# Thread: can't figure this inequality

1. ## can't figure this inequality

$\displaystyle (M^2+2MN-N^2 )^2<(M^2-2MN-N^2 )^2$

I went about it like this

$\displaystyle M^2+2MN-N^2<M^2-2MN-N^2\ \ \ \ \ \ \ \ (1)$
$\displaystyle M^2+2MN-N^2>-M^2+2MN+N^2\ \ \ \ \ \ (2)$

$\displaystyle (1)$

$\displaystyle M^2+2MN-N^2<M^2-2MN-N^2$

$\displaystyle 2MN<-2MN$

$\displaystyle \implies\ \ \ \ \ M<0<N\ \ \ or\ \ \ M>0>N$

$\displaystyle (2)$

$\displaystyle M^2+2MN-N^2>-M^2+2MN+N^2$

$\displaystyle M^2-N^2>-M^2+N^2$

$\displaystyle 2M^2-2N^2>0$

$\displaystyle (M+N)(M-N)>0$

so $\displaystyle (M+N)$ and $\displaystyle (M-N)$ have the same sign

$\displaystyle (M+N)\ \ and\ \ (M-N)\ \ >0\ \ \ \ (3)$

$\displaystyle (M+N)\ \ and\ \ (M-N)\ \ <0\ \ \ \ (4)$

$\displaystyle (3)$

$\displaystyle (M+N)\ \ and\ \ (M-N)\ \ >0$

$\displaystyle \implies\ \ \ \ \ \ M>|N|$

$\displaystyle (4)$

$\displaystyle (M+N)\ \ and\ \ (M-N)\ \ <0$

$\displaystyle \implies\ \ \ \ \ \ M<|N|$

so altogether I have

$\displaystyle M>0>N$
$\displaystyle M<0<N$
$\displaystyle M>|N|$
$\displaystyle M<|N|$

which seems to imply that all M and N fit apart from M=N

let M=2 N=1

$\displaystyle (M^2+2MN-N^2 )^2=7\ \ \ \ \(M^2-2MN-N^2 )^2=1$

but this has $\displaystyle (M^2+2MN-N^2 )^2>(M^2-2MN-N^2 )^2$

What am I doing wrong ?

2mn < -2mn
1 < -1

stop!

2mn<-2mn

m=2
n=-1
2mn=-4
-2mn=4
2mn<-2mn

4. ## Re: can't figure this inequality

Multiply out both sides and collect terms and you arrive at

$\displaystyle MN(M^{2}-N^{2}) < 0$

which clearly is not the case for all $\displaystyle M$ and $\displaystyle N.$

If $\displaystyle M$ and $\displaystyle N$ are both positive you would need to have $\displaystyle M<N.$

5. ## Re: can't figure this inequality

Thanks Bob, seems easier than the way I was going about it.

$\displaystyle MN(M^2-N^2)<0$

$\displaystyle M\ne0$

$\displaystyle M>0\ \ \ \implies \ \ \ 0<M<N\ \ or\ \ -M<N<0$

$\displaystyle M<0\ \ \ \implies \ \ \ N<M<0\ \ or\ \ -M>N>0$

Does this cover all of them ?