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Math Help - inequality:

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    83

    inequality:

    i need to show

     <br />
-1\leq \frac{-2b^2}{a^2+b^2}\leq0<br />

    now i have shown

     <br />
e^2=1+\frac{b^2}{a^2}\leq2<br />

    which gives me b^2\leq a^2

    clearly the expression is \leq 0 so i need to show

     <br />
-1\leq \frac{-2b^2}{a^2+b^2}

    this is simple enough by considering
     <br />
\frac{-2b^2}{a^2+b^2}+1 [*]<br />

     <br />
= \frac{a^2-b^2}{a^2+b^2}<br />

    as b^2\leq a^2

    a^2-b^2 > 0

    so [*]>0

    What i would like to do is not use the result and start from scratch to show

     <br />
-1\leq \frac{-2b^2}{a^2+b^2}

    using only
     <br />
e^2=1+\frac{b^2}{a^2}\leq2 <br />

    i cant seem to get the result, alwayus end up with an inequality round the wrong way so cant deduce the result. Maybe it cant be done this way which is why they gave the value -1 or im making a mistake:
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by jiboom View Post
    i need to show

     <br />
-1\leq \frac{-2b^2}{a^2+b^2}\leq0<br />

    now i have shown

     e^2=1+\frac{b^2}{a^2}\leq2

    which gives me b^2\leq a^2

    clearly the expression is \leq 0 so i need to show

     -1\leq \frac{-2b^2}{a^2+b^2}

    this is simple enough by considering
     <br />
\frac{-2b^2}{a^2+b^2}+1 [*]<br />

     <br />
= \frac{a^2-b^2}{a^2+b^2}<br />

    as b^2\leq a^2

    a^2-b^2 > 0

    so [*]>0

    What i would like to do is not use the result and start from scratch to show

     <br />
-1\leq \frac{-2b^2}{a^2+b^2}

    using only
     <br />
e^2=1+\frac{b^2}{a^2}\leq2 <br />

    i cant seem to get the result, alwayus end up with an inequality round the wrong way so cant deduce the result. Maybe it cant be done this way which is why they gave the value -1 or im making a mistake:
    I don't really get what you're stuck on...

    Using that result its clear that b^2 \leq a^2.

    Hence 2b^2 \leq a^2 + b^2.

    \frac{2b^2}{a^2 + b^2} \leq \frac{a^2 + b^2}{a^2 + b^2}= 1.

    Hence taking negatives (which we means we REVERSE the \leq sign) gives.

    \frac{-2b^2}{a^2 + b^2} \geq -1.


    (Remember, if a<b, -a>-b, example, 1<2 but -1>-2)
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  3. #3
    Member
    Joined
    Dec 2009
    Posts
    83
    thanks for reply.

    I missed that, was trying to start with

     <br />
b^2 \leq a^2<br />

     \frac{-2b^2}{a^2+b^2}>\frac{-2a^2}{a^2+b^2} = \frac{-2}{e^2}

    and

     e^2\leq 2
    so
    1 \leq \frac{2}{e^2}

    -1>\frac{-2}{e^2}

    maybe im being siily but does this imply result? im not sure it does as, for example i can have
    e^2=\frac{1}{2} so \frac{-2}{e^2} = -4

    and all i have is

     <br />
\frac{-2b^2}{a^2+b^2}>-4<br />
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