A nice inequalities question.

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.

(edited 9 years ago)

Reply 1

username1162972

OP

18

taken from c j bradleys book on inequalities.

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Reply 2

TenOfThem

18

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}

Reply 3

username1162972

OP

18

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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Reply 4

atsruser

11

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?

Reply 5

TenOfThem

18

Original post by atsruser

Are they looking more for an "inequality manipulation" type of solution?

No idea, I am not the OP

Reply 6

username1162972

OP

18

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.

Posted from TSR Mobile

A nice inequalities question.

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