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    Show that, if p>m>0 then
    p-m/p+m < x^2-2mx +p^2/x^2 +2mx+p^2<p+m/p-m
    for all real values of x
    feel free to post solutions and discuss.
    The signs are less then and equal to but i didnt know how to write them using my phone.
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    taken from c j bradleys book on inequalities.


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    (Original post by physicsmaths)
    Show that, if p>m>0 then
    p-m/p+m < x^2-2mx +p^2/x^2 +2mx+p^2<p+m/p-m
    for all real values of x
    feel free to post solutions and discuss.
    The signs are less then and equal to but i didnt know how to write them using my phone.
    \dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m}
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    (Original post by TenOfThem)
    \dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m}
    Thanks alot bro!


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    (Original post by TenOfThem)
    \dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m}
    This seems to work out fairly easily via a calculus approach. (Optimise the function in the middle - the upper and lower bounds seem to be as good as possible).

    Are they looking more for an "inequality manipulation" type of solution?
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    (Original post by atsruser)

    Are they looking more for an "inequality manipulation" type of solution?
    No idea, I am not the OP
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    (Original post by atsruser)
    This seems to work out fairly easily via a calculus approach. (Optimise the function in the middle - the upper and lower bounds seem to be as good as possible).

    Are they looking more for an "inequality manipulatiown" type of solution?
    i know the answer, this is just for others. Yep its a show that not a proof. A hint is they want some factorisation, ideally difference of squares.


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