16pq
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  4. #4
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    Sorry, dhiab, but I don't 100% understand your post - especially everything after step 1 (the expansion). E.g. why is 16pq suddenly \sqrt[4]{16p^4 q^4} and everything after that.

    Could you please explain further. Thanks, BG
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