inequality:

**jiboom** · April 5th 2010, 08:26 AM

i need to show

$ -1\leq \frac{-2b^2}{a^2+b^2}\leq0 $

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
$ \frac{-2b^2}{a^2+b^2}+1 [*] $

$ = \frac{a^2-b^2}{a^2+b^2} $

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

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

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

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:

**Deadstar** · April 5th 2010, 12:44 PM

Originally Posted by jiboom

i need to show

$ -1\leq \frac{-2b^2}{a^2+b^2}\leq0 $

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
$ \frac{-2b^2}{a^2+b^2}+1 [*] $

$ = \frac{a^2-b^2}{a^2+b^2} $

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

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

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

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)

**jiboom** · April 5th 2010, 01:43 PM

thanks for reply.

I missed that, was trying to start with

$ b^2 \leq a^2 $

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

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

$ \frac{-2b^2}{a^2+b^2}>-4 $

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